Surface #1 has work function ,and when bombarded with photons of wavelength emits photoelectrons with maximum energy .Surface #2 has work function ,and when bombarded by photons of wavelength emits photoelectrons with maximum energy .If ,then which of the following must be true?
C) for surface #1 is greater than for surface #2.
D) for surface #2 is greater than for surface #1.

Question

Surface #1 has work function  ,and when bombarded with photons of wavelength  emits photoelectrons with maximum energy  .Surface #2 has work function  ,and when bombarded by photons of wavelength  emits photoelectrons with maximum energy  .If  ,then which of the following must be true?
C) for surface #1 is greater than for surface #2. 
D) for surface #2 is greater than for surface #1.

Quizplus · Accepted Answer

The equation relating the maximum kinetic energy of a photoelectron and the work function of a surface is:
KEmax = hν - Φ
where h is Planck's constant, ν is the frequency of the incident photon, and Φ is the work function of the surface.

For Surface #1,
hν = hc/λ = (6.62607004 × 10^-34 J s)(2.9979 x 10^8 m/s)/(11eaa64f_8f99_f44f_a696_1deb2fa5bed9_TB5258_11) = 1.791 x 10^-16 J
KEmax = (11eaa64f_8f9a_1b60_a696_8d064645cf6c_TB5258_11)
Φ = (11eaa64f_8f99_cd3e_a696_b54c5af9a301_TB5258_11)

Solving for hν in terms of KEmax and Φ, we get:
hν = KEmax + Φ = (11eaa64f_8f9a_1b60_a696_8d064645cf6c_TB5258_11) + (11eaa64f_8f99_cd3e_a696_b54c5af9a301_TB5258_11) = (11eaa64f_8f9b_53e9_a696_d32469a93a71_TB5258_11)

For Surface #2,
hν = hc/λ = (6.62607004 × 10^-34 J s)(2.9979 x 10^8 m/s)/(11eaa64f_8f9a_6982_a696_351af9537f82_TB5258_11) = 1.124 x 10^-16 J
KEmax = (11eaa64f_8f9a_9093_a696_eb0ab6224284_TB5258_11)
Φ = (11eaa64f_8f9a_4271_a696_27bba5f2fc42_TB5258_11)

Solving for hν in terms of KEmax and Φ, we get:
hν = KEmax + Φ = (11eaa64f_8f9a_9093_a696_eb0ab6224284_TB5258_11) + (11eaa64f_8f9a_4271_a696_27bba5f2fc42_TB5258_11) = (11eaa64f_8f9b_7afa_a696_8921bf465bbe_TB5258_11)

Comparing the two values of hν, we see that (11eaa64f_8f9b_7afa_a696_8921bf465bbe_TB5258_11) is greater than (11eaa64f_8f9b_53e9_a696_d32469a93a71_TB5258_11). This means that the frequency of the incident photon required to produce photoelectrons with the same maximum kinetic energy is greater for Surface #2 than for Surface #1. Therefore, the threshold frequency for Surface #2 must be greater than the threshold frequency for Surface #1.

Choice D is the only option that correctly compares the threshold frequencies of the two surfaces.

Surface #1 Has Work Function ,And When Bombarded with Photons