Atomic Structure Question 427

Question: The threshold frequency of a metal is $ 1\times 10^{15}{s^{-1}}. $ The ratio of the maximum kinetic energies of the photoelectrons when the metal is irradiated with radiations of frequencies $ 1.5\times 10^{15}{s^{-1}} $ and $ 2.0\times 10^{15}{s^{-1}} $ respectively would be

Options:

A) 4 : 3

B) 1 : 2

C) 2 : 1

D) 3 : 4

Show Answer

Answer:

Correct Answer: B

Solution:

  • $ \frac{{{(KE)}_1}}{{{(KE)}_2}}=\frac{v_1-v_0}{v_2-v_0} $

$ =\frac{(1.5\times 10^{15}-1\times 10^{15})}{( 2.0\times 10^{15}-1\times 10^{15} )~}=\frac{1}{2} $



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