Nuclear Physics Fundamentals

Question 1

Question 1: 1. The following nuclear reaction equations are $\alpha$ decays

1. The following nuclear reaction equations are $\alpha$ decays

A. A. ${ } _ { 92 } ^ { 238 } \mathrm { U } \rightarrow { } _ { 90 } ^ { 234 } \mathrm { Th } + { } _ { 2 } ^ { 4 } \mathrm { He }$
B. B. ${ } _ { 1 } ^ { 2 } \mathrm { H } + { } _ { 1 } ^ { 3 } \mathrm { H } \rightarrow { } _ { 2 } ^ { 4 } \mathrm { He } + { } _ { 0 } ^ { 1 } \mathrm { n }$
C. C. ${ } _ { 90 } ^ { 234 } \mathrm { Th } \rightarrow { } _ { 91 } ^ { 234 } \mathrm {~Pa} + { } _ { - 1 } ^ { 0 } \mathrm { e }$
D. D. ${ } _ { 7 } ^ { 14 } \mathrm {~N} + { } _ { 2 } ^ { 4 } \mathrm { He } \rightarrow { } _ { 8 } ^ { 17 } \mathrm { O } + { } _ { 1 } ^ { 1 } \mathrm { H }$

Answer

Answer: A

Solution: The A reaction is $\alpha$ decay; the B and D reactions are the artificial transition equations of the nucleus; the C reaction is $\beta$ decay; so choose A.

Question 2

Question 2: 2. There is a small nuclear energy battery on the Jade Rabbit lunar rover to provide the electrical ...

2. There is a small nuclear energy battery on the Jade Rabbit lunar rover to provide the electrical energy needed when the rover is working in the dark for a long time, and the nuclear reaction equation in the nuclear energy battery is ${ } ^ { 94 } \mathrm { Pu } \rightarrow { } _ { 95 } ^ { 238 } \mathrm { Am } + \mathrm { X }$, and the following statements are correct ()

A. A. X is a neutron.
B. B. X is more ionizable than $\alpha$ particle
C. C. The nuclear reaction is a fission reaction with a heavy nucleus
D. D. The specific binding energy of ${ } ^ { 238 } \mathrm { Am }$ is greater than that of ${ } ^ { 234 } \mathrm { Pu }$.

Answer

Answer: D

Solution: A. According to the conservation of charge and mass, $X$ is an $\beta$ particle (i.e., an electron), so A is wrong; B. The $\beta$ particle is weaker in ionization than the $\alpha$ particle, so B is wrong; C. The reaction is a $\beta$ decay of a heavy nucleus, so C is wrong; D. The products of a nuclear reaction are more stable than the reactants, so ${ } ^ { 238 } \mathrm { Am }$ is more stable than ${ } ^ { 238 } \mathrm { Pu }$, so the specific binding energy of ${ } ^ { 238 } \mathrm { Am }$ is larger than that of ${ } ^ { 238 } \mathrm { Pu }$, so D is correct.

Question 3

Question 3: 3. Becquerel first discovered the phenomenon of natural radioactivity 120 years ago, and today the r...

3. Becquerel first discovered the phenomenon of natural radioactivity 120 years ago, and today the radioactivity of atomic nuclei is widely used in many fields. The following are radioactive decays ( )

A. A. ${ } _ { 90 } ^ { 234 } \mathrm { Th } \xrightarrow { 234 } \mathrm {~Pa} ^ { + } { } ^ { 0 } \mathrm { e }$
B. B. ${ } _ { 92 } ^ { 235 } \mathrm { U } { } ^ { + } { } _ { 0 } \mathrm { n } \xrightarrow [ 56 ] { 144 } \mathrm { Ba } { } ^ { + { } ^ { 89 } } \mathrm { Kr } + { } ^ { 31 } \mathrm { n }$
C. C. ${ } ^ { 14 } \mathrm {~N} { } ^ { + 4 } \mathrm { He } \xrightarrow { \rightarrow ^ { 17 } } \mathrm { O } { } ^ { + 1 } \mathrm { H }$
D. D. ${ } ^ { 4 } \mathrm { He } { } ^ { + { } _ { 13 } ^ { 27 } } \mathrm { Al } \xrightarrow { { } ^ { 30 } } \mathrm { P } { } ^ { + } { } _ { 0 } \mathrm { n }$

Answer

Answer: A

Solution: A. ${ } ^ { 234 } \mathrm { Th } \xrightarrow { 234 } \mathrm {~Pa} ^ { + 0 } { } ^ { 0 } \mathrm { e }$ is the equation for natural radiation phenomena. Therefore, A is correct; B.${ } ^ { 235 } \mathrm { U } { } ^ { 1 } { } _ { 0 } \rightarrow { } ^ { 144 } \mathrm { Ba } { } ^ { + 36 } \mathrm { Kr } + { } ^ { 39 } \mathrm { n }$ is the fission equation for a heavy nucleus. B. ${ } ^ { 235 } \mathrm { U } { } ^ { 1 } { } _ { 0 } \rightarrow { } ^ { 144 } \mathrm { Ba } { } ^ { + 36 } \mathrm { Kr } + { } ^ { 39 } \mathrm { n }$ is the fission equation for heavy nuclei; C. ${ } ^ { 14 } \mathrm {~N} { } ^ { + 4 } \mathrm { He } ^ { \rightarrow { } ^ { 17 } } \mathrm { O } { } ^ { + 1 } \mathrm { H }$ is the equation of artificial nuclear reaction. C. ${ } ^ { 14 } \mathrm {~N} { } ^ { + 4 } \mathrm { He } ^ { \rightarrow { } ^ { 17 } } \mathrm { O } { } ^ { + 1 } \mathrm { H }$ is an artificial nuclear reaction equation; D. ${ } ^ { 4 } \mathrm { He } { } ^ { + { } _ { 13 } ^ { 27 } } \mathrm { Al } \rightarrow ^ { 30 } \mathrm { P } { } ^ { + 1 } \mathrm { n }$ is an artificial nuclear reaction equation. D. ${ } ^ { 4 } \mathrm { He } { } ^ { + { } _ { 13 } ^ { 27 } } \mathrm { Al } \rightarrow ^ { 30 } \mathrm { P } { } ^ { + 1 } \mathrm { n }$ is an artificial nuclear reaction equation.

Question 4

Question 4: 4. When the radioactive element polonium (${ } ^ { 210 } \mathrm { Po }$) decays, it produces ${ } ^...

4. When the radioactive element polonium (${ } ^ { 210 } \mathrm { Po }$) decays, it produces ${ } ^ { 4 } \mathrm { He }$ and an unknown particle and emits ${ } ^ { \gamma }$ rays, and the nuclear reaction equation is ${ } _ { 84 } ^ { 210 } \mathrm { Po } \rightarrow { } _ { 82 } ^ { y } \mathrm { X } + { } _ { 2 } ^ { 4 } \mathrm { He } + \mathrm { \gamma }$, which is correctly summarized in the following statement. (usually used in the negative) have the possibility of

A. A. $\gamma$ rays are negatively charged
B. B. $y = 206$
C. C. Elevated temperatures favor decay
D. D. The specific binding energy of ${ } _ { 82 } ^ { y } \mathrm { X }$ nuclei is less than the specific binding energy of ${ } ^ { 210 } \mathrm { Po }$ nuclei.

Answer

Answer: B

Solution: A. $\gamma$ rays are electromagnetic waves and are not charged; B.From the conservation of mass number, $y = 210 - 4 = 206$ is an electromagnetic wave. B correct ; C. The half-life has nothing to do with the physical and chemical environment of the nucleus, C is wrong; D.The reaction gives off energy, the product is relatively more stable, and the specific binding energy increases, D error.

Question 5

Question 5: 5. The following experiments can reveal that atoms have a nuclear structure ( )

5. The following experiments can reveal that atoms have a nuclear structure ( )

A. A. photoelectric effect experiment
B. B. The discovery of Roentgen rays
C. C. $\alpha$ Particle Scattering Experiments
D. D. Discovery of hydrogen atom spectroscopy

Answer

Answer: C

Solution: The experiment that reveals that atoms have a nuclear structure is the $\alpha$ particle scattering experiment, so C is correct and ABD is wrong.

Question 6

Question 6: 6. The "DPRK nuclear crisis" has attracted global attention, centering on whether the DPRK's nuclear...

6. The "DPRK nuclear crisis" has attracted global attention, centering on whether the DPRK's nuclear power plants use light-water reactors or heavy-water reactors. Heavy Water Reactor (HWR) Heavy water reactors can produce plutonium ${ } ^ { 239 } \mathrm { Pu }$ for nuclear weapons while generating electricity. Plutonium ${ } ^ { 239 } \mathrm { Pu }$ can be produced by uranium ${ } ^ { 239 } \mathrm { U }$ decaying $n$ times $\beta$, then $\beta$ can be produced by uranium $n$. The $n$ is the $n$ of

A. A. 2
B. B. 239
C. C. 145
D. D. 92

Answer

Answer: A

Solution: The nuclear reaction equation is $$ { } _ { 94 } ^ { 239 } \mathrm { Pu } \rightarrow { } _ { 92 } ^ { 239 } \mathrm { U } + 2 { } _ { - 1 } ^ { 0 } \mathrm { e } $$ So $$ n = 2 $$

Question 7

Question 7: 7. Physicists obtain correct scientific knowledge and promote the development of physics through obs...

7. Physicists obtain correct scientific knowledge and promote the development of physics through observation and in-depth study of experiments. The following statements are consistent with the facts

A. A. Becquerel's discovery of natural radioactivity shows that atomic nuclei have complex structures
B. B. Bohr's introduction of quantum ideas into the atomic realm nicely explains the discrete features of all atomic spectra
C. C. Maxwell confirmed Hertz's theory of electromagnetic fields through a series of experiments
D. D. Rutherford proposed a model for the nuclear structure of the atom through his study of cathode rays

Answer

Answer: A

Solution: A. Becquerel discovered the phenomenon of natural radioactivity, indicating that the atomic nucleus has a complex structure, so A is correct; B.Bohr introduced the quantum concept into the atomic field and successfully explained the experimental law of hydrogen atom spectrum, so B is wrong; C. Maxwell predicted the existence of electromagnetic waves, Hertz confirmed through a series of experiments on the electromagnetic theory of light, so C error; D. Rutherford proposed the model of atomic nuclear structure through the study of $\alpha$ particle scattering experiments, so D is wrong.

Question 8

Question 8: The reaction equation for two deuterium fusions is ${ } ^ { 2 } \mathrm { H } + { } _ { 1 } ^ { 2 } ...

The reaction equation for two deuterium fusions is ${ } ^ { 2 } \mathrm { H } + { } _ { 1 } ^ { 2 } \mathrm { H } \rightarrow { } _ { 2 } ^ { 3 } \mathrm { He } + { } _ { 0 } ^ { 1 } \mathrm { n }$, where ${ } ^ { 2 } \mathrm { H }$ has a mass of 2.0136 u, ${ } _ { 0 } ^ { 1 } \mathrm { n }$ has a mass of ${ } ^ { 1.0087 u } , { } ^ { 3 } \mathrm { He }$ and ${ } ^ { 3.0150 u }$ has a mass of ${ } ^ { 1 \mathrm { u } }$. FORMULA_4]]. It is known that ${ } ^ { 1 \mathrm { u } }$ is equal to 931.5 MeV of energy, then

A. A. The fusion reaction process absorbs energy
B. B. The fusion reaction releases about 3.26 MeV of nuclear energy.
C. C. The specific binding energy of ${ } ^ { 2 } \mathrm { H }$ is greater than that of ${ } ^ { 3 } \mathrm { He }$.
D. D. ${ } ^ { 2 } \mathrm { H }$ has an average nuclear mass less than ${ } ^ { 3 } \mathrm { He }$ has an average nuclear mass

Answer

Answer: B

Solution: A. Total mass before the reaction : $2 \times 2.0136 \mathrm { u } = 4.0272 \mathrm { u }$, total mass after the reaction : $3.0150 \mathrm { u } + 1.0087 \mathrm { u } = 4.0237 \mathrm { u }$, and the mass loss $\Delta m = 0.0035 \mathrm { u } > 0$ indicates that energy is released, so A is wrong. B. Energy is released: $\Delta E = 0.0035 \mathrm { u } \times 931.5 \mathrm { MeV } / \mathrm { u } \approx 3.26 \mathrm { MeV }$, so B is correct. C. Fusion produces a more stable nucleus (higher specific binding energy), and the specific binding energy of helium-3 should be greater than that of deuterons, so C is wrong. D. The average nuclear mass of a deuteron is $2.0136 \mathrm { u } / 2 \approx 1.0068 \mathrm { u }$; the average nuclear mass of helium-3 is $3.0150 \mathrm { u } / 3 \approx 1.0050 \mathrm { u }$. The deuteron has a higher average nuclear mass, so D is wrong.

Question 9

Question 9: 9. The following statements are correct

9. The following statements are correct

A. A. Einstein first introduced the concept of quantization of energy
B. B. De Broglie proposed that physical particles have volatility
C. C. Bohr's proposed atomic theory rejected the nuclear structure of atoms
D. D. Becquerel's discovery of natural radiation shows that atoms have a complex structure

Answer

Answer: B

Solution: A. Planck first proposed the concept of quantization of energy, so A is wrong; B. De Broglie proposed that physical particles have volatility, so B is correct; C. Bohr proposed atomic model, but did not negate Rutherford's atomic nuclear structure, so C error; D. Becquerel found that the natural radiation phenomenon shows that the atomic nucleus has a complex structure, so D error.

Question 10

Question 10: 10. The following nuclear reaction equation is a $\alpha$ decay

10. The following nuclear reaction equation is a $\alpha$ decay

A. A. ${ } _ { 92 } ^ { 238 } \mathrm { U } \rightarrow { } _ { 90 } ^ { 234 } \mathrm { Th } + { } _ { 2 } ^ { 4 } \mathrm { He }$
B. B. ${ } _ { 90 } ^ { 234 } \mathrm { Th } \rightarrow { } _ { 91 } ^ { 234 } \mathrm {~Pa} + { } _ { - 1 } ^ { 0 } \mathrm { e }$
C. C. ${ } _ { 7 } ^ { 14 } \mathrm {~N} + { } _ { 2 } ^ { 4 } \mathrm { He } \rightarrow { } _ { 8 } ^ { 17 } \mathrm { O } + { } _ { 1 } ^ { 1 } \mathrm { H }$
D. D. ${ } _ { 5 } ^ { 11 } \mathrm {~B} + { } _ { 2 } ^ { 4 } \mathrm { He } \rightarrow { } _ { 7 } ^ { 14 } \mathrm {~N} + { } _ { 0 } ^ { 1 } \mathrm { n }$

Answer

Answer: A

Solution: A. The reacting nucleus has only one nucleus, and there are $\alpha$ particles in the generating nucleus, which is $\alpha$ decay; B. The reacting nucleus has only one nucleus, and there are $\beta$ particles in the generating nucleus, which is $\beta$ decay. The $\alpha$ particles in the reacting nucleus are used as "shells" to bombard the corresponding nucleus, which is a kind of artificial nuclear transformation, CD is wrong. Therefore, choose A.

Question 11

Question 11: 11. The nuclear reaction of a hydrogen bomb explosion is ( )

11. The nuclear reaction of a hydrogen bomb explosion is ( )

A. A. ${ } _ { 1 } ^ { 2 } \mathrm { H } + { } _ { 1 } ^ { 3 } \mathrm { H } \rightarrow { } _ { 2 } ^ { 4 } \mathrm { He } + { } _ { 0 } ^ { 1 } \mathrm { n }$
B. B. ${ } _ { 7 } ^ { 14 } \mathrm {~N} + { } _ { 2 } ^ { 4 } \mathrm { He } \rightarrow { } _ { 8 } ^ { 17 } \mathrm { O } + { } _ { 1 } ^ { 1 } \mathrm { H }$
C. C. ${ } _ { 92 } ^ { 235 } \mathrm { U } + { } _ { 0 } ^ { 1 } \mathrm { n } \rightarrow { } _ { 56 } ^ { 144 } \mathrm { Ba } + { } _ { 36 } ^ { 89 } \mathrm { Kr } + 3 { } _ { 0 } ^ { 1 } \mathrm { n }$
D. D. ${ } _ { 4 } ^ { 9 } \mathrm { Be } + { } _ { 2 } ^ { 4 } \mathrm { He } \rightarrow { } _ { 6 } ^ { 12 } \mathrm { C } + { } _ { 0 } ^ { 1 } n$

Answer

Answer: A

Solution: hydrogen bomb explosion nuclear reaction is a light fusion reaction, the question A is light fusion, BD are the artificial transformation of the atomic nucleus equation, C is a heavy fission reaction; so choose A.

Question 12

Question 12: 12. The following statements about Einstein's mass-energy relationship are correct ( )

12. The following statements about Einstein's mass-energy relationship are correct ( )

A. A. Only moving objects have energy, stationary objects have no mass-energy
B. B. A certain amount of mass always corresponds to a certain amount of energy.
C. C. The energy in $E = m c ^ { 2 }$ $E$ is actually the internal energy of the object.
D. D. From $\Delta E = \Delta m c ^ { 2 }$, mass and energy can be converted into each other

Answer

Answer: B

Solution: Whether at rest or in motion, an object will have energy, so A is wrong; Einstein's mass-energy equation $E = m c ^ { 2 }$, which simply states that there is a simple proportionality between the amount of energy an object has and its mass. The energy of an object increases As the energy of an object increases, the mass increases; as the energy decreases, the mass decreases, and E is not the internal energy of the object. The fact that $\Delta E = \Delta m c ^ { 2 }$ shows that mass and energy are quantitatively related does not mean that the two quantities can be transformed into each other, so B is correct and CD is wrong. The mass-energy equation $E = m c ^ { 2 }$ shows that the energy of an object is proportional to its mass, and the energy in a nuclear reaction can be calculated according to $\Delta E = \Delta m c ^ { 2 }$.

Question 13

Question 13: 13. The following statements are correct ( )

13. The following statements are correct ( )

A. A. The photoelectric effect and the diffraction of electrons indicate that particles are volatile
B. B. $\alpha$ Particle scattering experiments can be used to determine the charge of atomic nuclei and to estimate the radius of atomic nuclei
C. C. Hydrogen atom radiates a photon of energy decreases, the acceleration of the motion of the electrons outside the nucleus decreases
D. D. The greater the specific binding energy, the less firmly the nuclei are bound and the more unstable the nucleus is

Answer

Answer: B

Solution: The photoelectric effect explains the particle nature of light, and the diffraction of electrons shows that particles have fluctuations, option A is wrong; $\alpha$ Particle scattering experiments can be used to determine the charge of the nucleus and estimate the radius of the nucleus. Therefore, B is correct. According to Bohr's theory, the hydrogen atom radiates a photon of energy decreases, the orbital radius decreases, the Coulomb force increases, and the acceleration of the electron increases. Therefore, C is wrong. Than the binding energy is greater, said the nucleus in the nucleus of the nucleus more firmly combined, the more stable nucleus, option D error; Therefore, choose B.

Question 14

Question 14: 14. The following narratives are correct ( ) $\frac { \left( m _ { \mathrm { A } } + m _ { \math...

14. The following narratives are correct ( ) $\frac { \left( m _ { \mathrm { A } } + m _ { \mathrm { P } } - m _ { \mathrm { D } } \right) c ^ { 2 } } { 2 }$ ($c$ indicates the speed of light in a vacuum)

A. A. The photoelectric effect and the Compton effect provide insight into the particle side of light, the former showing that photons have momentum and the latter showing that photons have energy
B. B. When the photoelectric effect occurs, the maximum initial kinetic energy of the emitted photoelectron is proportional to the frequency of the incident light
C. C. $\beta$ The essence of decay is the transformation of a neutron in the nucleus of an atom into a proton and an electron
D. D. Knowing that the masses of the neutron, proton and deuteron are $m _ { \mathrm { A } } , m _ { \mathrm { P } , } m _ { \mathrm { D } }$, the binding energy of the deuteron is

Answer

Answer: C

Solution: A. The photoelectric effect and the Compton effect reveal the particle side of light in depth, but the former shows that photons have energy, and the latter shows that photons have momentum, A error; B. According to the photoelectric effect equation can be obtained $$ E _ { \mathrm { k } } = h v - W _ { 0 } $$ The maximum initial kinetic energy of a photoelectron is not proportional to the frequency of the incident light, B is wrong; C.$\beta$ The essence of the decay is $$ { } _ { 0 } ^ { 1 } \mathrm { n } \rightarrow { } _ { 1 } ^ { 1 } \mathrm { H } + { } _ { - 1 } ^ { 0 } \mathrm { e } $$ C Correct ; D.The neutron and proton combine to form a deuteron with a mass loss, and according to Einstein's mass-energy equation, the binding energy is $$ \Delta E = \left( m _ { \mathrm { A } } + m _ { \mathrm { p } } - m _ { \mathrm { D } } \right) C ^ { 2 } $$ D Error.

Question 15

Question 15: 15. According to the latest reports, the radioisotope iron ${ } ^ { 166 } \mathrm { Ho }$, which has...

15. According to the latest reports, the radioisotope iron ${ } ^ { 166 } \mathrm { Ho }$, which has important medical applications, the difference between the number of neutrons in the nucleus and the number of electrons outside the nucleus of an atom is

A. A. 32
B. B. 67
C. C. 99
D. D. 166

Answer

Answer: A

Solution: According to the relationship between the quantities in the nucleus, we can see that the number of electrons outside the nucleus = the number of protons = 67, and the number of neutrons is $166 - 67 = 99$, so the difference between the number of neutrons in the nucleus and the number of electrons outside the nucleus is $99 - 67 = 32$. A.32, option A meets the question; B.67, option B does not fit the question; C.99 , option C does not fit the question; D.166 , option D does not satisfy the question;

Question 16

Question 16: 16. A radioactive element has $7 / 8$ nuclei decayed in 21.6 days, the half-life of this element is

16. A radioactive element has $7 / 8$ nuclei decayed in 21.6 days, the half-life of this element is

A. A. 3.8 days
B. B. 7.2 days
C. C. 10.8 days
D. D. 21.6 days

Answer

Answer: B

Solution: According to the half-life formula: $m = M \left( \frac { 1 } { 2 } \right) ^ { \frac { t } { T } }$ can be obtained: $\frac { 1 } { 8 } M = M \left( \frac { 1 } { 2 } \right) ^ { \frac { t } { T } }$, so there are, $t = 3 T = 21.6$ days, $T = 7.2$ days, that is, half-life is 7.2 days, so the ACD is wrong, B correct.

Question 17

Question 17: 17. The following statements are correct

17. The following statements are correct

A. A. In $L C$ oscillating circuits, the capacitor is always charged when the current is increased
B. B. The mirage phenomenon is related to the total reflection of light, you can think of the air on the surface of the sea as composed of many horizontal air layers with different refractive indices, the closer to the surface of the sea, the lower the temperature of the air, the higher the density, the lower the refractive index
C. C. Heavy fission needs to be triggered by "thermal neutrons", and fission produces "fast neutrons", which need to be slowed down by inserting the cadmium rod a little deeper.
D. D. Cathode rays, $\beta$ rays, and photocurrents all contain electrons

Answer

Answer: D

Solution: A. In the $L C$ oscillating circuit, the magnetic field energy increases and the electric field energy decreases when the current increases, and the capacitor must be in a discharged state, so A is wrong; B. The temperature of the lower air on the surface of the sea is lower than that of the upper air, the density of the lower air is greater than that of the upper air, and the refractive index is greater than that of the upper air, so B is wrong; C. Cadmium rods in the role of fission is to absorb the neutrons produced by the reaction to control the rate of nuclear reaction, regulating the speed of reaction, so C error; D. Cathode rays, $\beta$ rays and photocurrent contain electrons, so D is correct;

Question 18

Question 18: 18. The energy radiated from a star comes from various thermonuclear reactions that take place insid...

18. The energy radiated from a star comes from various thermonuclear reactions that take place inside the star. It is known that the specific binding energy of the Yao nucleus ${ } ^ { 2 } \mathrm { H }$ is $E _ { 1 }$, and that of the helium nucleus ${ } ^ { 4 } \mathrm { He }$ is $E _ { 2 }$. The energy released by a thermonuclear reaction ${ } ^ { 2 } \mathrm { H } + { } _ { 1 } ^ { 2 } \mathrm { H } \rightarrow { } _ { 2 } ^ { 4 } \mathrm { He }$ can be expressed as

A. A. $E _ { 2 } - E _ { 1 }$
B. B. $E _ { 2 } - 2 E _ { 1 }$
C. C. $4 E _ { 2 } - 2 E _ { 1 }$
D. D. $4 E _ { 2 } - 4 E _ { 1 }$

Answer

Answer: D

Solution: The reaction gives off heat, and the specific binding energy of the deuteron is $E _ { 1 }$ and that of the helium nucleus is $E _ { 2 }$, and according to the specific binding energy equals the ratio of the binding energy to the number of nuclei, the nuclear energy released in the nuclear reaction is $$ \Delta E = 4 E _ { 2 } - \left( 2 E _ { 1 } + 2 E _ { 1 } \right) = 4 E _ { 2 } - 4 E _ { 1 } $$ Therefore, D is correct and ABC is incorrect.

Question 19

Question 19: The half-lives of the two radioactive elements are ${ } ^ { t _ { 0 } }$ and ${ } ^ { 2 t _ { 0 } }$...

The half-lives of the two radioactive elements are ${ } ^ { t _ { 0 } }$ and ${ } ^ { 2 t _ { 0 } }$, the total number of nuclei of the two elements is $N$ at $t = 0$, and the total number of nuclei that have not yet decayed is $\frac { N } { 3 }$ at $t = 4 t _ { 0 }$. FORMULA_4]], the total number of nuclei that have not yet decayed is $\frac { N } { 3 }$, and the total number of nuclei that have not yet decayed is $t = 4 t _ { 0 }$, and the total number of nuclei that have not yet decayed is $\frac { N } { 3 }$.

A. A. $\frac { N } { 12 }$
B. B. $\frac { N } { 9 }$
C. C. $\frac { N } { 8 }$
D. D. $\frac { N } { 6 }$

Answer

Answer: C

Solution: According to the meaning of the question, the number of nuclei of the element with half-life $t _ { 0 }$ is $x$, and the number of nuclei of the other element is $y$, and according to the meaning of the question, there are $$ x + y = N $$ After $2 t _ { 0 }$ we have $$ \frac { 1 } { 4 } x + \frac { 1 } { 2 } y = \frac { N } { 3 } $$ The association gives $$ x = \frac { 2 } { 3 } N , y = \frac { 1 } { 3 } N $$ At ${ } ^ { t = 4 t _ { 0 } }$, the element with $x$ has experienced 4 half-lives, and the element with $y$ has experienced 2 half-lives, so the total number of undecayed nuclei at this time is $$ n = \frac { 1 } { 2 ^ { 4 } } x + \frac { 1 } { 2 ^ { 2 } } y = \frac { N } { 8 } $$

Question 20

Question 20: 20. Cosmic rays interact with the Earth's atmosphere to produce neutrons, which strike nitrogen in t...

20. Cosmic rays interact with the Earth's atmosphere to produce neutrons, which strike nitrogen in the atmosphere and trigger a nuclear reaction that produces carbon 14. The nuclear reaction equation is ${ } ^ { 14 } \mathrm {~N} + { } _ { 0 } ^ { 1 } \mathrm { n } \rightarrow { } _ { 6 } ^ { 14 } \mathrm { C } + \mathrm { X }$, and carbon 14 is radioactive and is capable of spontaneously decaying to nitrogen in $\beta$. The reaction equation is ${ } ^ { 14 } \mathrm { C } \rightarrow { } _ { 7 } ^ { 14 } \mathrm {~N} + \mathrm { Y }$. Carbon 14 can be used to date substances. The half-life of carbon 14 is 5730 years.

A. A. X for electronic ${ } ^ { - 1 } \mathrm { e }$
B. B. Y for proton ${ } ^ { 1 } { } ^ { 1 } \mathrm { H }$
C. C. The amount of carbon 14 in a sample of ancient wood was $\frac { 1 } { 4 }$ that of a modern plant, suggesting that the ancient wood was roughly 5730 years old
D. D. The amount of carbon 14 in a sample of ancient wood was $\frac { 1 } { 4 }$ that of a modern plant, suggesting that the ancient wood was roughly 11,460 years old

Answer

Answer: D

Solution: AB. According to the conservation of mass and charge, X means proton and Y means electron, so AB is wrong; CD. According to the relationship between the residual mass and half-life of the radioactive material, it is satisfied that $$ m = m _ { 0 } \left( \frac { 1 } { 2 } \right) ^ { \frac { t } { T } } $$ Solve the following $$ t = 2 T = 11460 \text { 年 } $$ Therefore, D is correct and C is incorrect.

Question 21

Question 21: 21. The radioactive element radon (${ } ^ { 222 } \mathrm { Rn }$) decays into polonium ${ } ^ { 218...

21. The radioactive element radon (${ } ^ { 222 } \mathrm { Rn }$) decays into polonium ${ } ^ { 218 } \mathrm { Po }$ via $\alpha$, which has a half-life of 3.8 days; however, surveys have shown that, after a long geologic time, there are still naturally occurring ores in the Earth's crust that contain the radioactive element ${ } ^ { 222 } \mathrm { Rn }$. _FORMULA_3]] in the earth's crust after long geologic ages because ( )

A. A. The ${ } ^ { 222 } \mathrm { Rn }$ currently in the Earth's crust comes mainly from the decay of other radioactive elements
B. B. At the beginning of the Earth's formation, the amount of the element ${ } ^ { 222 } \mathrm { Rn }$ in the Earth's crust was sufficiently high
C. C. When the decay product ${ } ^ { 218 } { } ^ { 84 } \mathrm { Po }$ accumulates to a certain amount, the increase of ${ } _ { 84 } ^ { 218 } \mathrm { Po }$ slows down the decay process of ${ } ^ { 222 } \mathrm { Rn }$.
D. D. ${ } ^ { 222 } \mathrm { Rn }$ is found primarily in ores deep in the earth, where temperature and pressure alter its half-life

Answer

Answer: A

Solution: AB . Even if the content of elemental radon is high enough, there is almost no radon in the earth's crust after a long geological era, so the radon in the earth's crust at present mainly comes from the decay of other radioactive elements, so A is correct and B is wrong; CD.The length of the half-life of an element is determined by the factors of the atomic nucleus itself, and has nothing to do with the physical and chemical state in which the atom is located as well as the surrounding environment, temperature and pressure, so CD is wrong.

Question 22

Question 22: 22. Researchers have successfully used the energy produced by the decay of plutonium-238 as a source...

22. Researchers have successfully used the energy produced by the decay of plutonium-238 as a source of energy for the artificial heart, with the decay equation ${ } ^ { 94 } \mathrm { Pu } \rightarrow { } _ { 95 } ^ { 238 } \mathrm { Am } + \mathrm { X }$; some nuclear batteries utilize the energy released by the decay of uranium-232 to provide electricity, with the decay equation ${ } ^ { 232 } \mathrm { U } \rightarrow { } _ { 90 } ^ { 228 } \mathrm { Th } + \mathrm { Y }$, and the following statements are correct ([INLINE_FORMULA_1]). ) ![](/images/questions/phys-nuclear-physics/image-001.jpg)

A. A. Plutonium 238 undergoes $\alpha$ decay.
B. B. Decay of $\beta$ by uranium 232
C. C. Y has 4 more neutrons than X.
D. D. The specific binding energy of plutonium-238 is less than that of uranium-232.

Answer

Answer: D

Question 23

Question 23: 23. The following statements are correct ( )

23. The following statements are correct ( )

A. A. ${ } ^ { 235 } \mathrm { U } + { } ^ { 0 } \mathrm { n } \rightarrow { } ^ { 89 } \mathrm { Kr } + { } ^ { 144 } \mathrm { Ba } + 3 ^ { 0 } \mathrm { n }$ is fusion reaction
B. B. Radioactive elements are not radioactive when they form compounds with other elements.
C. C. Becquerel discovered the neutron by experiment, and Thomson discovered the proton by experiment
D. D. According to Borel's atomic model, the hydrogen atom can radiate up to 6 different frequencies of photons when it jumps from the excited state at quantum number $n = 4$ to the ground state

Answer

Answer: D

Solution: A. The reaction shown in A is a heavy fission reaction; B.Radioactive elements are still radioactive and their half-lives do not change when they form compounds with other elements; C. Chadwick discovered the neutron by experiment, Thomson discovered the electron by experiment, C error; D. According to Bohr's atomic model, the hydrogen atom can radiate at most $C _ { 4 } ^ { 2 } = 6$ when it leaps from the excited state of quantum number $n = 4$ to the ground state. D is correct;

Question 24

Question 24: 24. The specific binding energy of the nucleus of an atom is related to the mass number of the image...

24. The specific binding energy of the nucleus of an atom is related to the mass number of the image shown in the figure, the following statements are wrong ( ) ![](/images/questions/phys-nuclear-physics/image-002.jpg)

A. A. The binding energy of an atomic nucleus is equal to the minimum energy required to completely break it up into free nucleons
B. B. The higher the specific binding energy, the smaller the average mass of the nucleus and the more stable the nucleus is
C. C. ${ } ^ { 6 } \mathrm { Li }$ nuclei have a smaller specific binding energy than ${ } ^ { 4 } \mathrm { He }$ nuclei, so ${ } ^ { 6 } \mathrm { Li }$ nuclei are more stable than ${ } ^ { 4 } \mathrm { He }$ nuclei
D. D. ${ } ^ { 235 } \mathrm { U }$ Fission into an intermediate mass nucleus results in a mass loss, releasing energy

Answer

Answer: C

Solution: A. According to the definition of binding energy, the energy released when dispersed nuclei form the nucleus is called the binding energy of the nucleus, so the binding energy of the nucleus is equal to the minimum energy required to make it completely decompose into free nuclei, A is correct and does not meet the meaning of the question; B.The larger the specific binding energy, the smaller the average mass of the nucleus, and the more stable the nucleus is, B is correct and does not meet the meaning of the question; C.The ${ } ^ { 6 } \mathrm { Li }$ nucleus has a smaller specific binding energy than the ${ } ^ { 4 } \mathrm { He }$ nucleus, so the ${ } ^ { 6 } \mathrm { Li }$ nucleus is more unstable than the ${ } ^ { 4 } \mathrm { He }$ nucleus; D. ${ } ^ { 235 } \mathrm { U }$ fission into an intermediate-mass nucleus will result in a mass loss and release of energy, D is correct and does not meet the meaning of the question.

Question 25

Question 25: 25. The figure below is a schematic diagram of Chadwick's experiment to study the internal structure...

25. The figure below is a schematic diagram of Chadwick's experiment to study the internal structure of the atomic nucleus, by the naturally occurring radioactive element polonium (Po) emits $\alpha$ rays bombardment of beryllium will produce a stream of particles $a$, with a stream of particles $a$ after striking the paraffin wax will be hit by the particle flow $b$, it was known that () $b$ , it is known upon study that ( ) ![](/images/questions/phys-nuclear-physics/image-003.jpg)

A. A. $a$ for protons, $b$ for neutrons
B. B. $a$ for $\gamma$ rays, $b$ for neutrons
C. C. $a$ for neutrons, $b$ for $\gamma$ rays
D. D. $a$ for neutrons, $b$ for protons

Answer

Answer: D

Solution: When beryllium is bombarded with $\alpha$ rays from polonium, a penetrating neutral ray is emitted, which is called beryllium "radiation", i.e., neutron flow, and neutron bombardment of paraffin wax knocks protons out of hydrogen, i.e., forming a proton flow. Therefore, $a$ is a neutron and $b$ is a proton, so ABC is wrong and D is correct.

Question 26

Question 26: 26. By 2025, China's nuclear power construction will have entered a full-scale sprint. With the succ...

26. By 2025, China's nuclear power construction will have entered a full-scale sprint. With the successive commissioning of new commercial reactors, China's nuclear power technology is changing from a chaser to a frontrunner, accelerating the arrival of a new era of global clean energy. Qinshan Nuclear Power Plant is the first nuclear power plant researched, designed and built by China itself, and the "fuel" in its nuclear reactor is ${ } ^ { 235 } \mathrm { U }$, and the equation of a nuclear reaction is ${ } _ { 92 } ^ { 235 } \mathrm { U } + { } _ { 0 } ^ { 1 } \mathrm { n } \rightarrow { } _ { \mathrm { Z } } ^ { 88 } \mathrm { Sr } + { } _ { 54 } ^ { \mathrm { A } } \mathrm { Y } + 10 { } _ { 0 } ^ { 1 } \mathrm { n }$. The following statement is correct ( )

A. A. $Z = 38 , A = 137$
B. B. The neutrons produced by this reaction can directly trigger nuclear fission and are called "slow neutrons".
C. C. Neutron absorption by cadmium rods in nuclear reactors at nuclear power plants to control the reaction rate
D. D. One neutron can be eliminated on each side of this nuclear reaction equation, i.e. the nature of this nuclear reaction is decay

Answer

Answer: C

Solution: A. According to the conservation of charge we have $\mathrm { Z } + 54 = 92$ Solve for $Z = 38$ According to the conservation of mass number we have $235 + 1 = 88 + \mathrm { A } + 10$ Solve for $A = 138$, so $A$ is wrong; B. The neutron rate produced by the nuclear reaction is too fast, called fast neutrons, and uranium 235 "brush", uranium nucleus can not "catch" it, can not occur nuclear fission, so B error; C. If the number of neutrons is too much, the reaction speed is too fast, the use of cadmium rods to absorb neutrons, control the reaction speed, so C is correct; D. The reaction requires neutron bombardment of uranium 235, the formation of the excited state of uranium 236, thus splitting into several parts, uranium 235 will decay, but its half-life of $7.0 \times 10 ^ { 8 }$ years, the nuclear reactor is mainly utilized in the fission of uranium 235, so D error.

Question 27

Question 27: 27. Studies have shown that the nucleus of some elements may "capture" an electron from an extra-nuc...

27. Studies have shown that the nucleus of some elements may "capture" an electron from an extra-nuclear electron that is very close to it to form a new atom, a phenomenon known as "K-capture". For example, a beryllium nucleus (${ } ^ { 7 } \mathrm { Be }$) captures an electron from a K-layer electron orbital to form a lithium nucleus in an excited state and an energetic, uncharged neutrino $v$ with zero mass. The lithium nucleus in the excited state (${ } ^ { 7 } { } ^ { \mathrm { Li } }$) spontaneously emits $\gamma$ photons and returns to the ground state. The following statement is correct ( )

A. A. This reaction is a $\beta$ decay
B. B. No mass loss before or after the reaction
C. C. Beryllium has 4 neutrons and 3 protons in its nucleus.
D. D. The essence of the reaction is that a proton captures an electron and produces a neutron

Answer

Answer: D

Solution: AD . The nature of the reaction is that a proton captures an electron and then produces a neutron; it is not the release of an electron, so it is not a $\beta$ decay, so A is wrong and D is correct; B. The reaction releases energy, and there is a mass loss before and after the reaction, so B is wrong; C. The nucleus of the beryllium atom (${ } ^ { 7 } \mathrm { Be }$) has three neutrons and four protons, so C is wrong.

Question 28

Question 28: 28. The following statements are correct ( )

28. The following statements are correct ( )

A. A. Iodine 131 (i.e. ${ } ^ { 131 }$ I) is nuclear unstable and undergoes $\beta$ decay with a half-life of 8 days. After 8 days, the number of 30 radioactive iodine 131 decays must be 15.
B. B. A certain monochromatic light is used to irradiate a large number of hydrogen atoms in the $n = 2$ energy level in a container, and after absorbing this kind of electrons, the hydrogen atoms can emit three kinds of photons $\left( { } _ { 1 } > \lambda _ { 2 } > \lambda _ { 3 } \right)$ with wavelengths of ${ } ^ { \lambda _ { 1 } } , \lambda _ { 2 } , \lambda _ { 3 }$, respectively, and the wavelengths of the irradiated photons are $\lambda _ { 1 }$. FORMULA_3]]
C. C. $\alpha$ rays, $\beta$ rays, and $\gamma$ rays are all essentially electromagnetic waves, and $\gamma$ rays have the shortest wavelengths
D. D. When a certain monochromatic light irradiation of a metal surface can produce the photoelectric effect, the greater the wavelength of the incident light, the greater the maximum initial kinetic energy of the photoelectric effect produced

Answer

Answer: B

Solution: A. Half-life is a statistical law for a large number of nuclei, not applicable to a few particles, so A error; B. In the excited state of the hydrogen atom is not stable, can spontaneously to the low energy level jump and emit photons, hydrogen atoms only emit three different frequencies of the color light, know that the hydrogen atom in the $n = 3$ energy level. So a certain monochromatic light irradiates a large number of hydrogen atoms in the $n = 2$ energy level in a container, and leaps to the hydrogen atom in the $n = 3$ energy level. After absorbing this photon, the hydrogen atom is able to emit three photons with wavelengths of $\lambda _ { 1 } , \lambda _ { 2 } , \lambda _ { 3 }$ , ($\lambda _ { 1 } > \lambda _ { 2 } > \lambda _ { 3 }$) according to the photon frequency $$ v = \frac { c } { \lambda } $$ The smaller the photon wavelength, the greater the frequency. From ${ } ^ { n = 2 }$ leaps to $^ { n = 3 }$ then the wavelength of the irradiated light photon is ${ } ^ { \lambda _ { 1 } }$, so B is correct; C. The $\alpha$ ray is a helium nucleus, the $\beta$ The $\alpha$ rays are helium nuclei, $\beta$ rays are high-speed electrons, and only the $\gamma$ rays are essentially electromagnetic waves, so C is incorrect. $$ E _ { \mathrm { km } } = \frac { h c } { \lambda } - W _ { 0 } $$ D. According to the photoelectric effect equation $$ E _ { \mathrm { km } } = \frac { h c } { \lambda } - W _ { 0 } $$ can produce photoelectric effect, the incident light wavelength is larger, the maximum initial kinetic energy produced by the photoelectric effect is smaller. Therefore, D is wrong.

Question 29

Question 29: 29. Nuclear batteries are the ideal energy source for deep space probes, which are not affected by e...

29. Nuclear batteries are the ideal energy source for deep space probes, which are not affected by extremely cold or hot temperatures or disturbed by cosmic rays, and the plutonium-238 isotope temperature-difference batteries are based on the principle that the nuclear energy released when it decays is converted into electrical energy. The decay equation of plutonium-238 is ${ } ^ { 238 } \mathrm { Pu } \rightarrow { } _ { 92 } ^ { 234 } \mathrm { U } + \mathrm { X }$. The following statement is correct ( )

A. A. The rays emitted by the decay are a high-speed helium-nucleus stream, which is highly penetrating.
B. B. The average mass of nuclei in ${ } ^ { 238 } \mathrm { Pu }$ is less than the average mass of nuclei in ${ } ^ { 234 } \mathrm { U }$
C. C. The specific binding energy of ${ } ^ { 238 } \mathrm { Pu }$ is less than the specific binding energy of ${ } ^ { 234 } \mathrm { U }$.
D. D. The nucleus of ${ } ^ { 238 } \mathrm { Pu }$ has 4 more neutrons than the nucleus of ${ } ^ { 234 } \mathrm { U }$.

Answer

Answer: C

Solution: A. Helium nuclei are highly ionizable and weakly penetrating, so A is wrong; B. The average mass of nuclei in ${ } ^ { 238 } \mathrm { Pu }$ is smaller than that of nuclei in ${ } ^ { 234 } \mathrm { U }$ because of the mass loss due to energy release, so it is wrong in B. The average mass of helium nuclei is smaller than that of helium nuclei in ${ } ^ { 234 } \mathrm { U }$ because of the mass loss due to energy release; C. ${ } _ { 92 } ^ { 234 } \mathrm { U }$ is more stable than ${ } ^ { 238 } \mathrm { Pu }$, so the specific binding energy of ${ } ^ { 238 } \mathrm { Pu }$ is smaller than that of ${ } ^ { 234 } \mathrm { U }$, so C is correct; D. From the mass number and the nuclear charge number, it can be seen that the nucleus of ${ } ^ { 94 } \mathrm { Pu }$ has 2 more neutrons than that of ${ } ^ { 92 } \mathrm { U }$, so D is wrong.

Question 30

Question 30: 30. The following statements are correct ( )

30. The following statements are correct ( )

A. A. When a uranium nucleus undergoes $\alpha$ decay, it releases $\alpha$ particles and a certain amount of energy, and it is this spontaneous release of energy that is currently utilized in nuclear power plants.
B. B. If the photoelectric effect can occur by irradiating a metal with violet light, the photoelectric effect must not occur by using red light instead.
C. C. Hydrogen atoms absorb photons of a certain frequency when they jump from a higher energy level to a lower energy level
D. D. Both mechanical and electromagnetic waves have interference and diffraction properties

Answer

Answer: D

Solution: A, $\alpha$ decay of the energy released is relatively small, the current nuclear power plant is utilized by the fission of uranium nuclei release of energy, so A error. B, irradiation of a certain metal with violet light can be photoelectric effect can be known that the frequency of violet light is greater than the metal's limiting frequency, the frequency of red light is less than the frequency of violet light, irradiation with red light does not necessarily produce the photoelectric effect, so B error; C, the hydrogen atom outside the nuclear electron from a higher energy level jump to a lower energy level, the energy is reduced to release a certain frequency of photons. C, the hydrogen atom's extra-nuclear electrons from the higher energy level jump to the lower energy level, the energy decreases, to release a certain frequency of photons, so C is wrong. D, interference and diffraction are Porter has the nature of mechanical and electromagnetic waves have interference, diffraction characteristics, so D is correct. Therefore, choose D. [Eyesight] This question examines the photoelectric effect, Bohr theory and other basic knowledge points, the key to familiarize themselves with the textbook, more accumulation, keep in mind these basic knowledge points.

Question 31

Question 31: 31. In medicine, $\gamma$ rays produced by cobalt 60 are commonly used to treat patients with malign...

31. In medicine, $\gamma$ rays produced by cobalt 60 are commonly used to treat patients with malignant tumors. The decay mode of cobalt 60 Cobalt 60 has a decay program of $\mathrm { Co } \rightarrow { } _ { 28 } ^ { 60 } \mathrm { Ni } + \mathrm { X }$ and a half-life of 5.27 years. The following statement is true

A. A. The half-life of cobalt 60 decreases at elevated ambient temperatures.
B. B. X particle is a proton ${ } ^ { 1 } \mathrm { H }$
C. C. The X particle is released when a proton in the nucleus of a cobalt 60 atom is converted into a neutron.
D. D. 0.1 g of cobalt 60 after 5.27 years, 0.05 g remains undecayed

Answer

Answer: D

Solution: A. The external environment does not affect the half-life of radioactive elements; BC. BC.By the conservation of mass, the X particle is ${ } ^ { - 1 } \mathrm { e }$, and the decay is ${ } ^ { \beta }$, in which a neutron in the nucleus of the atom is converted into an electron and a proton, so BC is wrong; D. From the half-life formula, the undecayed mass is $$ m = \frac { 1 } { 2 } m _ { 0 } = 0.05 \mathrm {~g} $$ Therefore, D is correct.

Question 32

Question 32: 32. In the following four nuclear reaction equations, X denotes a proton

32. In the following four nuclear reaction equations, X denotes a proton

A. A. ${ } _ { 15 } ^ { 30 } \mathrm { P } \rightarrow { } _ { 14 } ^ { 30 } \mathrm { Si } + \mathrm { X }$
B. B. ${ } _ { 92 } ^ { 238 } U \rightarrow { } _ { 90 } ^ { 234 } T h + X$
C. C. ${ } _ { 13 } ^ { 27 } \mathrm { Al } + { } _ { 0 } ^ { 1 } \mathrm { n } \rightarrow { } _ { 12 } ^ { 27 } \mathrm { Mg } + \mathrm { X }$
D. D. ${ } _ { 13 } ^ { 27 } \mathrm { Al } + { } _ { 2 } ^ { 4 } \mathrm { He } \rightarrow { } _ { 15 } ^ { 30 } \mathrm { P } + \mathrm { X }$

Answer

Answer: C

Solution: From the equation $A$, $X$ has a mass number equal to zero and a charge of 1, so $X$ is a positron and $A$ is incorrect; $B$ in $B$ is a positron and $X$ is an electron. [INLINE_FORMULA_5]] In $B$, $X$ has a mass of 4 and a charge of 2, so X is a helium nucleus, B is wrong; in C, X has a mass of 1 and a charge of 1, so X is a proton, C is correct; in D, X has a mass of 1 and a charge of 0, so X is a neutron, D is wrong; therefore, C is chosen. Point: This question examines the conservation of mass and charge before and after a nuclear reaction equation; at the same time, it is important to remember the symbols of various elementary particles, such as protons, neutrons, $\alpha$ particles, and positives and negatives electrons, and so on.

Question 33

Question 33: 33. The following statements are true about modern physics

33. The following statements are true about modern physics

A. A. The higher the binding energy, the more strongly the nucleons are bound within the nucleus and the more stable the nucleus is
B. B. One nuclear reaction equation for uranium fission is ${ } ^ { 235 } \mathrm { U } \rightarrow { } _ { 56 } ^ { 141 } \mathrm { Ba } + { } _ { 36 } ^ { 92 } \mathrm { Kr } + 2 { } _ { 0 } ^ { 1 } \mathrm { n }$
C. C. X is a neutron in ${ } _ { 92 } ^ { 238 } \mathrm { U } \rightarrow { } _ { 90 } ^ { 234 } \mathrm { Th } + \mathrm { X }$ and the nuclear reaction type is $\beta$ Decay
D. D. Nuclear energy must be released when nuclei with low average binding energies combine or break up into nuclei with high average binding energies.

Answer

Answer: D

Solution: A. The larger the specific binding energy, the more firmly the nucleons in the nucleus are bound, and the more stable the nucleus is, option A is wrong; B. In the nuclear reaction equation of uranium fission, the reactants must have neutrons, option B is wrong; C. X in ${ } _ { 92 } ^ { 238 } \mathrm { U } \rightarrow { } _ { 90 } ^ { 234 } \mathrm { Th } + \mathrm { X }$ has a mass of 4 and an electric charge of 2. It is a $\alpha$ particle, and the type of nuclear reaction is $\alpha$ decay; D. The average binding energy is the energy released by each nucleon when it combines with a nucleus to form a nucleus. The larger the average binding energy, the more stable the nucleus is, and the nucleus with a small average binding energy must release nuclear energy when it combines with or decomposes into a nucleus with a large average binding energy and a mass loss. Therefore, D is correct.

Question 34

Question 34: 34. ${ } _ { 61 } ^ { 147 } \mathrm { Pm }$ can be used to make thickness gauges and radioisotope ba...

34. ${ } _ { 61 } ^ { 147 } \mathrm { Pm }$ can be used to make thickness gauges and radioisotope batteries when $\beta$ decays, and the half-life of ${ } ^ { 147 } \mathrm { Pm }$ is known to be 2.64 years.

A. A. ${ } _ { 61 } ^ { 147 } \mathrm { Pm }$ is used as a thickness gauge to take advantage of the ionization properties of $\beta$ rays.
B. B. The charge of the new nucleus after ${ } _ { 61 } ^ { 147 } \mathrm { Pm }$ decay of $\beta$ is 62
C. C. 100 ${ } ^ { 147 } \mathrm { Pm }$ nuclei after 2.64 years 50 remain
D. D. Increases isotope cell life when temperatures are lowered

Answer

Answer: B

Solution: A. ${ } ^ { 147 } \mathrm { Pm }$ is used as a thickness gauge to take advantage of the penetrating ability of $\beta$ rays, so A is wrong; B. ${ } ^ { 147 } \mathrm { Pm }$ decays and releases an electron ${ } ^ { - 1 } \mathrm { e }$, and the charge of the new nucleus is 62 because of the conservation of charge before and after the decay, so B is correct; C. Half-life is the result of a large number of nuclei statistics, for some nuclei do not hold, so C is wrong; D. Half-life is determined by the nucleus itself, the temperature will not affect the half-life, so D error. D. The half-life is determined by the nucleus itself, and the temperature does not affect the half-life.

Question 35

Question 35: 35. In order to explain the nature of light, Newton proposed the particle theory of light, and today...

35. In order to explain the nature of light, Newton proposed the particle theory of light, and today it is recognized that light has wave-particle duality. ![](/images/questions/phys-nuclear-physics/image-004.jpg)

A. A. (1) (2)
B. B. (2) (3)
C. C. (3) (4)
D. D. (2) (4)

Answer

Answer: B

Solution: (1) The experiment was a $\alpha$ particle scattering experiment, on the basis of which Rutherford proposed the doctrine of the nuclear structure of the atom, which has nothing to do with the nature of light. (2) Interference is a unique property of waves, so the two-hole interference experiment shows that light has volatility. (3) this experiment is the photoelectric effect experiment, indicating that light has a particle nature (4) three rays in the electric field deflection experiment, can determine the electrical nature of the rays, can not explain the nature of light so choose B.

Question 36

Question 36: 36. According to Planck, the founder of quantum theory, the history of science is not only a sequenc...

36. According to Planck, the founder of quantum theory, the history of science is not only a sequence of facts, rules and consequent mathematical descriptions; it is also a history of concepts. The following description is correct

A. A. The study of the spectra of hydrogen atoms led to the modeling of the nuclear structure of the atom
B. B. Half-life is related to both the amount and time of the substance, and can be used to determine geologic age, biological age, etc.
C. C. The rays produced in natural radiation phenomena can all be deflected in electric or magnetic fields
D. D. Light nuclear energy undergoes fusion by overcoming the Coulomb repulsion, not the nuclear force.

Answer

Answer: D

Solution: A.$\alpha$ The experiment of particle bombardment of gold foil led to the modeling of the nuclear structure of the atom, so A is wrong; B.The half-life is determined by the internal factors of the nucleus and has nothing to do with the amount of matter or time, so B is wrong; C. The $Y$ rays produced by the phenomenon of natural radioactivity are electromagnetic waves, which are not electrically charged and cannot be deflected in electric or magnetic fields, so C is wrong; D. Light nuclear energy fusion conditions is to overcome the Coulomb repulsion, rather than overcome the nuclear force, so D is correct.

Question 37

Question 37: 37. Nuclear power, as a clean, low-carbon, safe, efficient, stable and low-carbon energy source that...

37. Nuclear power, as a clean, low-carbon, safe, efficient, stable and low-carbon energy source that could be developed on a large scale, played an important role in promoting green development and helping to realize the "double carbon" goal. According to statistics, a million kilowatts of electric power of nuclear power plants and coal-fired power plants, compared with the annual carbon dioxide emissions can be reduced by more than 6 million tons, can be seen that nuclear energy is one of the larger emission reduction effect of energy. The following statements about nuclear reactions are correct

A. A. Heavy fission releases energy while light fusion absorbs it.
B. B. The equation for nuclear fission occurring in a nuclear reactor is ${ } ^ { 235 } \mathrm { U } + { } _ { 0 } ^ { 1 } n \rightarrow { } _ { 56 } ^ { 144 } \mathrm { Ba } + { } _ { 36 } ^ { 89 } \mathrm { Kr } + 2 { } _ { 0 } ^ { 1 } n$
C. C. A ${ } ^ { 56 } \mathrm { Ba }$ nucleus from nuclear fission has 56 protons and 88 neutrons.
D. D. The half-life of ${ } ^ { 235 } \mathrm { U }$ is 700 million years, 100 ${ } ^ { 92 } \mathrm { U }$ nuclei have not decayed in 50 of them after 700 million years.

Answer

Answer: C

Solution: A. When nuclear fusion releases energy, there is a loss of mass, so A is wrong; B.The equation for nuclear fission in a nuclear reactor is $$ { } _ { 92 } ^ { 235 } \mathrm { U } + { } _ { 0 } ^ { 1 } n \rightarrow { } _ { 56 } ^ { 144 } \mathrm { Ba } + { } _ { 36 } ^ { 89 } \mathrm { Kr } + 3 _ { 0 } ^ { 1 } n $$ Therefore, B is wrong; C. There are 56 protons and 88 neutrons in a ${ } ^ { 56 } \mathrm { Ba }$ nucleus produced by nuclear fission, so C is correct; D. Half-life is a statistical law for a large number of particles, and 100 $^ { 92 } { } ^ { 235 } \mathrm { U }$ nuclei don't conform to the half-life law, so D is wrong. Therefore, C is correct.

Question 38

Question 38: 38. On August 24, 2023, local time, the Fukushima Daiichi nuclear power plant in Japan initiated the...

38. On August 24, 2023, local time, the Fukushima Daiichi nuclear power plant in Japan initiated the discharge of nuclear contaminated water into the sea. Radioactive elements in the contaminated water pose an immeasurable threat to human society and the health of the marine ecosystem, and one of the nuclear reactions is The half-life of ${ } _ { 38 } ^ { 90 } \mathrm { Sr } \rightarrow { } _ { 39 } ^ { 90 } \mathrm { Y } + \mathrm { X } \quad , \quad { } _ { 38 } ^ { 90 } \mathrm { Sr }$ is 28 years, the following statements are correct

A. A. X is a neutron
B. B. In seawater, the half-life of ${ } ^ { 90 } \mathrm { Sr }$ decreases
C. C. The specific binding energy of ${ } ^ { 99 } \mathrm { Y }$ is greater than that of ${ } ^ { 90 } \mathrm { Sr }$.
D. D. 50 ${ } ^ { 90 } \mathrm { Sr }$ nuclei after 28 years, only 25 ${ } ^ { 90 } \mathrm { Sr }$ nuclei left undecayed

Answer

Answer: C

Solution: A. From the nuclear reaction ${ } ^ { 98 } \mathrm { Sr } \rightarrow { } _ { 39 } ^ { 90 } \mathrm { Y } + \mathrm { X }$, according to the conservation of mass and charge, we can know that X is an electron, so A is wrong; B. In seawater, the half-life of ${ } ^ { 90 } \mathrm { Sr }$ remains unchanged, so B is wrong; C. The larger the specific binding energy is, the more stable the nucleus is. The specific binding energy of ${ } ^ { 39 } \mathrm { Y }$ is larger than that of ${ } ^ { 90 } \mathrm { Sr }$, so C is correct; D. Half-life is a statistical law for a large number of particles and does not apply to a few nuclei, so D is wrong.

Question 39

Question 39: 39. In about 5 billion years the Sun will no longer be able to sustain its current fusion and will b...

39. In about 5 billion years the Sun will no longer be able to sustain its current fusion and will begin gravitationally collapsing to increase its temperature until its core is hot enough to cause the following new fusion process: ${ } ^ { 4 } \mathrm { He } + { } _ { 4 } ^ { 8 } \mathrm { Be } \rightarrow { } _ { 6 } ^ { 12 } \mathrm { C }$, then ( ) ![](/images/questions/phys-nuclear-physics/image-005.jpg) atomic nucleus mass number

A. A. ${ } ^ { 4 } \mathrm { He }$ is more stable than ${ } ^ { 12 } \mathrm { C }$.
B. B. The average nuclear mass of ${ } _ { 2 } ^ { 4 } \mathrm { He }$ is larger than that of ${ } ^ { 12 } \mathrm { C }$.
C. C. The binding energy of ${ } _ { 2 } ^ { 4 } \mathrm { He }$ is larger than that of ${ } ^ { 12 } \mathrm { C }$.
D. D. This fusion process requires the absorption of energy

Answer

Answer: B

Solution: A. The size of the specific binding energy reflects the degree of stability of the atomic nucleus, the larger the specific binding energy is, the more firmly the nuclei are bound, the more stable the nucleus is. The ${ } ^ { 12 } \mathrm { C }$ is more stable than ${ } ^ { 4 } \mathrm { He }$, so A is wrong; BCD. From the figure, we can see that the binding energy of ${ } ^ { 12 } \mathrm { C }$ is larger than that of ${ } ^ { 4 } \mathrm { He }$, and light fusion releases energy, the specific binding energy increases, and there is a mass loss, so the average mass of the nucleus decreases, so B is correct and CD is wrong;

Question 40

Question 40: 40. When U-235 is bombarded by neutrons, barium and krypton are produced, while three nuclei Z are r...

40. When U-235 is bombarded by neutrons, barium and krypton are produced, while three nuclei Z are released, and the equation of nuclear fission is ${ } _ { 92 } ^ { 235 } \mathrm { U } + { } _ { 0 } ^ { 1 } \mathrm { n } \rightarrow { } _ { 56 } ^ { 144 } \mathrm { Ba } + { } _ { 36 } ^ { 89 } \mathrm { Kr } + 3 \mathrm { Z }$, then ( ) High School Physics Assignment, October 30, 2025

A. A. Z is a proton.
B. B. Conservation of nucleon number during fission, no mass loss
C. C. The specific binding energy of ${ } ^ { 144 } \mathrm { Ba }$ is greater than that of ${ } ^ { 235 } \mathrm { U }$.
D. D. The purpose of using moderators in nuclear reactors is to slow down nuclear reactions

Answer

Answer: C

Solution: A. Let the mass number of Z be $m$ and the number of protons be $n$, then according to the conservation of mass number and the conservation of the number of protons, there are $$ \begin{gathered} 235 + 1 = 144 + 89 + 3 m \\ 92 = 56 + 36 + 3 n \end{gathered} $$ The solution is $$ m = 1 , n = 0 $$ Then we can see that Z is a neutron, so A is wrong; B. Although the number of nuclei is conserved in the fission process, the average mass of nuclei before and after the reaction is different, so there is a mass loss in the fission process, so B is wrong; C. The new nucleus generated after fission of a heavy nucleus is more stable, and the larger the specific binding energy is, the more stable the nucleus is, so ${ } _ { 56 } ^ { 144 } \mathrm { Ba }$'s Therefore, the specific binding energy of ${ } _ { 56 } ^ { 144 } \mathrm { Ba }$ is greater than that of ${ } _ { 92 } ^ { 235 } \mathrm { U }$, so C is correct; D. The purpose of using moderators in nuclear reactors is to slow down the fast neutrons so that the neutrons can have a chain reaction with the uranium nucleus, not to slow down the speed of the nuclear reaction, but to increase the utilization rate of neutrons, so D is wrong.

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