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 :
\[\lambda =\frac{\hbar }{\sqrt{2mE}}\];\[\frac{\lambda '}{\lambda }=\sqrt{\frac{E}{E'}}\]Þ\[\frac{E}{E}={{\left( \frac{0.5}{1} \right)}^{2}}\]Þ\[E'=\frac{E}{0.25}=4E\] The energy should be added to decrease wavelength. \[=E'-E=3E\]You need to login to perform this action.
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