Atomic Structure Question 359
Question: Photoelectric emission is observed from a surface for frequencies $ v_1 $ and $ v_2 $ of the incident radiation $ (v_1-v_2) $ . If the maximum kinetic energies of the photoelectrons in the two cases are in the ratio 1: k then the threshold frequency $ v_1 $ is given by
Options:
A) $ \frac{v_2-v_1}{k-1} $
B) $ \frac{kv_1-v_2}{k-1} $
C) $ \frac{kv_2-v_1}{k-1} $
D) $ \frac{v_2-v_1}{k} $
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Answer:
Correct Answer: B
Solution:
- When frequency is $ v_1 $ , $ hv_1=hv_0+\frac{1}{2}mu_1^{2} $ …(i) When frequencyis $ v_2 $ , $ hv_2=hv_0+\frac{1}{2}mu_2^{2} $ …(ii) $ \because $ $ \frac{1}{2}mu_1^{2}-\frac{1}{k}( \frac{1}{2}mu_2^{2} ) $
$ \therefore $ from Eq. (i) $ hv_1=hv_0+\frac{1}{2k}mu_2^{2} $ …(iii) or $ \frac{1}{2}mu_2^{2}=khv_1-khv_0 $ …(iv) From Eqs. (ii) and (iv) $ hv_2=hv_0+khv_1-khv_0 $ or $ v_0(1-k)=v{ _2}-kv_1 $ or $ v_0=\frac{kv_1-v_2}{k-1} $
BETA
