A) 1V
B) \[\text{3}\text{.75}\,\text{V}\]
C) 3.2 V
D) 0.75 V
Correct Answer: B
Solution :
Key Idea: Use following formula to find the value of stopping potential. \[{{W}_{0}}=e{{V}^{0}}=\frac{hc}{\lambda }\] where\[c=\]velocity of light \[=3.0\times {{10}^{8}}\,m/s\] \[\,\lambda =\]wavelength \[=3000\overset{\text{o}}{\mathop{\text{A}}}\,=3000\times {{10}^{-10}}m\] \[{{W}_{0}}=\]work function \[=1\,eV\] \[{{W}_{0}}=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{3000\times {{10}^{-10}}}\] \[=6.6\times {{10}^{-19}}J\] \[=0.6\times {{10}^{-18}}J\] \[\therefore \] \[{{W}_{0}}=e{{V}^{0}}=0.6\times {{10}^{-18}}J\] \[\therefore \] \[{{V}^{0}}=\frac{0.6\times {{10}^{-18}}}{e}\] \[=\frac{0.6\times {{10}^{-18}}}{1.6\times {{10}^{-19}}}=3.75\,V.\]
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