Dual Nature Of Light Question 357
Question: Find the frequency of light which ejects electrons from a metal surface fully stopped by a retarding potential of 3V. The photoelectric effect begin in this metal at frequency of $ 6\times 10^{14} $ per second.
Options:
A) $ 1.32\times 10^{15}Hz $
B) $ 3.28\times 10^{14}Hz $
C) $ 6.22\times 10^{15}Hz $
D) $ 2.22\times 10^{11}Hz $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] The work function of the metal $ W_{0}=hf_{0}=(6.63\times {{10}^{-34}})\times (6\times 10^{14}) $ $ =3.98\times 10^{19}J $ If f is the frequency of the light, then $ hc=W_{0}+eV_{0} $ $ =\frac{3.98\times {{10}^{-19}}+1.6\times {{10}^{-19}}\times 3}{6.63\times {{10}^{-34}}}=1.32\times 10^{15}Hz $
BETA
