A) 10 m/sec
B) \[1\times {{10}^{3}}\]m/sec
C) \[1\times {{10}^{4}}m/\sec \]m/sec
D) \[1\times {{10}^{6}}m/\sec \]m/sec
Correct Answer: D
Solution :
\[E={{W}_{0}}+{{K}_{\max }};\ E=\frac{12375}{3000}=4.125\ eV\] \[\Rightarrow {{K}_{\max }}=E-{{W}_{0}}=4.125\ eV-1\ eV=3.125\ eV\] \[\Rightarrow \frac{1}{2}mv_{\max }^{2}=3.125\times 1.6\times {{10}^{-19}}\ J\] \[\Rightarrow {{v}_{\max }}=\sqrt{\frac{2\times 3.125\times 1.6\times {{10}^{-19}}}{9.1\times {{10}^{-31}}}}\]\[=1\times {{10}^{6}}m/s\]
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