A) \[1\]
B) \[3\]
C) \[9\]
D) \[6\]
Correct Answer: C
Solution :
: As \[\frac{1}{2}m\upsilon _{\max }^{2}=e{{V}_{S}}\] where \[{{\upsilon }_{\max }}\]is the maximum velocity of the electron and \[{{V}_{S}}\] is the stopping potential. \[{{V}_{S}}=\frac{1}{2}\frac{m}{e}\upsilon _{\max }^{2}\] Here, \[\frac{e}{m}=1.8\times {{10}^{11}}C\,k{{g}^{-1}},\,{{\upsilon }_{\max }}=1.8\times {{10}^{6}}m{{s}^{-1}}\] \[\therefore \] \[{{V}_{S}}=\frac{1}{2}\times \frac{1}{1.8\times {{10}^{11}}}\times {{(1.8\times {{10}^{6}})}^{2}}=9V\]
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