A) four times the initial energy
B) equal to the initial energy
C) twice the initial energy
D) thrice the initial energy
Correct Answer: D
Solution :
de-Broglie wavelength \[\lambda =\frac{h}{\sqrt{2mE}}\] \[\therefore \] \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\sqrt{\frac{{{E}_{2}}}{{{E}_{1}}}}\] \[\Rightarrow \] \[\frac{1\times {{10}^{-9}}}{0.5\times {{10}^{-9}}}=\sqrt{\frac{{{E}_{2}}}{{{E}_{1}}}}\] \[\Rightarrow \] \[2=\sqrt{\frac{{{E}_{2}}}{{{E}_{1}}}}\] \[\Rightarrow \] \[\frac{{{E}_{2}}}{{{E}_{1}}}=4\] \[\therefore \] \[{{E}_{2}}=4{{E}_{1}}\] \[\therefore \] Energy to be added \[={{E}_{2}}-{{E}_{1}}\] \[=4{{E}_{1}}-{{E}_{1}}=3{{E}_{1}}\]
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