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Modern Physics - Practice Questions (28)

Question 1: When a metal with cutoff frequency $\frac { 3 } { 4 } v$ is irradiated with monochromatic light of f...

When a metal with cutoff frequency $\frac { 3 } { 4 } v$ is irradiated with monochromatic light of frequency ${ } _ { v }$, the maximum initial kinetic energy of the escaping photoelectrons is (Planck's constant is $h$).

  • A. A. $\frac { 1 } { 4 } h v$
  • B. B. $\frac { 1 } { 2 } h v$
  • C. C. $\frac { 3 } { 4 } h v$
  • D. D. $h v$

Answer: A

Solution: [Knowledge Point]Einstein photoelectric effect equation [Detailed Explanation]According to the Einstein photoelectric effect equation there are $E _ { \mathrm { k } } = h v - W _ { 0 } = h v - h \cdot \frac { 3 } { 4 } v = \frac { 1 } { 4 } h v$

Question 2: Extreme ultraviolet light source used in EUV lithography, CUHK makes breakthrough in extreme ultravi...

Extreme ultraviolet light source used in EUV lithography, CUHK makes breakthrough in extreme ultraviolet light source research. A light source emits extreme ultraviolet light at a frequency of ${ } ^ { 2 \times 10 ^ { 16 } } \mathrm {~Hz}$. The Planck's constant is ${ } ^ { 6.6 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm {~s} }$. The energy of a photon of this light is

  • A. A. $2.0 \times 10 ^ { - 18 } \mathrm {~J}$
  • B. B. ${ } ^ { 1.32 \times 10 ^ { - 18 } \mathrm {~J} }$
  • C. C. $3.3 \times 10 ^ { - 14 } \mathrm {~J}$
  • D. D. ${ } ^ { 1.32 \times 10 ^ { - 17 } \mathrm {~J} }$

Answer: D

Solution: [Knowledge Points] Einstein photon theory [Detailed explanation] The energy of the light photon is $$ \varepsilon = h v = 6.6 \times 10 ^ { - 34 } \times 2 \times 10 ^ { 16 } \mathrm {~J} = 1.32 \times 10 ^ { - 17 } \mathrm {~J} $$

Question 3: When a metal is irradiated with monochromatic light of frequency [[INLINE_${ } ^ { v }$, the maximum...

When a metal is irradiated with monochromatic light of frequency [[INLINE_${ } ^ { v }$, the maximum initial momentum of escaping photoelectrons is [[INLINE_$p$, and when the metal is irradiated with monochromatic light of frequency [[INLINE_$2 ^ { v }$, the maximum initial momentum of escaping photoelectrons is [[INLINE_$2 p$.$2 p$,then the cutoff frequency of the metal) [[INLINE_${ } _ { 0 }$ is equal to

  • A. A. $\frac { 1 } { 2 } v$
  • B. B. $\frac { 1 } { 3 } v$
  • C. C. $\frac { 2 } { 3 } v$
  • D. D. $\frac { 1 } { 4 } v$

Answer: C

Solution: Knowledge point] Einstein photoelectric effect equation [Detailed Explanation]According to the meaning of the question $$ \begin{gathered} E _ { \mathrm { k } } = \frac { 1 } { 2 } m v ^ { 2 } \\ p = m v \end{gathered} $$ can be obtained $$ E _ { \mathrm { k } } = \frac { p ^ { 2 } } { 2 m } $$ Combined with the photoelectric effect equation, we can see that $$ h v - h v _ { 0 } = \frac { 1 } { 2 } m v ^ { 2 } = \frac { p ^ { 2 } } { 2 m } $$ The same reasoning leads to $$ 2 h v - h v _ { 0 } = \frac { ( 2 p ) ^ { 2 } } { 2 m } $$ Solving for $$ v _ { 0 } = \frac { 2 } { 3 } v $$

Question 4: Among the phenomena described in the following groups, the one that shows the particle nature of lig...

Among the phenomena described in the following groups, the one that shows the particle nature of light is

  • A. A. photoelectric effect phenomenon
  • B. B. The phenomenon of interference of light
  • C. C. Diffraction of light
  • D. D. Refraction of light

Answer: A

Solution: [Knowledge points] photoelectric effect phenomenon and its interpretation, diffraction conditions, the conditions of double-slit interference, light refraction phenomenon A. The photoelectric effect phenomenon shows that light has a particle, A correct; BCD. light interference, diffraction, refraction phenomenon illustrates the volatility of light, can not explain the particle nature of light, BCD error.

Question 5: The formula $\varepsilon = h v$ can be used to directly calculate

The formula $\varepsilon = h v$ can be used to directly calculate

  • A. A. amps
  • B. B. input voltage
  • C. C. energies
  • D. D. resistive

Answer: C

Solution: Einstein's theory of photons The formula $\varepsilon = h v$ is the photon energy formula, which was proposed by Planck based on blackbody radiation.

Question 6: In studying the scattering of X-rays by graphite, Compton found that in the scattered X-rays, except...

In studying the scattering of X-rays by graphite, Compton found that in the scattered X-rays, except for those with an incident wavelength ${ } ^ { \lambda _ { 0 } }$, there were no significant differences between the scattered X-rays and the incident wavelength. Compton found that in the scattered X-rays, in addition to components with the same wavelength as the incident wavelength ${ } ^ { \lambda _ { 0 } }$, there were components with wavelengths greater than ${ } ^ { \lambda _ { 0 } }$. These components with wavelengths greater than ${ } ^ { \lambda _ { 0 } }$ are compared to the incident X-rays.)

  • A. A. Increased energy
  • B. B. Increase in momentum
  • C. C. Decrease in wave speed
  • D. D. frequency reduction

Answer: D

Solution: Einstein's photon theory, the momentum of the photon and its equation [Detailed explanation] these wavelengths greater than ${ } ^ { \lambda _ { 0 } }$ components and the incident X-rays compared to the medium of propagation is unchanged, the wave speed $v$ is unchanged, the energy and momentum of the X-ray photon is respectively $$ \begin{gathered} \varepsilon = h v = h \frac { v } { \lambda } \\ p = \frac { h } { \lambda } \end{gathered} $$ Therefore, the frequency $v$ decreases, the energy $\varepsilon$ decreases, and the momentum $p$ decreases, so ABC is incorrect and D is correct.

Question 7: Of the four monochromatic colors of light, red, yellow, green and violet, the one with the least ene...

Of the four monochromatic colors of light, red, yellow, green and violet, the one with the least energy is

  • A. A. ultraviolet light photon
  • B. B. red light photon
  • C. C. green light photon
  • D. D. yellow light photon

Answer: B

Solution: [Knowledge Points] Einstein photon theory [Detailed explanation] The energy of each photon is $$ E 0 = h v = h \frac { c } { \lambda } $$ Red, yellow, green, violet four monochromatic light, the frequency of red light is the smallest, the wavelength is the longest, the energy of the photon is the smallest, the energy of the photon is the smallest, the energy of the photon is the smallest, the energy of the photon is the smallest.

Question 8: The "full femtosecond" myopia correction surgery uses a laser with a wavelength of $\lambda = 1.053 ...

The "full femtosecond" myopia correction surgery uses a laser with a wavelength of $\lambda = 1.053 \times 10 ^ { - 6 } \mathrm {~m}$, which is known as Planck's constant $h = 6.63 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm {~s}$ and the speed of light propagation in vacuum $c = 3 \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s }$. The Planck constant $h = 6.63 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm {~s}$ is known, and light travels at $c = 3 \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s }$ in a vacuum, so the energy of the photons in the laser is approximately

  • A. A. $1.9 \times 10 ^ { - 18 } \mathrm {~J}$
  • B. B. $1.9 \times 10 ^ { - 19 } \mathrm {~J}$
  • C. C. $2.2 \times 10 ^ { - 18 } \mathrm {~J}$
  • D. D. $2.2 \times 10 ^ { - 19 } \mathrm {~J}$

Answer: B

Solution: [Knowledge Points] Einstein's photon theory [Detailed Explanation] By $$ \begin{aligned} & \varepsilon = h v \\ & v = \frac { c } { \lambda } \end{aligned} $$ we get $$ \varepsilon \approx 1.9 \times 10 ^ { - 19 } \mathrm {~J} $$

Question 9: The following physical quantities are vectors

The following physical quantities are vectors

  • A. A. cut-off frequency
  • B. B. Benelux
  • C. C. energy level difference
  • D. D. The amount of change in momentum

Answer: D

Solution: [Knowledge points] photoelectric effect of the limit frequency, scalar and vector [Detailed explanation] A. Cut-off frequency is only the size and direction of the scalar, so A error; B. Specific charge is only the size of the scalar without direction, so B error; C. The energy level difference is a scalar quantity with only magnitude and no direction, so C is wrong; D. The change of momentum is a vector that has both magnitude and direction, so D is correct.

Question 10: With a monochromatic light irradiation of a metal, the maximum initial kinetic energy of photoelectr...

With a monochromatic light irradiation of a metal, the maximum initial kinetic energy of photoelectrons produced is $E _ { k }$ , the number of photoelectrons emitted per unit time is $n$ , if the intensity of the incident light is increased, the

  • A. A. $E _ { k }$ increase, $n$ increase
  • B. B. $E _ { k }$ increased, $n$ unchanged
  • C. C. $E _ { k }$ UNCHANGED, $n$ UNCHANGED
  • D. D. $E _ { k }$ unchanged, $n$ increased

Answer: D

Solution: [Knowledge Point]Einstein photoelectric effect equation [Detailed explanation] According to the photoelectric effect equation $$ E _ { \mathrm { k } } = h v - W _ { 0 } $$ According to the photoelectric effect equation $$ E _ { \mathrm { k } } = h v - W _ { 0 } $$, it can be seen that the initial kinetic energy is related to the frequency of the light, and has nothing to do with the intensity of the light. The maximum initial kinetic energy of photoelectrons has nothing to do with the intensity of incident light, which increases with the increase in the frequency of incident light, greater than the limiting frequency of light irradiation of metal, the intensity of the photocurrent (reflecting the number of photoelectrons emitted per unit of time) is directly proportional to the intensity of the incident light.

Question 11: When a metal surface is irradiated with light of frequency $v$, the maximum initial kinetic energy o...

When a metal surface is irradiated with light of frequency $v$, the maximum initial kinetic energy of the escaping photoelectrons is $E k$; if the metal is irradiated with light of frequency $3 v$, the maximum initial kinetic energy of the escaping photoelectrons is $3 v$.

  • A. A. $3 E k$
  • B. B. $2 E k$
  • C. C. $3 h v - E k$
  • D. D. $2 h v + E k$

Answer: D

Solution: Knowledge point] Einstein photoelectric effect equation [Detailed Explanation]From the Einstein photoelectric effect equation can be known $$ \begin{gathered} E k = h v - W 0 \\ E k ^ { \prime } = h \cdot 3 v - W _ { 0 } \end{gathered} $$ So $$ E k ^ { \prime } = 2 h v + E k $$ D is correct.

Question 12: With red light irradiation of a photoelectric tube photoelectric effect, measured photoelectrons of ...

With red light irradiation of a photoelectric tube photoelectric effect, measured photoelectrons of the maximum initial kinetic energy of $E _ { 1 }$; if you change to violet light irradiation of the photoelectric tube, measured photoelectrons of the maximum initial kinetic energy of $E _ { 2 }$, then

  • A. A. $E _ { 2 } = E _ { 1 }$
  • B. B. $E _ { 2 } > E _ { 1 }$
  • C. C. $E _ { 2 } < E _ { 1 }$
  • D. D. indeterminate

Answer: B

Solution: Knowledge point] The maximum initial kinetic energy of photoelectrons [Detailed Explanation]According to the photoelectric effect equation $$ E _ { \mathrm { k } } = h v - W _ { 0 } $$ know that the incident light frequency increases, the maximum initial kinetic energy of photoelectrons escaping from the surface of the metal increases, that is $$ E _ { 2 } > E _ { 1 } $$ Therefore, B is correct, ACD is wrong.

Question 13: When a metal surface is irradiated with monochromatic light, whether or not photoelectrons escape de...

When a metal surface is irradiated with monochromatic light, whether or not photoelectrons escape depends on the incident light.

  • A. A. frequency
  • B. B. dissociation
  • C. C. exposure time
  • D. D. photon number

Answer: A

Solution: Knowledge point] The limiting frequency of the photoelectric effect [Detailed explanation] The photoelectric effect occurs when the frequency of the incident light is greater than the limiting frequency of the metal.

Question 14: Red, orange, yellow, green four monochromatic light, photon energy is the largest

Red, orange, yellow, green four monochromatic light, photon energy is the largest

  • A. A. red light
  • B. B. orange light
  • C. C. yellow light
  • D. D. green light

Answer: D

Solution: [Knowledge] Einstein photon theory [Detailed explanation] red, orange, yellow, green four monochromatic light, green light wavelength is the smallest, the largest frequency, according to the $$ E = h v $$ According to $$ E = h v $$, the photon energy of green light is the largest, so D is correct and ABC is wrong.

Question 15: In the photoelectric effect experiment, with a frequency of $1.5 v$ A light and frequency of $v$ B l...

In the photoelectric effect experiment, with a frequency of $1.5 v$ A light and frequency of $v$ B light were irradiated with the same metal, A light irradiation of the curb voltage is B light irradiation curb voltage of 2 times, then the cut-off frequency of the photoelectric effect of the metal for )

  • A. A. $\frac { 1 } { 3 } v$
  • B. B. $\frac { 1 } { 2 } v$
  • C. C. $\frac { 2 } { 3 } v$
  • D. D. $\frac { 3 } { 4 } v$

Answer: B

Solution: Knowledge point] The limit frequency of photoelectric effect [Detailed explanation] According to the photoelectric effect equation, there are $$ \mathrm { e } U _ { \mathrm { CA } } = h \times 1.5 v - h v _ { 0 } , \mathrm { e } U _ { \mathrm { CB } } = h \times v - h v _ { 0 } $$ can be obtained by associating $$ v _ { 0 } = \frac { 1 } { 2 } v $$

Question 16: Known as a metal of the escape work of 4.63 eV, a group of hydrogen atoms in the ground state ${ } ^...

Known as a metal of the escape work of 4.63 eV, a group of hydrogen atoms in the ground state ${ } ^ { ( n = 1 ) }$ absorbed energy is excited, the photons radiated only three photons can make the metal photoelectric effect, then the maximum energy absorbed by the ground state of the hydrogen atoms is | $n$ | $E / \mathrm { eV }$ | $E / \mathrm { eV }$ | $E / \mathrm { eV }$ | :---: | :---: | | $\infty$ | 0 | | 4 3 - - | 0.85 | | 4 3- - | 0.85 | 2 - 1.51 | | 2 - 3.40 | 2 - 3.40 | 2 - 3.40 | | $1 - - 13.60$

  • A. A. 13.60 eV
  • B. B. 12.75 eV
  • C. C. 12.09 eV
  • D. D. 10.20 eV

Answer: B

Solution: [Knowledge] Bohr theory of the hydrogen atom spectral interpretation [Detailed explanation] hydrogen atom is excited by the radiation of photons in three photons can make the metal photoelectric effect, the energy of the three photons is greater than 4.63 eV, the hydrogen atom is excited to 4 energy levels, the maximum energy absorbed by the hydrogen atom in the ground state is 12.75 eV.

Question 17: Photons, or light quanta, are particles that transmit electromagnetic interactions. Photons have ene...

Photons, or light quanta, are particles that transmit electromagnetic interactions. Photons have energy, and if the frequency of light is ${ } _ { 0 }$ $= 2.0 \times 10 ^ { 15 } \mathrm {~Hz}$ ,known Planck's constant $h = 6.6 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm {~s}$ ,then the energy of the photon is ()

  • A. A. $1.32 \times 10 ^ { - 19 } \mathrm {~J}$
  • B. B. $1.32 \times 10 ^ { - 18 } \mathrm {~J}$
  • C. C. $3.3 \times 10 ^ { - 19 } \mathrm {~J}$
  • D. D. $3.3 \times 10 ^ { - 49 } \mathrm {~J}$

Answer: B

Solution: [KNOWLEDGE POINT] Bohr's theoretical explanation of the spectrum of the hydrogen atom [Details] The energy of the photon sought ${ } ^ { E = h v _ { 0 } } = 1.32 \times 10 ^ { - 18 } \mathrm {~J}$

Question 18: $H _ { \alpha } , H _ { \beta } , H _ { \gamma } , H _ { \delta }$ is the visible part of the spectr...

$H _ { \alpha } , H _ { \beta } , H _ { \gamma } , H _ { \delta }$ is the visible part of the spectrum of a hydrogen atom with four lines, which are generated by jumps from the $n = 3 , 4$ , 5, and 6 energy levels to the $n = 2$ energy level, respectively. 5, 6 energy levels to the $n = 2$ energy level. The longest wavelength of the 4 lines is ().

  • A. A. $\mathrm { H } _ { \alpha }$
  • B. B. $\mathrm { H } _ { \beta }$
  • C. C. $\mathrm { H } _ { \mathrm { r } }$
  • D. D. $\mathrm { H } _ { \delta }$

Answer: A

Solution: [Knowledge Point]The basic assumptions of Bohr's atomic theory [Detailed Explanation]According to the frequency conditions of Bohr's atomic model there are $$ h \frac { c } { \lambda } = E _ { \mathrm { n } } - E _ { 2 } $$ i.e. $$ \lambda = \frac { h c } { E _ { \mathrm { n } } - E _ { 2 } } $$ Therefore, when $n$ is the smallest, it corresponds to the longest wavelength, i.e., the wavelength of the spectrum produced when the hydrogen atom jumps from the $n = 3$ energy level to the $n = 2$ energy level. The longest wavelength is ${ } ^ { \mathrm { H } _ { \alpha } }$.

Question 19: The nucleus of an atom consists of protons and neutrons, and the size of the diameter of the nucleus...

The nucleus of an atom consists of protons and neutrons, and the size of the diameter of the nucleus of an atom may fall in which interval of the following ruler diagram ( ) | $10 ^ { - 17 } \mathrm {~m}$ | $10 ^ { - 15 } \mathrm {~m} \quad 10 ^ { - 13 } \mathrm {~m} \quad 10 ^ { - 11 } \mathrm {~m} \quad 10 ^ { - 9 } \mathrm {~m}$ | | | | | | | :--- | :---: | :---: | :---: | :---: | :---: | :---: | | | $A$ | $B$ | $C$ | $D$ | | |

  • A. A. A
  • B. B. B
  • C. C. C
  • D. D. D

Answer: B

Solution: Knowledge point] null The nucleus of an atom is extremely small, with a diameter between $10 ^ { - 15 } \sim 10 ^ { - 14 } \mathrm {~m}$ and a volume of only a few hundred billionths of the volume of the atom.

Question 20: A model of the nuclear structure of the atom was proposed through the study of which of the followin...

A model of the nuclear structure of the atom was proposed through the study of which of the following experimental phenomena

  • A. A. photoelectric effect phenomenon
  • B. B. $\alpha$ Particle Scattering Experiments
  • C. C. black body radiation experiment
  • D. D. natural radiation phenomenon

Answer: B

Solution: Knowledge Points] Nuclear Structure Model of the Atom Rutherford proposed a model of the nuclear structure of the atom by studying the phenomenon of $\alpha$ particle scattering.

Question 21: How many different frequencies of photons may be excited by a large number of hydrogen atoms during ...

How many different frequencies of photons may be excited by a large number of hydrogen atoms during the jump from the $n = 5$ energy level to the $n = 2$ energy level?

  • A. A. 3
  • B. B. 6
  • C. C. 10
  • D. D. 12

Answer: B

Solution: Bohr theory explains the spectrum of the hydrogen atom. [Detailed Explanation] In the process of a large number of hydrogen atoms jumping from the $n = 5$ energy level to the $n = 2$ energy level, there are $5 \rightarrow 4$, $4 \rightarrow 3,3 \rightarrow 2,5 \rightarrow 3,5 \rightarrow 2,4 \rightarrow 2$ and [[INLINE_, and we can see that there are six kinds of light in total. $4 \rightarrow 3,3 \rightarrow 2,5 \rightarrow 3,5 \rightarrow 2,4 \rightarrow 2$ , so there are 6 kinds of light.

Question 22: Through the experimental setup shown in the figure, Rutherford established a model of the nuclear st...

Through the experimental setup shown in the figure, Rutherford established a model of the nuclear structure of the atom. During the experiment, if the fluorescent screen and microscope are placed at positions 1,2,3 respectively. the position at which the minimum number of particles can be observed is ( ) ![](/images/questions/phys-modern-physics/012e47f3969b.jpg)

  • A. A. 1
  • B. B. 2
  • C. C. 3
  • D. D. as many as

Answer: A

Solution: The $\alpha$ Particle Scattering Experiment In Rutherford's $\alpha$ particle scattering experiment, most of the $\alpha$ particles passed through the original path, a few $\alpha$ particles were deflected at a large angle, and very few rebounded, according to the image, the microscope is placed in position 1 to show the rebound phenomenon. According to the image, when the microscope is placed at position 1, it shows the phenomenon of rebound, so we can see that the number of particles observed at position 1 is the lowest, i.e., the position where the lowest number of particles can be observed is 1 .

Question 23: If a neutron hits ${ } _ { 6 } ^ { 14 } \mathrm { C } _ { \text {produces } } { } ^ { 15 } \mathrm {...

If a neutron hits ${ } _ { 6 } ^ { 14 } \mathrm { C } _ { \text {produces } } { } ^ { 15 } \mathrm {~N}$ and its nuclear reaction equation is ${ } _ { 6 } ^ { 14 } \mathrm { C } + { } _ { 0 } ^ { 1 } \mathrm { n } \rightarrow { } _ { 7 } ^ { 15 } \mathrm {~N} + \mathrm { X }$, then X is which of the following particles?

  • A. A. ${ } _ { 1 } ^ { 1 } \mathrm { H }$
  • B. B. ${ } _ { 0 } ^ { 1 } \mathrm { n }$
  • C. C. ${ } _ { 1 } ^ { 0 } \mathrm { e }$
  • D. D. ${ } _ { - 1 } ^ { 0 } \mathrm { e }$

Answer: D

Solution: [Knowledge Point]Writing of nuclear reaction equations [Detailed Explanation]According to the conservation of mass number, the mass number of X is $14 + 1 - 15 = 0$ According to the conservation of charge number, the charge number of $X$ is $6 + 0 - 7 = - 1$. X is ${ } ^ { - 1 } \mathrm { e }$ .

Question 24: Regarding the nuclear reaction ${ } ^ { 54 } \mathrm { Cr } + { } _ { 95 } ^ { 243 } \mathrm { Am } ...

Regarding the nuclear reaction ${ } ^ { 54 } \mathrm { Cr } + { } _ { 95 } ^ { 243 } \mathrm { Am } \rightarrow { } _ { 119 } ^ { A } \mathrm { X } + 2 { } _ { 0 } ^ { 1 } \mathrm { n }$ , the following options are correct ( )

  • A. A. $A = 293$
  • B. B. A = 294
  • C. C. A = 295
  • D. D. A = 296

Answer: C

Solution: [Knowledge Point]Writing of nuclear reaction equations [Detailed Explanation]According to the conservation of mass number: the total mass number before the reaction is $54 + 243 = 297$ After the reaction, the total mass is $\mathrm { A } + 2 \times 1 = A + 2$. From conservation we have $297 = A + 2$ Solving for $A = 295$ .

Question 25: Becquerel discovered the phenomenon of natural radioactivity, revealing that the atomic nucleus has ...

Becquerel discovered the phenomenon of natural radioactivity, revealing that the atomic nucleus has a complex structure. The following nuclear reactions represent nuclear decay ( )

  • A. A. ${ } _ { 92 } ^ { 238 } \mathrm { U } \rightarrow { } _ { 90 } ^ { 234 } \mathrm { Th } + { } _ { 2 } ^ { 4 } \mathrm { He }$
  • B. B. ${ } _ { 92 } ^ { 235 } \mathrm { U } + { } _ { 0 } ^ { 1 } \mathrm { n } \rightarrow { } _ { 56 } ^ { 144 } \mathrm { Ba } + { } _ { 36 } ^ { 89 } \mathrm { Kr } + 3 { } _ { 0 } ^ { 1 } \mathrm { n }$
  • C. C. ${ } _ { 1 } ^ { 2 } \mathrm { H } \boldsymbol { + } _ { 1 } ^ { 3 } \mathrm { H } \rightarrow { } _ { 2 } ^ { 4 } \mathrm { He } \boldsymbol { + } _ { 0 } ^ { 1 } \mathrm { n }$
  • D. D. ${ } _ { 4 } ^ { 9 } \mathrm { Be } + { } _ { 2 } ^ { 4 } \mathrm { He } \rightarrow { } _ { 6 } ^ { 12 } \mathrm { C } + { } _ { 0 } ^ { 1 } \mathrm { n }$

Answer: A

Solution: Characteristics and nature of $\alpha$ decay and how to write its equations. Option A is $\alpha$ decay, which meets the definition of nuclear decay; Option B is a nuclear fission reaction; Option C is a light fusion reaction; Option D is an artificial nuclear reaction in which neutrons are found.

Question 26: One of the nuclear reaction equations that produces ${ } ^ { 18 } \mathrm {~F}$ nuclei in scientific...

One of the nuclear reaction equations that produces ${ } ^ { 18 } \mathrm {~F}$ nuclei in scientific and technological research is ${ } ^ { 18 } \mathrm { O } + { } _ { 1 } ^ { 1 } \mathrm { H } \rightarrow { } _ { 9 } ^ { 18 } \mathrm {~F} + \mathrm { X }$ , where X is ( )

  • A. A. ${ } _ { - 1 } ^ { 0 } \mathrm { e }$
  • B. B. ${ } _ { 0 } ^ { 1 } n$
  • C. C. ${ } _ { 1 } ^ { 0 } \mathrm { e }$
  • D. D. ${ } _ { 2 } ^ { 4 } \mathrm { He }$

Answer: B

Solution: Knowledge Points] The writing of nuclear reaction equations [Detailed Explanation]According to the nuclear reaction equation of the number of mass and charge conservation : The total mass number on the left is $18 + 1 = 19$ The total charge number on the left $8 + 1 = 9$ The total mass number on the right is $18 + A _ { \mathrm { x } }$ ,which needs to satisfy $18 + A _ { \mathrm { x } } = 19$ Therefore $A _ { \mathrm { x } } = 1$ . The total number of charges on the right side is $9 + Z _ { X }$ ,which needs to satisfy $9 + Z _ { X } = 9$. Therefore $Z _ { \mathrm { x } } = 0$ The particle with mass 1 and charge 0 is ${ } ^ { 1 } \mathrm { n }$ (neutron).

Question 27: Americium 243 (${ } ^ { 243 } \mathrm { Am }$) is an artificially obtained radioactive element that ...

Americium 243 (${ } ^ { 243 } \mathrm { Am }$) is an artificially obtained radioactive element that is highly radioactive and chemically active. If americium 243 decay equation is: ${ } ^ { 243 } \mathrm { Am } \rightarrow { } _ { a } ^ { 239 } \mathrm { X } + { } _ { 2 } ^ { b } \mathrm { Y } , \mathrm { X } , \mathrm { Y }$ represents two different elemental symbols, then ( )

  • A. A. $a = 97 , b = 2$
  • B. B. $a = 93 , b = 4$
  • C. C. $a = 97 , b = 4$
  • D. D. $a = 93 , b = 2$

Answer: B

Solution: Characteristics and nature of $\alpha$ decay and how to write its equation. [Explanation] According to the decay equation ${ } ^ { 243 } \mathrm { Am } \rightarrow { } _ { a } ^ { 239 } \mathrm { X } + { } _ { 2 } ^ { b } \mathrm { Y }$, the number of masses and charges are conserved. According to the decay equation $243 = 239 + b , 95 = a + 2$, the number of masses and charges are conserved. Solve for $a = 93 , b = 4$.

Question 28: Becquerel first discovered the phenomenon of natural radioactivity 120 years ago, and today the radi...

Becquerel first discovered the phenomenon of natural radioactivity 120 years ago, and today the radioactivity of atomic nuclei is used in a wide range of applications. The following are radioactive decays

  • A. A. ${ } _ { 6 } ^ { 14 } \mathrm { C } \rightarrow { } _ { 7 } ^ { 14 } \mathrm {~N} + { } _ { - 1 } ^ { 0 } \mathrm { e }$
  • B. B. ${ } _ { 92 } ^ { 235 } U + { } _ { 0 } ^ { 1 } n \rightarrow { } _ { 53 } ^ { 139 } I + { } _ { 39 } ^ { 95 } Y + 2 { } _ { 0 } ^ { 1 } n$
  • C. C. ${ } _ { 1 } ^ { 2 } \mathrm { H } + { } _ { 1 } ^ { 3 } \mathrm { H } \rightarrow { } _ { 2 } ^ { 4 } \mathrm { He } + { } _ { 0 } ^ { 1 } \mathrm { n }$
  • D. D. . ${ } _ { 2 } ^ { 4 } \mathrm { He } + { } _ { 13 } ^ { 27 } \mathrm { Al } \rightarrow { } _ { 15 } ^ { 30 } \mathrm { P } + { } _ { 0 } ^ { 1 } \mathrm { n }$ ## Reference Answer | Question No. | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | | Answers | A | D | C | D | C | C | A | C | D | B | B | | Question No. | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | | Answers | D | D | D | D | B | A | D | B | B | B | A | Question No. | Question No. | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 | Answer | B | 29 | 30 | Question No. | Answers | B | B | B | A | D | C | A | B | B | B | A |

Answer: A

Solution: Characteristics and nature of $\alpha$ decay and how to write its equations. A is the equation of $\beta$ decay, B is the fission of heavy nuclei, C is the fusion of light nuclei, and D is the artificial conversion of atomic nuclei, so A is correct.
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