A) \[\text{3V}\]
B) 4 V
C) \[\text{2 V}\]
D) \[~\text{2}.\text{5 V}\]
Correct Answer: C
Solution :
In photoelectric effect, energy is conserved. In photoelectric effect, Einstein's equation is given by Photon energy = kinetic energy of electron + work function i.e. \[hv=e{{V}_{s}}+h{{v}_{0}}\] where, \[{{V}_{s}}\] is the stopping potential and \[{{v}_{0}}\] is the threshold frequency \[\therefore \] \[{{V}_{s}}=\frac{h}{e}(v-{{v}_{0}})\] ? (i) Given, \[{{v}_{0}}=3.3\times {{10}^{14}}Hz,v=8.2\times {{10}^{14}}Hz\] \[h=6.6\times {{10}^{-34}}J-s,e=1.6\times {{10}^{-19}}C\] Substituting these values in Eq. (i), we get \[{{V}_{s}}=\frac{6.6\times {{10}^{-34}}({{8.210}^{14}}-3.3\times {{10}^{14}})}{1.6\times {{10}^{-19}}}\] \[{{V}_{s}}=\frac{6.6\times 4.9}{1.6}\times {{10}^{-1}}=2V\] \[\Rightarrow \] \[{{V}_{s}}=2V\]
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