A) \[1\,\,V\]
B) \[3.75\,\,V\]
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\[c=velocity\,\,of\,\,light=3.0\times {{10}^{8}}m/s\] \[\lambda =wavelength=3000{\AA}=3000\times {{10}^{-10}}m\] \[{{W}_{0}}=\]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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