A) 2V
B) 4 V
C) 6 V
D) 8 V
Correct Answer: A
Solution :
Key Idea: Kinetic energy of photoelectron is \[e{{V}_{0}}\] where \[{{V}_{0}}\] is stopping potential. From Einsteins photoelectric equation \[{{E}_{k}}=\frac{1}{2}mv_{\max }^{2}=hv-W\] where \[{{E}_{k}}\] is maximum kinetic energy of electron, v is frequency and W is work function. \[\therefore \] \[\frac{1}{2}mv_{\max }^{2}=4eV-2eV=2eV\] but \[\frac{1}{2}mv_{\max }^{2}=e{{V}_{0}}\] where \[{{V}_{0}}\] is stopping potential. Thus, \[e{{V}_{0}}=2eV\] \[\Rightarrow \] \[{{V}_{0}}=2V\]
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