A) 30V
B) 45V
C) 59V
D) Information is insufficient
Correct Answer: B
Solution :
Given, \[{{v}_{\max }}=4\times {{10}^{8}}cm/s=4\times {{10}^{6}}m/s\] \[\therefore \] \[{{K}_{\max }}=\frac{1}{2}mc_{\max }^{2}=\frac{1}{2}\times 9\times {{10}^{-31}}\times {{(4\times {{10}^{6}})}^{2}}\] \[=7.2\times {{10}^{-18}}J=45eV\] Hence, stopping potential \[|{{V}_{0}}|=\frac{{{K}_{\max }}}{e}=\frac{45\,eV}{e}\] \[=45V\]
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