A) \[3\times {{10}^{5}}\]
B) \[{{K}_{A}}={{K}_{B}}/2\]
C) \[4\times {{10}^{4}}\]
D) \[3.65\times {{10}^{7}}\]
Correct Answer: B
Solution :
By using\[E={{W}_{0}}+\frac{1}{2}mv_{\max }^{2}\]; where \[E=\frac{12375}{4558}=2.71\ eV\] \[\Rightarrow 2.71\ eV=2.5\ eV+\frac{1}{2}\times 9.1\times {{10}^{-31}}\times v_{\max }^{2}\] \[\Rightarrow 0.21\times 1.6\times {{10}^{-19}}=\frac{1}{2}\times 9.1\times {{10}^{-31}}\times v_{\max }^{2}\] \[\Rightarrow {{v}_{\max }}=2.65\times {{10}^{5}}\ m/s\]
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