Dual Nature Of Light Question 401
Question: Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The wavelength of the emitted electron is:
Options:
A) $ <2.8\times {{10}^{-9}}m $
B) $ \ge 2.8\times {{10}^{-9}}m $
C) $ \le 2.8\times {{10}^{-12}}m $
D) $ <2.8\times {{10}^{-10}}m $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Work function $ \phi $ of metal = 2.28 eV Wavelength of light $ X=500nm=500\times {{10}^{-9}}m $ $ \lambda =500nm=500\times {{10}^{-9}}m $ $ K{E_{\max }}=\frac{hc}{\lambda }-\phi $ $ K{E_{\max }}=\frac{6.6\times {{10}^{-34}}\times 3\times 10^{8}}{5\times {{10}^{-7}}\times 1.6\times {{10}^{-19}}}-2.28 $ $ K{E_{\max }}=2.48-2.20.2ev $ $ {\lambda_{\min }}=\frac{h}{p}=\frac{h}{\sqrt{2m{{( KE )}{\max }}}} $ $ {\lambda{\min }}=\frac{25}{9}\times {{10}^{-9}}=2.80\times {{10}^{-9}}nm $
$ \therefore \lambda \ge 2.8\times {{10}^{-9}}m $
BETA
