Atomic Structure Question 197

Question: A certain metal when irradiated to light (v = 3.2 $ \times 10^{16} $ Hz) emits photoelectrons with twice kinetic energy as did photoelectrons when the same metal is irradiated by ligh1 (v = 2.0 $ \times 10^{16} $ Hz) The $ v_0 $ (threshold frequency) of metal is

Options:

A) $ 1.2\times 10^{14}Hz $

B) $ 8\times 10^{15}Hz $

C) $ 1.2\times 10^{16}Hz $

D) $ 4\times 10^{12}Hz $

Show Answer

Answer:

Correct Answer: B

Solution:

  • $ KE=hv_0-hv_0 $

$ hv_1-hv_0=2(hv_2-hv_0) $

$ v_0=2(v_2-v_1) $

$ =2(2.0\times 10^{16})-(3.2\times 10^{16}) $

$ =8\times 10^{15}{s^{-1}}=8\times 10^{15}HZ $



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