Atomic Structure Question 296
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} $
BETA
