A) 4.9 V
B) 3.0 V
C) 2.0 V
D) 1.0 V
Correct Answer: C
Solution :
Einsteins photoelectric equation is \[KE=hv-\text{o}{{\text{ }\!\!|\!\!\text{ }}_{0}}\] \[e{{V}_{s}}=h(v-{{v}_{0}})\] Where \[{{V}_{s}}\] is threshold or cut-off voltage. \[{{V}_{s}}=\frac{h}{e}(v-{{v}_{0}})\] Here,\[h=6.6\times {{10}^{-34}}J-s,e=1.6\times {{10}^{-19}}C,\] \[v=8.2\times {{10}^{14}}\,Hz,\,{{v}_{0}}=3.3\times {{10}^{14}}\,Hz\] Substituting the values in the above relation, we have \[{{V}_{s}}=\frac{6.6\times {{10}^{-34}}}{1.6\times {{10}^{-19}}}(8.2-3.3)\times {{10}^{14}}\] \[=2\,V\]
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