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