| (Given : Planck's constant\[\left( h \right)=6.63\times {{10}^{34}}Js,\]electron charge \[e=1.6\times {{10}^{19}}C\]) |
[JEE Main Held on 12-4-2019 Morning]
A) 1.95 eV
B) 1.82 eV
C) 1.66 eV
D) 2.12 eV
Correct Answer: C
Solution :
\[h\text{v=}\phi \,\text{+}\,\text{e}{{\text{v}}_{0}}\] \[{{\text{v}}_{0}}=\frac{h\text{v}}{e}-\frac{\phi }{e}\]\[{{\text{v}}_{0}}\]is zero for \[\text{v}=4\times {{10}^{14}}Hz\] \[0=\frac{h\text{v}}{e}-\frac{\phi }{e}\Rightarrow \phi =h\text{v}\] \[=\frac{6.63\times {{10}^{-34}}\times 4\times {{10}^{14}}}{1.6\times {{10}^{-19}}}=1.66e\text{v}.\]
You need to login to perform this action.
You will be redirected in
3 sec
