Photoelectric Effect

Question 1

Question 1: 1. A cesium atom with a wavelength of $\lambda$ is irradiated by yellow light with an escape functio...

1. A cesium atom with a wavelength of $\lambda$ is irradiated by yellow light with an escape function of $W$, causing the photoelectric effect to occur. It is known that Planck's constant is $h$, the speed of light in vacuum is $c$, and the charge of an electron is $e$. Then the corresponding curb voltage is

A. A. $\frac { h c } { \lambda e }$
B. B. $\frac { h c } { \lambda e } - \frac { W } { e }$
C. C. $\frac { h \lambda } { c e } - \frac { W } { e }$
D. D. $\frac { h \lambda } { c e }$

Answer

Answer: B

Solution: From the photoelectric effect equation $$ E _ { \mathrm { km } } = h v - W , v = \frac { c } { \lambda } , E _ { \mathrm { km } } = U _ { \mathrm { c } } e $$ Solve for $$ U _ { \mathrm { c } } = \frac { h c } { \lambda e } - \frac { W } { e } $$

Question 2

Question 2: 2. One of the following statements about the phenomenon of light is true

2. One of the following statements about the phenomenon of light is true

A. A. The surface of a soap bubble in sunlight shows colored streaks, which is a diffraction of light
B. B. A TV remote control uses infrared pulse signals to change channels.
C. C. The photoelectric effect cannot occur when a beam of light hits a metal because the wavelength of the light is too short.
D. D. In a two-slit interference experiment, if you only increase the spacing between the two slits, the fringe spacing must become larger.

Answer

Answer: B

Solution: A. The surface of a soap bubble in the sunlight shows colored stripes, which is the interference phenomenon of light. Therefore, A is wrong; B.The remote control uses infrared light, which has a long wavelength, to change the channel, so B is correct; C. a beam of light irradiation to a metal can not occur on the photoelectric effect, because the frequency of the beam of light is too small, the wavelength is too long, so C error; D. According to the double-slit interference fringe spacing public $x = \frac { L } { d } \lambda$, only the double-slit spacing increases, the fringe spacing must become smaller, so D error.

Question 3

Question 3: 3. which physical method, idea, or principle is shared by the following three statements (1) Borel a...

3. which physical method, idea, or principle is shared by the following three statements (1) Borel assumed that the orbitals of electrons are discrete and cannot run in arbitrary orbitals (2) in order to explain the phenomenon of the photoelectric effect, Einstein believed that light itself is composed of an indivisible energy, these energy is called photons (3) Planck believes that the energy of the vibrating charged particles can only be an integer multiple of a minimum energy value $\varepsilon$, which can not be subdivided into the smallest energy value $\varepsilon$ is called an energon.

A. A. The concept of quantization of the microcosm
B. B. Conservation of Energy Perspective
C. C. Physical approach to equivalent substitution
D. D. Physical particles have volatility

Answer

Answer: A

Question 4

Question 4: 4. About the red, green and violet beams of monochromatic light, the following statements are true

4. About the red, green and violet beams of monochromatic light, the following statements are true

A. A. Red light frequency max.
B. B. The wavelength of red light is minimized in air
C. C. The energy of red light photons is greater than the energy of green light photons
D. D. Minimum spacing between two adjacent bright stripes of violet light seen with the same double-slit interference device

Answer

Answer: D

Solution: AB . Red light has the longest wavelength and the smallest frequency, so AB is wrong; C. The frequency of red light is less than the frequency of green light is small, so the energy of red light photons is less than the energy of green light photons, so C error; D. According to the formula $$ \Delta x = \frac { L } { d } \lambda $$ The shortest wavelength of violet light, with the same double-slit interference device to see the violet light adjacent to the two bright fringe spacing minimum, so D is correct.

Question 5

Question 5: 5. The following statements are incorrect

5. The following statements are incorrect

A. A. Nuclear reaction equation: "$X$" in ${ } _ { 4 } ^ { 9 } \mathrm { Be } + { } _ { 2 } ^ { 4 } \mathrm { He } \rightarrow { } _ { 6 } ^ { 12 } \mathrm { C } + \mathrm { X }$ is neutrons
B. B. In the photoelectric effect, the magnitude of the saturation current is related to the frequency of the incident light, independent of the intensity of the incident light
C. C. The release of nuclear energy must be accompanied by a loss of mass, since mass and energy are related.
D. D. Bohr introduced quantum ideas into the realm of the atom, and his theory was able to explain the characteristics of the spectrum of the hydrogen atom

Answer

Answer: B

Solution: A. According to the conservation of mass number and charge number, we know that the X charge number in the equation of the nuclear reaction is 0, the mass number is 1, and it is a neutron, so A is correct and does not meet the meaning of the question; B. The photoelectric effect occurs, increase the intensity of incident light irradiating the photoelectric tube, the photocurrent in the circuit must increase, then the size of the saturated photocurrent depends on the intensity of the incident light, so B is wrong, in line with the meaning of the question; C. According to Einstein's mass-energy equation $\Delta E = \Delta m c ^ { 2 }$, it can be seen that the nuclear reaction that releases nuclear energy must be accompanied by a mass loss, so C is correct and does not meet the meaning of the question; D. Bohr introduced the quantum concept into the atomic field, can well explain the hydrogen atomic spectrum, but can not explain the helium atomic spectral characteristics, which is the limitation of Bohr's theory, so D is correct, does not meet the meaning of the question;

Question 6

Question 6: 6. The following statements about modern physics are correct

6. The following statements about modern physics are correct

A. A. Rutherford proposed that the interior of the atomic nucleus consisted of nucleons through $\alpha$ particle scattering experiments
B. B. For the same metal, the limiting frequency is constant, independent of both the frequency of the incident light and the intensity of the light
C. C. The hydrogen atom's extra-nuclear electrons absorb photons of a certain frequency when they jump from a higher energy level to a lower energy level
D. D. When a ${ } ^ { 234 } \mathrm { Th }$ nucleus undergoes ${ } ^ { \beta }$ decay, the new nucleus has the same number of neutrons and the same mass number compared to the original nucleus

Answer

Answer: B

Solution: Rutherford's $\alpha$ particle scattering experiment shows that atoms have a nuclear structure, so A is wrong; for the same metal, its limiting frequency is constant, independent of the frequency of the incident light and the intensity of the light, so B is correct; from Bohr's theory, it can be seen that the hydrogen atom's electrons outside the nucleus have to release a certain frequency of photons when they are leapt from a higher energy level to a lower one, so C is incorrect. Wrong; Root According to the essence of $\beta$ decay, in the process of $\beta$ decay, one of the neutrons in the nucleus is transformed into a proton and an electron, so the mass number of the new nucleus remains unchanged, and the number of neutrons is reduced by one, so D is wrong. The number of neutrons decreases by one. Rutherford proposed the model of atomic nuclear structure on the basis of $\alpha$ particle scattering experiments; the limiting frequency of a metal is determined by the metal itself; the change of kinetic energy and the change of potential energy of hydrogen atom according to Bohr's theory; the new nucleus is one neutron fewer than the original nucleus when $\beta$ decays. When $\beta$ decays, the new nucleus has one fewer neutron than the original nucleus and the mass number remains the same.

Question 7

Question 7: 7. "Raman scattering" refers to a certain frequency of light irradiation to the surface of the sampl...

7. "Raman scattering" refers to a certain frequency of light irradiation to the surface of the sample, the molecules in the substance and the photons undergo energy transfer, scattering different frequencies of light. If the photon will be part of the energy transfer to the molecules in the reflection, the

A. A. Light travels faster
B. B. The wavelength of the photon becomes smaller
C. C. The frequency of the photon becomes larger
D. D. The momentum of the photon becomes smaller

Answer

Answer: D

Solution: A. The speed of propagation of photons is unchanged, A error; B. As the photon transfers part of its energy to the molecule, the energy of the photon decreases, the frequency becomes smaller and the wavelength should become larger, B error; C. As the photon transfers part of its energy to the molecule, the energy of the photon decreases and its frequency becomes smaller, C is wrong; D. According to the photon's momentum formula $$ p = \frac { h } { \lambda } $$ D. According to the formula $$ p = \frac { h } { \lambda } $$, when the Planck's constant $h$ does not change, the wavelength $\lambda$ becomes larger, and its momentum becomes smaller, D is correct.

Question 8

Question 8: 8. It is known that with a frequency of $\gamma$ monochromatic light irradiation of a metal surface,...

8. It is known that with a frequency of $\gamma$ monochromatic light irradiation of a metal surface, the maximum initial kinetic energy of the escaping photoelectrons is $E$, known as Planck's constant is $h$, then the limiting frequency of the photoelectric effect to occur in this metal should be

A. A. $\gamma$-Eh
B. B. $\gamma + E h$
C. C. $\gamma - \frac { E } { h }$
D. D. $\gamma ^ { + } \frac { E } { h }$

Answer

Answer: C

Solution: According to the photoelectric effect equation :$W _ { 0 } = h v - E$ and because :$W _ { 0 } = h v _ { 0 }$, the limiting frequency can be found to be $v _ { 0 } = v - \frac { E } { h }$, so A, B and D are wrong; C is correct.

Question 9

Question 9: 9. For the photoelectric effect phenomenon, we found that there is a cut-off frequency, for a metal,...

9. For the photoelectric effect phenomenon, we found that there is a cut-off frequency, for a metal, in the green light irradiation can occur photoelectric effect, then in the yellow light irradiation, whether the photoelectric effect can occur

A. A. might
B. B. surely
C. C. I'm sure you won't.
D. D. Related to the intensity of the yellow light

Answer

Answer: A

Solution: Since the frequency of yellow light is smaller than that of green light, the photoelectric effect may occur under the irradiation of yellow light according to $E _ { k m } = h v - W _ { 0 }$, so A is consistent with the meaning of the question and BCD is not consistent with the meaning of the question.

Question 10

Question 10: 10. A photolithography machine is the core device for making chips, and its main function is to use ...

10. A photolithography machine is the core device for making chips, and its main function is to use light to print graphics on a mask plate onto a silicon wafer. As shown in the figure, the DUV photolithography machine uses deep ultraviolet light with a wavelength of 193 nm. In order to improve the ability to project fine maps, a liquid is filled between the photoresist and the projection objective to improve the resolution, and the following statements are true when compared to the absence of added liquid ![](/images/questions/phys-photoelectric-effect/image-001.jpg) A ![](/images/questions/phys-photoelectric-effect/image-002.jpg) B

A. A. Deep ultraviolet light travels faster when it enters a liquid
B. B. Deep UV takes longer to travel the same distance in liquids
C. C. Deep UV photons are more energetic in liquids
D. D. Deep UV diffracts more readily in liquids and improves resolution

Answer

Answer: B

Solution: A. Light propagation speed $c$ in a vacuum is greater than the propagation speed $v$ in the medium, deep ultraviolet light into the liquid propagation speed becomes smaller, A error; B. Set propagation $L$ distance, the time in a vacuum $$ t = \frac { L } { c } $$ Time in liquid $$ \begin{aligned} & t ^ { \prime } = \frac { L } { v } \\ & t ^ { \prime } > t \end{aligned} $$ Therefore, B is correct; C. Deep ultraviolet light into the liquid frequency is unchanged, according to $E = h v$ can be photon energy remains unchanged, C is wrong; D. Deep ultraviolet into the liquid frequency is unchanged, the propagation speed becomes smaller, the wavelength becomes shorter, less likely to be obvious diffraction, D error.

Question 11

Question 11: 11. Planck's constant is known to be $h = 6.63 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm {~s}$...

11. Planck's constant is known to be $h = 6.63 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm {~s}$, meta-charge $e = 1.60 \times 10 ^ { - 19 } \mathrm { C }$, and as shown in the figure is an image of the variation of the suppression voltage $U _ { c }$ of the calcium metal with the frequency of the incident light $v$, the image of $v _ { 0 }$ has a value of approximately () ![](/images/questions/phys-photoelectric-effect/image-003.jpg)

A. A. $7.7 \times 10 ^ { 14 }$
B. B. $1.3 \times 10 ^ { 15 }$
C. C. $2.1 \times 10 ^ { 33 }$
D. D. $4.8 \times 10 ^ { 33 }$

Answer

Answer: A

Solution: ground $$ h v - W _ { 0 } = e U _ { c } $$ When $v = 0$ $$ U _ { c } = \frac { - W _ { 0 } } { e } = - 3.20 \mathrm {~V} $$ The solution is $$ W _ { 0 } = 5.12 \times 10 ^ { - 19 } \mathrm {~J} $$ When $U _ { c } = 0$ $$ v _ { 0 } = \frac { W _ { 0 } } { h } = 7.7 \times 10 ^ { 14 } \mathrm {~Hz} $$

Question 12

Question 12: 12. The following are correct about the achievements of physicists and the scientific theories they ...

12. The following are correct about the achievements of physicists and the scientific theories they have developed ( ).

A. A. Chadwick derived the existence of neutrons within atoms through $\alpha$ particle scattering experiments
B. B. Rutherford's theory of the nuclear structure of the atom assumes that the positive charge of the atom is evenly distributed throughout the atom
C. C. Planck proposed that electron orbitals and atomic energy are quantized based on the spectrally discrete properties of the hydrogen atom
D. D. Einstein's equation for the photoelectric effect concludes that the maximum initial kinetic energy of photoelectrons is independent of the time of light exposure

Answer

Answer: D

Solution: A. $\alpha$ The particle scattering experiment was done by Rutherford, and Chadwick discovered the neutron by studying the new particles produced by paraffin wax exposed to "beryllium rays"; B. Rutherford's theory of the nuclear structure of the atom states that all the positive charges of an atom are in a very small nucleus, B. Wrong; C. Bohr proposed that the electron orbitals and atomic energy are quantized according to the spectrally discrete properties of the hydrogen atom, C error; D. Einstein's photoelectric effect equation concludes that the maximum initial kinetic energy of photoelectrons is related to the frequency of light and escape work, independent of the time of light, D is correct.

Question 13

Question 13: 13. The escape work of the metal tungsten is known to be 4.54 eV, and the relationship between the e...

13. The escape work of the metal tungsten is known to be 4.54 eV, and the relationship between the energy of the hydrogen atom in the $n$ energy level and the energy in the ground state is $E _ { n } = \frac { E _ { 1 } } { n ^ { 2 } } , n = 2,3,4 , \cdots$, where the energy of the hydrogen atom in the ground state is $E _ { 1 } = - 13.6 \mathrm { eV }$. The following statement is correct ( )

A. A. A group of hydrogen atoms in the $n = 3$ energy level leaps to a lower energy level, radiating up to 4 monochromatic colors of light
B. B. Hydrogen atoms in the $n = 3$ energy level jump directly to the ground state, radiating monochromatic light that can cause the photoelectric effect in tungsten metal
C. C. A hydrogen atom in the $n = 3$ energy level jumps to the $n = 2$ energy level, radiating monochromatic light that can cause the photoelectric effect in tungsten metal.
D. D. A hydrogen atom in the ground state jumps to the $n = 4$ energy level and emits photons

Answer

Answer: B

Solution: A. A group of hydrogen atoms in the $n = 3$ energy level leaps to a lower energy level and can radiate at most $\mathrm { C } _ { 3 } ^ { 2 } = 3$ monochromatic light, so A is wrong; B. The light from the $$ E _ { n } = \frac { E _ { 1 } } { n ^ { 2 } } $$ the energy of a hydrogen atom in the $n = 3$ energy level is known to be $$ E _ { 3 } = \frac { E _ { 1 } } { 3 ^ { 2 } } = - 1.51 \mathrm { eV } $$ A hydrogen atom in the $n = 3$ energy level jumps directly to the ground state and radiates a photon with energy $$ \Delta E _ { 31 } = ( - 1.51 \mathrm { eV } ) - ( - 13.6 \mathrm { eV } ) = 12.09 \mathrm { eV } $$ is greater than the escape work of tungsten metal, 4.54 eV, and the photoelectric effect can occur, so B is correct; C. The energy of a hydrogen atom in the $n = 2$ energy level is greater than that of a tungsten metal. [BLOCK_FORMULA_3]] The energy of a photon radiated by a hydrogen atom in the $n = 3$ energy level jumping to the $n = 2$ energy level $$ \Delta E _ { 32 } = ( - 1.51 \mathrm { eV } ) - ( - 3.4 \mathrm { eV } ) = 1.89 \mathrm { eV } $$ The photon energy radiated is less than the escape work of tungsten metal, 4.54 eV, so the photoelectric effect cannot occur, so C is wrong; D. The hydrogen atom in the ground state leaps to the $n = 4$ energy level and needs to absorb photons, so D is wrong.

Question 14

Question 14: 14. Physicists through in-depth observation and study of experiments, to obtain correct scientific k...

14. Physicists through in-depth observation and study of experiments, to obtain correct scientific knowledge, to promote the development of physics, the following statements are consistent with the facts ( )

A. A. The photoelectric effect shows that light is particulate, and the Compton effect shows that light is fluctuating
B. B. Rutherford used artificial transformations to discover the proton and predict the neutron
C. C. Bohr's atomic theory successfully explained the laws of atomic luminescence
D. D. Becquerel discovered the existence of nuclei in atoms by studying the phenomenon of natural radiation

Answer

Answer: B

Solution: A. The photoelectric effect and the Compton effect show that light is particulate, so A is wrong; B. Rutherford discovered the proton and predicted the existence of the neutron by means of artificial transformation, so B is correct; C. Borel's atomic theory successfully explained the experimental laws of the hydrogen atom spectrum, rather than the laws of atomic luminescence, so C wrong; D. Rutherford discovered the existence of nuclei in atoms through the scattering of $\alpha$ particles, so D is wrong;

Question 15

Question 15: 15. There is a new photoelectric effect quantum material with an escape work of $W _ { 0 }$. When ul...

15. There is a new photoelectric effect quantum material with an escape work of $W _ { 0 }$. When ultraviolet light irradiates the material, only a coherent photoelectron beam with a single kinetic energy and momentum is produced. When this electron beam is irradiated with a double slit spaced at $d$, interference fringes are formed on an observation screen spaced at $L$ from the slit, and the spacing of the fringes is measured to be $\Delta x$. The electron mass is $m$, the Planck constant is $h$, and the speed of light is $c$.

A. A. Momentum of the electron $p _ { e } = \frac { h L } { d \Delta x }$
B. B. Kinetic energy of an electron $E _ { k } = \frac { h L ^ { 2 } } { 2 m d ^ { 2 } \Delta x ^ { 2 } }$
C. C. Energy of photons $E = W _ { o } + \frac { h c L } { d \Delta x }$
D. D. Momentum of photons $p = \frac { h ^ { 2 } L ^ { 2 } } { 2 c m d ^ { 2 } \Delta x ^ { 2 } }$

Answer

Answer: A

Solution: A. According to the formula for stripe spacing $\Delta x = \frac { L } { d } \lambda$ the wavelength can be obtained as $\lambda = \frac { \Delta x d } { L }$ According to the de Broglie wave formula $p _ { \mathrm { e } } = \frac { h } { \lambda }$ the momentum of the electron is $p _ { \mathrm { e } } = \frac { h L } { d \Delta x }$. Therefore, A is correct; B. According to the relationship between kinetic energy and momentum $E _ { \mathrm { k } } = \frac { p ^ { 2 } } { 2 m }$ B. According to the relationship between kinetic energy and momentum $E _ { \mathrm { k } } = \frac { p ^ { 2 } } { 2 m }$, the momentum of the electron is $E _ { \mathrm { k } } = \frac { h ^ { 2 } L ^ { 2 } } { 2 m d ^ { 2 } ( \Delta x ) ^ { 2 } }$. Therefore, B is wrong; C. The energy of a photon $E = W _ { o } + E _ { k } = W _ { o } + \frac { h ^ { 2 } L ^ { 2 } } { 2 m d ^ { 2 } ( \Delta x ) ^ { 2 } }$ C. The energy of the photon $E = W _ { o } + E _ { k } = W _ { o } + \frac { h ^ { 2 } L ^ { 2 } } { 2 m d ^ { 2 } ( \Delta x ) ^ { 2 } }$ is wrong; D. The momentum of the photon $p = m c$ Energy of the photon $E = m c ^ { 2 }$ The energy of the photon $E = m c ^ { 2 }$ is $E = m c ^ { 2 }$. Then the momentum of the photon $p = \frac { W _ { 0 } } { c } + \frac { h ^ { 2 } L ^ { 2 } } { 2 m d ^ { 2 } ( \Delta x ) ^ { 2 } }$ Therefore, D is wrong.

Question 16

Question 16: 16. 1897 British physicist Joseph John Thomson in the study of cathode rays discovered the electron,...

16. 1897 British physicist Joseph John Thomson in the study of cathode rays discovered the electron, which is the first human discovery of elementary particles. The following statements about electrons are correct

A. A. The discovery of the electron shows that the nucleus has an internal structure
B. B. $\beta$ The rays may also be a stream of electrons formed by the ionization of electrons outside the nucleus of an atom, which has a moderate penetrating power
C. C. In the photoelectric effect experiment, the escaping photoelectrons come from the free electrons in the metal
D. D. Rutherford's model of the nuclear structure of the atom assumes that the orbital radii of the electrons outside the nucleus are quantized

Answer

Answer: C

Solution: A. The discovery of the electron shows that atoms have an internal structure and cannot show that the nucleus has an internal structure. Atoms are made up of a nucleus and electrons outside the nucleus. Therefore, A is incorrect. B. $\beta$-rays are formed by electrons released when neutrons decay into protons inside the nucleus, and have nothing to do with electrons outside the nucleus. Therefore, B is wrong. C. According to the definition of the photoelectric effect phenomenon can be known photoelectric effect experiments, the escaping photoelectrons from the free electrons in the metal. Therefore, C is correct. D. Bohr theory that the radius of the electron orbital is quantized, Rutherford's model of the nuclear structure of the atom that there is a very small nucleus in the center of the atom, called the nucleus, the atom's entire positive charge and almost all the mass is concentrated in the nucleus, the negatively charged electrons in the space outside the nucleus around the nucleus rotating, so D is wrong.

Question 17

Question 17: 17. The end of the nineteenth century to the early twentieth century, some physicists on the study o...

17. The end of the nineteenth century to the early twentieth century, some physicists on the study of certain physical phenomena directly contribute to the establishment and development of "modern atomic physics", on the following four diagrams involving physical knowledge is correctly stated ![](/images/questions/phys-photoelectric-effect/image-004.jpg) Figure 1 ![](/images/questions/phys-photoelectric-effect/image-005.jpg) Figure 2 ![](/images/questions/phys-photoelectric-effect/image-006.jpg) Figure 3 ![](/images/questions/phys-photoelectric-effect/image-007.jpg) Figure 4 ## Read

A. A. Figure 1 is the experimental law of blackbody radiation, Einstein in order to explain this experimental law, the first proposed "energy son" generalization
B. B. The emergence of strong lasers has made it possible for an electron to absorb several photons in a very short time, which has been confirmed by experiments. As shown in Fig. 2, if a zinc plate is irradiated with light of wavelength $\lambda$, the needle of the detector does not deflect; if the plate is irradiated with a strong laser of wavelength $\lambda$, the needle of the detector may deflect!
C. C. As shown in Figure 3, a hydrogen atom in the $n = 4$ energy level jumps to a lower energy level and can emit up to six different frequencies of light
D. D. Figure 4 shows the deflection of three types of rays produced in the natural radiation phenomenon in an electric field, where the ray represented by line (3) has the strongest penetrating ability

Answer

Answer: B

Solution: The electrons of the zinc plate can absorb many photons in a very short time and escape, and the needle of the solid checker may be deflected, B is correct; a hydrogen atom in the $n = 4$ energy level leaps to a lower energy level, and it can emit up to three kinds of light with different frequencies, i.e., from $n = 4$ to $n = 3$ to 2 and finally to 1, C is wrong; from the figure, we know that the electric current is not as strong as that of the laser, and the laser is not as strong as that of the zinc plate. From the figure, we can see that the direction of the electric field is horizontally to the right, and the particle (3) is deflected to the right, so it is positively charged, and it is the $\alpha$ particle, which has the strongest ionizing effect, and it is D wrong.

Question 18

Question 18: 18. Now a photoelectric effect experiment with a photoelectric tube, when irradiated with a certain ...

18. Now a photoelectric effect experiment with a photoelectric tube, when irradiated with a certain frequency of light, there is a photocurrent, the following statement is correct

A. A. Keeping the frequency of the incident light constant, the intensity of the incident light becomes larger and the saturation photocurrent becomes larger
B. B. The saturation photocurrent must become larger as the frequency of the incident light becomes higher
C. C. The frequency of the incident light becomes higher, the light intensity remains constant, and the maximum initial kinetic energy of the photoelectron remains constant
D. D. Keeping the intensity of the incident light constant and decreasing the frequency of the incident light, a photocurrent is always generated

Answer

Answer: A

Solution: A. According to the conclusion of the photoelectric effect experiment, keep the frequency of the irradiated light unchanged, the intensity of the irradiated light becomes larger, the saturation current becomes larger, so A is correct; B. According to the photoelectric effect equation $\mathrm { E } _ { \mathrm { km } } = \mathrm { hv } - \mathrm { W } _ { 0 }$, the frequency of irradiated light becomes higher, the maximum initial kinetic energy of photoelectrons becomes larger, and the saturation photocurrent is not necessarily larger, so B is wrong; C. The frequency of the incident light becomes higher, the light intensity remains unchanged, and the maximum initial kinetic energy of the photoelectron becomes larger, option C is wrong; D. Keep the incident light intensity unchanged, constantly reducing the frequency of incident light, when the frequency of incident light is less than the cut-off frequency of the metal, the photoelectric effect will not occur, there will be no photocurrent, option D error;

Question 19

Question 19: 19. The emitted power of a laser is $P$, the wavelength of the emitted laser light in a medium with ...

19. The emitted power of a laser is $P$, the wavelength of the emitted laser light in a medium with a refractive index of $n$ is $\lambda$, and it is known that light travels in a vacuum with a speed of $c$, and that Planck's constant is $h$. [INLINE_FORMULA_4]] , then the number of photons that can be emitted by this laser per second is

A. A. $\frac { h c } { \lambda n P }$
B. B. $\frac { \lambda n P } { h c }$
C. C. $\frac { \lambda P } { n h c }$
D. D. $\frac { \lambda P } { h c }$

Answer

Answer: B

Solution: Let the wavelength of laser light in vacuum be ${ } ^ { \lambda _ { 0 } }$, then we have $$ n = \frac { \lambda _ { 0 } } { \lambda } $$ The energy of this laser photon is $$ E = h v = \frac { h c } { \lambda _ { 0 } } $$ Therefore, the number of photons that this laser can emit per second is $$ N = \frac { P } { E } $$ The joint solution is $$ N = \frac { \lambda n P } { h c } $$

Question 20

Question 20: 21. The figure shows the energy level diagram of the hydrogen atom. A group of hydrogen atoms in the...

21. The figure shows the energy level diagram of the hydrogen atom. A group of hydrogen atoms in the excited state of $n = 4$, in the process of jumping to a lower energy level to the outward emission of photons, can be radiated out of six different frequencies of light, of which there are a number of frequencies of light can be made to escape the work of 6.20 eV of the metal plate photoelectric effect ( ) | $n$ | $E / \mathrm { eV }$ <br> $\infty$ <br> 4 <br> 3 <br> 2 | | :---: | :---: | | - 1.51 | - - 3.5 | - - | - 3.40 | 1 $\_\_\_\_$ -13.6

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

Answer

Answer: B

Solution: A group of hydrogen atoms in the excited state of $n = 4$ emits photons outward as it leaps to a lower energy level, and can emit photons of six different energies: Leaping from $n = 4$ to $n = 1$ emits photons with an energy of 12.75 eV. From $n = 4$ to $n = 2$, the energy of the radiated photon is 2.55 eV. From $n = 4$ to $n = 3$, the radiated photon energy is 0.66 eV. From $n = 3$ to $n = 1$, the photon energy radiated is 12.09 eV. From $n = 3$ to $n = 2$, the radiated photon energy is 1.89 eV. From $n = 2$ to $n = 1$, the photon energy is 10.2 eV. It can be seen that there are three kinds of photon energy is greater than the escape work of the metal plate 6.20 eV, so there are three kinds of frequency of light can make the metal plate photoelectric effect, so choose B.

Question 21

Question 21: 22. The following phenomenon illustrates the particulate nature of light

22. The following phenomenon illustrates the particulate nature of light

A. A. Electrons are emitted when a zinc plate is irradiated with ultraviolet light.
B. B. White light irradiation of soap film shows colorful patterns
C. C. Camera lenses with transparency enhancement film are lavender in color.
D. D. Poisson Brightening

Answer

Answer: A

Solution: A. When the zinc plate is irradiated with ultraviolet light, electrons are emitted, which is a phenomenon of the photoelectric effect, indicating the particle nature of light, option A is correct; B. White light irradiation of soap film showing color patterns, which is the film interference phenomenon, indicating that the volatility of light, option B is wrong; C. The camera lens affixed with a transmittance-enhancing film shows a lavender color, which is the interference phenomenon of light, indicating that the volatility of light, option C is wrong; D. Poisson bright spots, which is the diffraction phenomenon of light, indicating the fluctuation of light, option D error;

Question 22

Question 22: 23. As shown in the figure, with the wire to the checker connected to the zinc plate, when irradiati...

23. As shown in the figure, with the wire to the checker connected to the zinc plate, when irradiating the zinc plate with ultraviolet light, the phenomenon that occurs correctly is ![](/images/questions/phys-photoelectric-effect/image-008.jpg)

A. A. There are photons escaping from the zinc plate.
B. B. There are electrons escaping from the zinc plate
C. C. A negatively charged electrical detector
D. D. Zinc plate is negatively charged

Answer

Answer: B

Solution: When irradiating the zinc plate with ultraviolet light, the photoelectric effect occurs, and electrons escape from the zinc plate, which loses electrons and is positively charged, so the checker is positively charged and opens at a certain angle.

Question 23

Question 23: 24. The following important physics experiments are correctly described as follows

24. The following important physics experiments are correctly described as follows

A. A. $\alpha$ Particle scattering experiments are the experimental basis for the theory of the nuclear structure of the atom
B. B. $\alpha$ Particle scattering experiments propose that positive charges are uniformly distributed throughout the atom
C. C. Discovery of electrons reveals nucleus has more complex structure
D. D. The photoelectric effect proves that light is both fluctuating and particulate.

Answer

Answer: A

Solution: A.$\alpha$ The particle scattering experiment is the experimental basis of the theory of the nuclear structure of the atom, option A is correct; B.$\alpha$ The particle scattering experiment proposes that the positive charges are all concentrated on the nucleus, not evenly distributed throughout the atom, option B is wrong; C. The discovery of the electron revealed that atoms have a more complex structure, option C is wrong; D. The photoelectric effect proves that light has a particle nature, option D is wrong.

Question 24

Question 24: 25. Now use the device shown in Figure A to do experiments on the properties of light, $a , b$ two m...

25. Now use the device shown in Figure A to do experiments on the properties of light, $a , b$ two monochromatic light on the light screen to get the pattern shown in Figure B, the following statements are correct ( ) ![](/images/questions/phys-photoelectric-effect/image-009.jpg)

A. A. P is a single slit, the frequency of $a$ light is higher than the frequency of $b$ light
B. B. P is a double slit, the frequency of $a$ light is lower than the frequency of $b$ light
C. C. In a vacuum, $a$ light travels faster than $b$ light.
D. D. $a$ light has a smaller photon energy than $b$ light

Answer

Answer: D

Solution: B. Single-slit diffraction pattern of the center of the brightest stripes of the widest and brightest, to the sides of the narrowing, double-slit interference pattern is the dark and light stripes, stripes spacing is equal, the width of the stripes is equal, so P for the single-slit, so B error; A. The longer the wavelength of light is, the wider the central bright stripe of single-slit diffraction is, so the wavelength of $a$ light is larger than the wavelength of $b$ light, and the frequency of $a$ light is lower than that of $b$ light, so it's A wrong. (a); C. The propagation speed of all colors of light in vacuum is equal, so C is wrong; D. Since the frequency of $a$ light is lower than that of $b$ light, it can be seen that the photon energy of $a$ light is smaller than that of $b$ light from $\varepsilon = h v$. D is correct. Therefore, D is correct.

Question 25

Question 25: 26. Lesions in the eye cause changes in the refractive index of light in different parts of the eye....

26. Lesions in the eye cause changes in the refractive index of light in different parts of the eye. A glass sphere is now used to simulate an eyeball to study the effect on light propagation. The glass sphere is made of two hemispheres with different refractive indices and radii of $R$ that are put together left and right, with a collocation surface of ${ } ^ { M N }$, a center of the sphere of ${ } ^ { O } , ~ Q$ as the midpoint of ${ } ^ { M O }$, and a center of the sphere at ${ } ^ { M O }$, with a center of the sphere at ${ } ^ { P Q }$ perpendicular to the center of the sphere. FORMULA_4]] is perpendicular to ${ } ^ { M N }$ and crosses the left hemisphere at point $P$. A beam of complex-colored light enters from point $P$ at an angle of incidence of $60 ^ { \circ }$ and splits into two beams of $a , b$. If the refractive index of $a$ light in the left and right hemispheres is $\sqrt { 3 }$ and ${ } ^ { \sqrt { 2 } }$, respectively, and the speed of light in a vacuum is $c$, then the following statements are correct ( ) ![](/images/questions/phys-photoelectric-effect/image-010.jpg)

A. A. $a$ The angle of ejection of light from the right hemisphere is also $60 ^ { \circ }$
B. B. $a$ The time it takes for light to travel through this glass sphere is $\frac { R } { 2 c } ( \sqrt { 6 } + 3 )$
C. C. Double-slit interference occurs through the same device, and the $b$ stripes of light are widely spaced apart
D. D. When irradiating the same phototube, $a$ light causes the maximum initial kinetic energy of the escaping photoelectrons to be large.

Answer

Answer: B

Solution: A. According to the question, the angle of refraction of $a$ light in the left hemisphere is equal to the angle of incidence in the right hemisphere, and since $a$ light in the left and right hemispheres has different refractive indices, the angle of emission of $a$ light from the right hemisphere is not equal to the angle of $60 ^ { \circ }$. FORMULA_5]], so A is wrong; B. According to the meaning of the question, from the formula $n = \frac { c } { v }$, $a$ the propagation speed of light in the left hemisphere is $v _ { 1 } = \frac { c } { \sqrt { 3 } } = \frac { \sqrt { 3 } } { 3 } c a$ The speed of propagation of light in the right hemisphere is $$ v _ { 2 } = \frac { c } { \sqrt { 2 } } = \frac { \sqrt { 2 } } { 2 } c $$ From the geometric relationship, $a$ light travels in both the left and right hemispheres at a distance of $\frac { \sqrt { 3 } } { 2 } R$ , and the $a$ time for the light to propagate in the glass sphere is is $$ t = \frac { \frac { \sqrt { 3 } } { 2 } R } { \frac { \sqrt { 3 } } { 3 } c } + \frac { \frac { \sqrt { 3 } } { 2 } R } { \frac { \sqrt { 2 } } { 2 } c } = \frac { 3 R } { 2 c } + \frac { \sqrt { 6 } R } { 2 c } = \frac { R } { 2 c } ( 3 + \sqrt { 6 } ) $$ Therefore, B is correct; CD. According to the law of refraction, from the figure, the refractive index of $a$ light in the left hemisphere is smaller than that of $b$ light in the left hemisphere, the frequency of $a$ light is smaller than that of ${ } ^ { b }$ light. frequency, the wavelength of ${ } ^ { a }$ light is greater than the wavelength of ${ } ^ { b }$ light, and it is known that the maximum initial kinetic energy of the photoelectrons escaping from $a$ light is small when irradiating the same photocell from $a$ light. INLINE_FORMULA_20]] can be known that when double-slit interference occurs through the same device, the $b$ light has a small spacing between the stripes, so CD is wrong.

Question 26

Question 26: 27. The surface of a metal is irradiated with a certain monochromatic light and photoelectrons fly o...

27. The surface of a metal is irradiated with a certain monochromatic light and photoelectrons fly out of the metal surface. The following statement is correct

A. A. If the intensity of the irradiated light is weakened, the time for the photoelectrons to fly out must become longer
B. B. If the frequency of the irradiated light decreases, the number of photoelectrons flying out per unit time must decrease
C. C. If the frequency of the irradiated light decreases, no more photoelectrons may be flown out
D. D. If the frequency of the irradiated light remains unchanged and the intensity is reduced to half of the original, the kinetic energy of the flying photoelectrons may be halved.

Answer

Answer: C

Solution: AD. The intensity of light does not affect the energy of photoelectrons, only affects the number of photoelectrons emitted per unit of time, does not affect the time of photoelectrons flying out, then if only reduce the intensity of the flying out of photoelectrons, the energy remains unchanged, the number of decreases, so the AD is wrong; BC. The photoelectric effect occurs when the frequency of the incident light is greater than the cut-off frequency. If the incident light frequency is smaller than the cut-off frequency, the photoelectric effect can not occur, no photoelectrons fly out; if the incident light frequency is still greater than the cut-off frequency, the photoelectric effect can occur, due to the intensity of the light is unchanged, the number of photoelectrons per unit of time increased, so B is wrong, C is correct.

Question 27

Question 27: 28. The following statements about optical phenomena are correct

28. The following statements about optical phenomena are correct

A. A. Checking the flatness of the surface of a workpiece with a standard plane utilizes the phenomenon of light reflection.
B. B. The colorful streaks of oil film on the water surface in sunlight is a diffraction phenomenon of light
C. C. Camera lenses with permeable films utilize the principle of the photoelectric effect.
D. D. In the Tang poem "The pool is clear, the water is shallow, the lotus moves, the fish scatter", "the water is shallow" is due to the refraction of light.

Answer

Answer: D

Solution: A. Check the flatness of the surface of the workpiece with a standard plane using the air film between the two workpieces to form the interference of light, that is, film interference phenomenon, so A error; B. Sunlight on the water surface of the oil film showing colored stripes is the light of the film interference phenomenon, so B error; C. Camera lens plus transparency film is through the interference of light, reducing the intensity of reflected light, thereby increasing the intensity of transmitted light, so C error; D. People see the bottom of the water is light from the water into the air diagonally refracted to form an imaginary image, belonging to the refraction of light, like the location of the actual object than shallow, so it looks like the bottom of the water is shallow, in fact, very deep, so D is correct.

Question 28

Question 28: 29. When a tungsten surface is irradiated with ultraviolet light at a wavelength of 200 nm, the maxi...

29. When a tungsten surface is irradiated with ultraviolet light at a wavelength of 200 nm, the maximum initial kinetic energy of an electron escaping the surface is $4.7 \times 10 ^ { - 19 } \mathrm {~J}$. The Planck constant is $6.63 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm {~s}$ and the speed of light in a vacuum is $3.00 \times 10 ^ { 8 } \mathrm {~m} \cdot \mathrm {~s} ^ { - 1 }$. The lowest frequency of monochromatic light that can cause the photoelectric effect in tungsten is approximately

A. A. $5.0 \times 10 ^ { 4 } \mathrm {~Hz}$
B. B. $8.0 \times 10 ^ { 14 } \mathrm {~Hz}$
C. C. $1.0 \times 10 ^ { 15 } \mathrm {~Hz}$
D. D. $1.2 \times 10 ^ { 15 } \mathrm {~Hz}$

Answer

Answer: B

Solution: According to the photoelectric effect equation: $E _ { K m } = h \gamma - W _ { O }$; the relationship between the speed of light, wavelength, and frequency is: $\gamma = \frac { c } { \lambda }$, and substituting the data into the above equation, we have $$ W _ { 0 } = h \gamma - E _ { K m } = 6.63 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm {~s} \times \frac { 3 \times 10 ^ { 8 } } { 300 \times 10 ^ { - 9 } } \mathrm {~s} ^ { - 1 } - 1.28 \times 10 ^ { - 19 } \mathrm {~J} = 5.35 \times 10 ^ { - 19 } \mathrm {~J} $$ According to the escape work $W _ { 0 } = h \gamma _ { 0 }$, we have $$ \gamma _ { 0 } = \frac { W _ { 0 } } { h } = \frac { 5.35 \times 10 ^ { - 19 } } { 6.63 \times 10 ^ { - 34 } } \mathrm {~Hz} \approx 8 \times 10 ^ { 14 } \mathrm {~Hz} $$

Question 29

Question 29: 30. The following statements are correct

30. The following statements are correct

A. A. If the photoelectric effect occurs when a metal is irradiated with violet light, the photoelectric effect must occur when the metal is irradiated with green light.
B. B. The large deflection of a few $\alpha$ particles in the $\alpha$ particle scattering experiment was one of the main bases for Rutherford's conjecture of a model of the nuclear structure of the atom
C. C. From Bohr's theory, the hydrogen atom's extra-nuclear electrons have to release photons of a certain frequency when they jump from a higher energy level to a lower one, while the electrons' kinetic energy decreases and their potential energy increases
D. D. When an atomic nucleus undergoes $\alpha$ decay, the number of neutrons in the new nucleus is reduced by 4 compared to the original nucleus

Answer

Answer: B

Question 30

Question 30: 31. In an experiment to study the laws of the photoelectric effect, a monochromatic light with a fre...

31. In an experiment to study the laws of the photoelectric effect, a monochromatic light with a frequency of $v$ is used to irradiate a phototube, and the measured suppression voltage is $U$.

A. A. If only the intensity of this monochromatic light is increased, the suppression voltage must be greater than $U$
B. B. If the monochromatic light with frequency $2 ^ { v }$ is used instead, the suppression voltage must be greater than $2 U$.
C. C. If a monochromatic light with frequency $2 ^ { v }$ is used, the saturation photocurrent must be twice as high.
D. D. If we irradiate with monochromatic light of frequency $0.5 v$, it must be impossible to irradiate photoelectrons.

Answer

Answer: B

Solution: A. According to the question $$ e U = E _ { k m } = h v - W _ { 0 } $$ It can be seen that the suppression voltage is independent of the intensity of the monochromatic light, if only increase the intensity of the monochromatic light, the suppression voltage will not increase, so A error ; B. If the frequency of $2 v$ is used to irradiate the monochromatic light, then there are $$ 2 h v - W _ { 0 } = 2 h v - 2 W _ { 0 } + W _ { 0 } = 2 e U + W _ { 0 } > 2 e U $$ It can be seen that if a monochromatic light with frequency $2 v$ is used instead, the suppression voltage must be greater than $2 U$, so B is correct; C. After the photoelectric effect occurs, the size of the saturation photocurrent depends on the intensity of the incident light and has nothing to do with the frequency of the incident light, so if the monochromatic light irradiation with a frequency of $2 v$ is used instead, the saturation photocurrent remains unchanged, so C is wrong; D. According to the conditions of the photoelectric effect can be known, the frequency of the incident light is greater than the limiting frequency in order to occur photoelectric effect, if the frequency of $0.5 v$ monochromatic light irradiation, the photoelectric effect may also occur, so D error;

Question 31

Question 31: 32. The following is true about modern physics

32. The following is true about modern physics

A. A. The maximum initial kinetic energy of an escaping photoelectron in the photoelectric phenomenon is proportional to the frequency of the incident light.
B. B. According to Bohr's model, the kinetic energy of the electrons outside the nucleus increases when the excited state hydrogen atom jumps to the ground state
C. C. $\beta$ Decay is a strong interaction process
D. D. The greater the average mass of the nucleons in an atomic nucleus, the more stable the nucleus is

Answer

Answer: B

Solution: A. According to the photoelectric effect equation $E _ { \mathrm { k } } = k - W _ { 0 }$, it can be seen that the maximum initial kinetic energy of the escaping photoelectrons is linearly related to the frequency of the incident light, not proportional, so A is wrong; B. Hydrogen atom from the excited state jump to the ground state, the electron orbital radius $r$ decreases, by the Coulomb force to provide centripetal force to get $\frac { k e ^ { 2 } } { r ^ { 2 } } = \frac { m v ^ { 2 } } { r }$ The kinetic energy of the electron can be $E _ { \mathrm { k } } = \frac { k e ^ { 2 } } { 2 r }$ When $r$ decreases, the kinetic energy of the electrons outside the nucleus increases, so B is correct; C. $\beta$ decay is a weak interaction process, so C is wrong; D. The stability of the atomic nucleus is determined by the specific binding energy. The larger the average mass of the nucleus is, the lower the specific binding energy is, and the more unstable the nucleus is, so D is wrong.

Question 32

Question 32: 33. a, b two charts is the energy level diagram of the hydrogen atom, the arrow in the figure shows ...

33. a, b two charts is the energy level diagram of the hydrogen atom, the arrow in the figure shows the direction of the electrons outside the nucleus in the two energy levels between the leap; in the photoelectric effect experiment, respectively, with blue light and different intensities of yellow light to study the relationship between the photocurrent and voltage, the resulting images were as shown in the two images of c and d. A B figure, the electron in the jump photon absorption is Figure; C D figure, can correctly represent the relationship between photocurrent and voltage is $\_\_\_\_$Figure ![](/images/questions/phys-photoelectric-effect/image-011.jpg) A ![](/images/questions/phys-photoelectric-effect/image-012.jpg) B ![](/images/questions/phys-photoelectric-effect/image-013.jpg) C ![](/images/questions/phys-photoelectric-effect/image-014.jpg) D

A. A. A C
B. B. B C
C. C. A D
D. D. Z Ding

Answer

Answer: D

Solution: The electron absorbs photons during the leap, the energy becomes larger, and leaps from the low energy level to the high energy level, so it is B. The frequency of blue light is larger than the frequency of yellow light, according to the equation $E _ { k m } = h v - W _ { 0 } , e U _ { c } = E _ { k m }$, it can be seen that the suppression voltage of the blue light with a large frequency ${ } ^ { U }$ is larger than the suppression voltage of the yellow light, so it is D. So D is correct and ABC is wrong.

Question 33

Question 33: 34. irradiate a metal with two monochromatic light of frequency $v$ and $2 v$, the ratio of the maxi...

34. irradiate a metal with two monochromatic light of frequency $v$ and $2 v$, the ratio of the maximum initial kinetic energies of the escaping photoelectrons is $1 : 3$, and the known Planck constant is $h$, and the speed of light in a vacuum is $c$, and the electron charge is $e$. The speed of light in a vacuum is $c$ and the electron charge is $e$. The following statement is correct

A. A. If the metal is irradiated with monochromatic light of frequency $2 v$, the number of photoelectrons escaping per unit time must be high
B. B. The photoelectric effect does not occur when the metal is irradiated with monochromatic light of frequency $\frac { 1 } { 4 } ^ { v }$.
C. C. Two monochromatic lights, A and B, irradiate the metal, and as long as the intensity of the light is the same, the curb voltage of the corresponding photocurrent is the same
D. D. The escape work for this metal is $\frac { 1 } { 4 } h v$

Answer

Answer: B

Solution: A. The number of photoelectrons escaping per unit time is related to the intensity of light, and since the relationship between the intensity of light is unknown, option A is wrong ; BD, photon energy were: $E _ { 1 } = h v _ { \text {和 } } E _ { 2 } = 2 h v$, according to Einstein photoelectric effect equation can be known photoelectron maximum Initial kinetic energy: $E _ { k 1 } = h v - W _ { \text {和 } } E _ { k 2 } = 2 h v - W$, the maximum initial kinetic energy of the escaping photoelectron ratio is $1 : 3$, can be obtained by association with the escaping work: $W = \frac { 1 } { 2 } h v$, with a frequency of $\frac { 1 } { 4 } h v$ monochromatic light irradiation of the metal can not occur The photoelectric effect cannot occur when the metal is irradiated with monochromatic light at a frequency of $\frac { 1 } { 4 } h v$; C, two different frequencies of light, the maximum initial kinetic energy of photoelectrons is different, by the kinetic energy theorem can be known corresponding to the curb voltage is different, so option C is wrong;

Question 34

Question 34: 37. Photoelectric tube is a light signal into an electrical signal device, in the communications, me...

37. Photoelectric tube is a light signal into an electrical signal device, in the communications, medical, security monitoring and other fields are widely used. The photoelectric tube into the circuit shown in the figure, with a frequency of $v$ light irradiation K plate, adjust the slide of the slide rheostat P, when the sensitive ammeter G is 0, the voltmeter V of the number of $U$, the voltage is usually also known as the inhibitory voltage. The Planck constant is $h$ and the electron charge is $e$, the following statements are correct ( ) ![](/images/questions/phys-photoelectric-effect/image-015.jpg)

A. A. The initial kinetic energy of a photoelectron escaping from the K-plate is inversely proportional to the restraining voltage
B. B. If the intensity of the incident light is increased, the curb voltage increases
C. C. The escape work of the K-plate material is $h ^ { v } - e U$
D. D. If only increase the frequency of the incident light, so that G is 0, then you need to adjust the left slide P

Answer

Answer: C

Solution: A. From the kinetic energy theorem $$ - e U = 0 - E _ { \mathrm { km } } $$ Therefore, the maximum initial kinetic energy of a photoelectron escaping from the K-pole is proportional to the suppression voltage $U$, so A is wrong; B. By the Einstein photoelectric effect equation [BLOCK_FORMULA_1]] Simplify to get $$ e U = h v - W _ { 0 } $$ The suppression voltage is independent of the intensity of the incident light, so B is wrong; C.By $$ e U = h v - W _ { 0 } $$ it can be seen that $$ W _ { 0 } = h v - e U $$ Therefore, C is correct; D. By $$ e U = h v - W _ { 0 } $$ D. By $$ e U = h v - W _ { 0 } $$, it can be seen that increasing the frequency of the incident light, the suppression voltage also increases, if the $G$ is 0, you need to adjust the slide $P$ to the right, so D is wrong.

Question 35

Question 35: 38. Silicon photovoltaic cells are devices that utilize the principle of the photovoltaic effect, an...

38. Silicon photovoltaic cells are devices that utilize the principle of the photovoltaic effect, and the following statements are correct

A. A. A silicon photovoltaic cell is a device that converts electrical energy into light energy
B. B. Any electrons in a silicon photovoltaic cell that have absorbed photon energy can escape
C. C. The maximum initial kinetic energy of an escaping photoelectron is related to the frequency of the incident light
D. D. The phenomenon of the photoelectric effect shows that light is volatile

Answer

Answer: C

Solution: A. Silicon photovoltaic cell is a device that converts light energy into electricity, so A is wrong; B. Silicon photovoltaic cell absorbed photon energy in the electron only photon energy is greater than the escape work to escape, so B error; C. By $E _ { \mathrm { k } } = h v - W _ { 0 }$, it can be seen that the maximum initial kinetic energy of photoelectrons is related to the frequency of the incident light, with the increase in the frequency of the incident light and increase, so C is correct; D. The photoelectric effect phenomenon shows that light has a particle, so D is wrong.

Question 36

Question 36: 39. It is known that the photon energy range of visible light is about $1.62 \mathrm { eV } : 3.11 \...

39. It is known that the photon energy range of visible light is about $1.62 \mathrm { eV } : 3.11 \mathrm { eV }$, and the energy level diagram of the hydrogen atom is shown in the figure, and a large number of atoms are in the [INLINE_FORMULA_1]] range. $n = 4$ Energy level of the hydrogen atom to the lower energy level jump, the following statements are correct ( ) | $n$ | $E / \mathrm { eV }$ | | :--- | :--- | | $\infty$ | 0 | | 5 | - 0.54 | 4 | - 0.85 | - 3 | - 1.85 | - 3 | - 1.85 3 | - 1.51 | [INLINE_FORMULA_4]] | 0 | 5 | - 0.54 | 4 | - 0.85 | 3 | - 1.51 | | 2 | - 3.4 | 3.4 | 1

A. A. It radiates three frequencies of light
B. B. The shortest wavelength of light is radiated when jumping to the $n = 3$ energy level
C. C. When jumping to the $n = 3$ energy level, the radiated light will have a significant thermal effect
D. D. When jumping to the $n = 3$ energy level, the light energy radiated causes the photoelectric effect on aluminum with an escape work of 4.2 eV.

Answer

Answer: C

Solution: A. When a large number of hydrogen atoms in the $n = 4$ energy level jump to a lower energy level, the number of photons that can be radiated with different frequencies is $$ C _ { 4 } ^ { 2 } = 6 $$ A Error ; B. by $$ \Delta E = h v = h \frac { c } { \lambda } $$ B. From $$ \Delta E = h v = h \frac { c } { \lambda } $$, it can be seen that the photon frequency is the lowest and the wavelength is the longest when it jumps to the $n = 3$ energy level; C. To $n = 3$ energy level leap, the photon energy of radiation is 0.66 eV, less than red light, so the wavelength is greater than the red light, belongs to the infrared, has a significant thermal effect, C is correct; D. To $n = 3$ energy level jump, the energy of the radiated photon is less than the escape work, so it can't make the aluminum photoelectric effect, D error.

Question 37

Question 37: 40. A student used the device shown in the figure to study the phenomenon of photoelectric effect. A...

40. A student used the device shown in the figure to study the phenomenon of photoelectric effect. A monochromatic light irradiation photoelectric tube cathode K, the photoelectric effect phenomenon, close the switch $S$, in the anode $A$ and the cathode $K$ between the reversal of the voltage, through the adjustment of the slide rheostat slider gradually increase the voltage until the ammeter in the The current is exactly zero, and the voltage value $U$ displayed by the voltmeter at this time is called the curb voltage. Now the frequency of ${ } ^ { v _ { 1 } }$ and ${ } ^ { v _ { 2 } }$ monochromatic light irradiation of the cathode, the measured voltage of the curb are $U _ { 1 }$ and $U _ { 2 }$, and the mass of the electron is $m$, the mass of the electron is $e$. FORMULA_8]] and the charge is $e$, then the following relation is correct ( ) ![](/images/questions/phys-photoelectric-effect/image-016.jpg) High School Physics Assignment, October 30, 2025

A. A. Planck constant $h = \frac { e \left( U _ { 2 } - U _ { 1 } \right) } { v _ { 1 } - v _ { 2 } }$
B. B. Cutoff frequency of cathode K metal $v _ { 0 } = \frac { v _ { 1 } U _ { 2 } - v _ { 2 } U _ { 1 } } { U _ { 1 } - U _ { 2 } }$
C. C. Cathode K metal escape work $W _ { 0 } = \frac { e \left( v _ { 2 } U _ { 1 } - v _ { 1 } U _ { 2 } \right) } { v _ { 1 } - v _ { 2 } }$
D. D. Maximum initial velocity of electrons at cathode K when irradiated by monochromatic light with frequency $v _ { 2 }$ $v _ { \mathrm { m } 2 } = \sqrt { \frac { 2 e U _ { 1 } } { m } }$

Answer

Answer: C

Photoelectric Effect

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