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 \[\lambda =\frac{h}{\sqrt{2mE}}\] \[\therefore \] \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\sqrt{\frac{{{E}_{2}}}{{{E}_{1}}}}\Rightarrow \frac{1\times {{10}^{-9}}}{0.5\times {{10}^{-9}}}=\sqrt{\frac{{{E}_{2}}}{{{E}_{1}}}}\] \[\Rightarrow \] \[2=\sqrt{\frac{{{E}_{2}}}{{{E}_{1}}}}\Rightarrow \frac{{{E}_{2}}}{E1}=4\] \[\therefore \] \[{{E}_{2}}=4{{E}_{1}}\] \[\therefore \]Energy to be added \[={{E}_{2}}=-{{E}_{1}}\] \[=4{{E}_{1}}-{{E}_{1}}=3{{E}_{1}}\]
You need to login to perform this action.
You will be redirected in
3 sec
