A) 1.1 V
B) 1.8 V
C) 1.26 V
D) 0.8 V
Correct Answer: A
Solution :
Using the relation, \[e{{V}_{s}}=\frac{hc}{\lambda }-\text{o }\!\!|\!\!\text{ }\] \[{{V}_{s}}=\]stopping potential \[\text{o }\!\!|\!\!\text{ =}\]work function of the substance \[{{V}_{s}}=\frac{hc}{e\lambda }-\frac{\text{o }\!\!|\!\!\text{ }}{e}\] \[=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{1.6\times {{10}^{-19}}\times 4000\times {{10}^{-10}}}-\frac{2\times 1.6\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}\] \[=3.1-2\] \[=1.1\,V\]
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