Topic Menu

Cyber-chess a chess instruction site

Photoelectric Effect - Multiple Choice Questions

Q1. The intensity of a monochromatic light source is increased.

Which of the following is correct?

If the intensity increases more photons hit the surface each second. Therefore the number of photoelectrons released will increase.

The energy of the emitted photon depends on its frequency (E = hf). Increasing the intensity has no effect on frequency - so energy is unchanged.

Choice C is the correct answer
Energy of an emitted photon
Number of photons emitted per second
A
increases
increases
B
increases
unchanged
C
unchanged
increases
D
unchanged
unchanged

 

Q2. A beam of light of wavelength λ is incident on a clean metal surface and photoelectrons are emitted.

The wavelength of the light is halved but energy incident per second is kept the same.

Which row in the table is correct?

If the wavelength is halved the frequency doubles (c = fλ)

If the frequency doubles the energy of the photons doubles (E=hf), and as the energy is being kept constant, the intensity of the light must have been reduced, thereby reducing the number of photons incident on the surface. That will reduce the number of photoelectrons emitted per second.

The increase in energy given to each electron when the freeing occurs will result in excess KE of released photoelectrons.

Choice C is the correct answer.
Maximum kinetic energy of the emitted photoelectrons
Number of photoelectrons emitted per second
A
Increases
Unchanged
B
Decreases
Increases
C
Increases
Decreases
D
Decreases
Unchanged

 

Q3. Photons of wavelength 290 nm are incident on a metal plate.

The work function of the metal is 4.1 eV

What is the maximum kinetic energy of the emitted electrons?

A

0.19 eV

B

4.3 eV
C

6.9 eV
D

8.4 eV

 

c = fλ

f = c/λ

hf = hc/λ = φ + Ek (MAX)

Ek (MAX) = hc/λ - φ

φ = 4.1 eV

φ = 4.1 x 1.6 x 10-19 J

φ = 6.56 x 10-19 J

∴ Ek(MAX) = 6.63 x 10-34 x 3.0 x 108/(290 x 10-9) - 6.56 x 10-19

Ek(MAX) = (6.86 - 6.56) x 10-19

Ek(MAX) = 0.3 x 10-19J

Ek(MAX) = 0.3 x 10-19/1.6 x 10-19 eV

Ek(MAX) = 0.19 eV

Choice A

Q4. When light of a certain frequency greater than the threshold frequency of a metal is directed at the metal, photoelectrons are emitted from the surface.

The power of the light incident on the metal surface is doubled.

Which row shows the effect on the maximum kinetic energy and the number of photoelectrons emitted per second?

Maximum kinetic energy
Number of photoelectrons emitted per second
A
remains unchanged

remains unchanged
B
doubles

remains unchanged
C
remains unchanged

doubles
D
doubles

doubles

 

Choice C

Q5. Line X on the graphs below shows how the maximum kinetic energy of emitted photoelectrons varies with the frequency of incident radiation for a particular metal.

Which graph shows the results for a metal Y that has a higher work function than X?

The lines should be parallel.... so only A is possible

The work function is the intercept on the Y axis.... only in A would the intercept from the Y line be a greater value than the X one.