A) 2 V
B) 3 V
C) 5 V
D) 1 V
Correct Answer: A
Solution :
[a] \[K.E.=hv-h{{v}_{th}}=e{{V}_{0}}\,\,({{V}_{0}}=cut\,of\,voltage)\] \[\Rightarrow \]\[{{V}_{0}}=\frac{h}{e}(8.2\times {{10}^{14}}-3.3\times {{10}^{14}})\] \[=\frac{6.6\times {{10}^{-34}}\times 4.9\times {{10}^{14}}}{1.6\times {{10}^{-19}}}\approx 2\,V\].You need to login to perform this action.
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