Dual Nature Of Light Question 400
Question: The threshold frequency for a metallic surfaces corresponds on an energy of 6.2 eV and the stopping potential for a radiation incident on this surface is 5V. The incident radiation lies in
Options:
A) ultra-violet region
B) infra-red region
C) visible region
D) X-ray region
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ \phi =6.2eV=6.2\times 1.6\times {{10}^{-19}}J $ $ V=5volt $ $ \frac{hc}{\lambda }-\phi =eV_{0}\Rightarrow \lambda =\frac{hc}{\phi +eV_{0}} $ $ =\frac{6.6\times {{10}^{-34}}\times 3\times 10^{8}}{1.6\times {{10}^{-19}}( 6.2+5 )} $ This range lies in ultra violet range.
BETA
