Atomic Structure Question 12
A certain metal when irradiated with light (v = 3.2 $ \times 10^{16} $ Hz) emits photoelectrons with twice kinetic energy as did photoelectrons when the same metal is irradiated by light (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=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 $
BETA
