A) Four times the initial energy
B) Thrice the initial energy
C) Equal to the initial energy
D) Twice the initial energy
Correct Answer: B
Solution :
\[\lambda =\frac{h}{\sqrt{2mE}}\Rightarrow \lambda \propto \frac{1}{\sqrt{E}}\Rightarrow \frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\sqrt{\frac{{{E}_{2}}}{{{E}_{1}}}}\] \[\Rightarrow \frac{{{10}^{-10}}}{0.5\times {{10}^{-10}}}=\sqrt{\frac{{{E}_{2}}}{{{E}_{1}}}}\Rightarrow {{E}_{2}}=4{{E}_{1}}\] Hence added energy \[={{E}_{2}}-{{E}_{1}}=3{{E}_{1}}\]
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