A) \[5:1\]
B) \[25:4\]
C) \[20:1\]
D) \[5:4\]
Correct Answer: B
Solution :
\[K.E.=\frac{1}{2}m{{v}^{2}}=\frac{{{(P)}^{2}}}{2m}\] \[P=\frac{h}{\lambda }\] \[K.E.=\frac{h}{2m{{\lambda }^{2}}}\] \[\therefore \frac{{{(K.E.)}_{A}}}{{{(K.E.)}_{B}}}=\frac{{{m}_{B}}\lambda _{B}^{2}}{{{m}_{A}}\lambda _{A}^{2}}=\frac{{{m}_{B}}}{4{{m}_{B}}}.\frac{{{(5)}^{2}}}{{{(1)}^{2}}}=\frac{25}{4}\]
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