A) 2 V
B) 3 V
C) 5 V
D) 1 V
Correct Answer: A
Solution :
\[{{V}_{0}}=\frac{E-V}{e}\] \[=\frac{h(v-{{v}_{0}})}{e}\] \[=\frac{6.62\times {{10}^{-34}}(8.2\times {{10}^{14}})(8.2\times {{10}^{14}})-3.3\times {{10}^{14}})}{1.6\times {{10}^{-19}}}\]\[=\frac{6.62\times {{10}^{-34}}}{1.6}\times 4.9\times {{10}^{33}}\] \[=\frac{6.62\times 4.9\times {{10}^{-1}}}{1.6}\] \[{{V}_{0}}=2\,volt\]
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