A) \[\text{4 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{15}}}\,{{\text{H}}_{\text{Z}}}\]
B) \[\text{5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{15}}}\,{{\text{H}}_{\text{Z}}}\]
C) \[\text{1}\text{.6 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{15}}}\,{{\text{H}}_{\text{Z}}}\]
D) \[\text{2}\text{.5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{15}}}\,{{\text{H}}_{\text{Z}}}\]
Correct Answer: C
Solution :
Energy released from emission of electron \[E=(-3.4)-(-13.6)\] \[=10.2\,eV\] From photo electric equation. Work function \[\phi =E-eV=hv\] \[v=\frac{E-eV}{h}\] \[=\frac{(10.2-3.57)e}{6.67\times {{10}^{-34}}}\] \[v=\frac{6.63\times 1.6\times {{10}^{-19}}}{6.67\times {{10}^{-34}}}\] \[=1.6\times {{10}^{15}}Hz\]You need to login to perform this action.
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