A) 2
B) 3
C) 4
D) 6
Correct Answer: C
Solution :
Specific charge of electron, \[\frac{e}{m}=1.8\times {{10}^{11}}C\,k{{g}^{-1}}\] Maximum kinetic energy of photoelectron \[\frac{1}{2}mv_{\max }^{2}=e{{V}_{s}}\] where\[{{V}_{s}}\]is the stopping potential. \[\Rightarrow \] \[{{V}_{s}}=\frac{mv_{\max }^{2}}{2e}=\frac{v_{\max }^{2}}{2(e/m)}\] \[=\frac{{{(1.2\times {{10}^{6}})}^{2}}}{2\times 1.8\times {{10}^{11}}}\] \[=0.4\times 10=4\,V\]You need to login to perform this action.
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