A) \[2\times {{10}^{6}}\,m/s\]
B) \[2\times {{10}^{7}}\,m/s\]
C) \[8\times {{10}^{5}}\,m/s\]
D) \[8\times {{10}^{6}}\,m/s\]
Correct Answer: D
Solution :
From \[E={{W}_{0}}+\frac{1}{2}mv_{\max }^{2}\] Þ \[2h{{\nu }_{0}}=h{{\nu }_{0}}+\frac{1}{2}mv_{1}^{2}\] Þ \[h{{\nu }_{0}}=\frac{1}{2}mv_{1}^{2}\] .....(i) and \[5h{{\nu }_{0}}=h{{\nu }_{0}}+\frac{1}{2}mv_{2}^{2}\] Þ \[4h{{\nu }_{0}}=\frac{1}{2}mv_{2}^{2}\] .....(ii) Dividing equation (ii) by (i) \[{{\left( \frac{{{v}_{2}}}{{{v}_{1}}} \right)}^{2}}=\frac{4}{1}\] Þ \[{{v}_{2}}=2{{v}_{1}}=2\times 4\times {{10}^{6}}=8\times {{10}^{6}}m/s\]You need to login to perform this action.
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