Em Waves Question 82
Question: Given that a photon of light of wavelength 10,000 $ \overset{\text{o}}{\mathop{\text{A}}}, $ has an energy equal to 1.23 eV. When light of wavelength 5000 $ \overset{\text{o}}{\mathop{\text{A}}}, $ and intensity $ I _{0} $ falls on a photoelectric cell, the saturation current is $ 0.40\times {{10}^{-6}} $ $ \overset{\text{o}}{\mathop{\text{A}}}, $ and the stopping potential is 1.36 V; then the work function is
Options:
A) $ 0.43,eV $
B) $ 1.10,eV $
C) $ 1.36,eV $
D) $ 2.47,eV $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Since $ E\propto \frac{1}{\lambda } $ , so energy corresponding to 5000 $ \overset{\text{o}}{\mathop{\text{A}}}, $ is $ E=2.46eV $ Now, $ hv-W=eV _{s} $ Or $ 2.46eV-W=1.36eV $
$ W=(2.46-1.36)eV=1.1eV $
BETA
