A) 1:4
B) 1:2
C) 1:1
D) 2:1
Correct Answer: B
Solution :
de-Broglie wavelength \[\lambda =\frac{h}{mv}\] Here, \[{{\lambda }_{e}}=\frac{h}{{{m}_{e}}\frac{c}{2}}\]and \[{{\lambda }_{p}}=\frac{h}{{{m}_{p}}c}\] Given, \[{{\lambda }_{e}}={{\lambda }_{p}}\] So, \[\frac{h}{{{m}_{e}}\frac{c}{2}}=\frac{h}{{{m}_{p}}c}\] \[\frac{{{m}_{e}}}{{{m}_{p}}}=2\] Ratio of KE \[\frac{{{K}_{e}}}{{{K}_{p}}}=\frac{\frac{1}{2}{{m}_{e}}v_{e}^{2}}{\frac{1}{2}{{m}_{p}}v_{p}^{2}}\] \[\frac{{{K}_{e}}}{{{K}_{p}}}=\frac{1}{2}\]You need to login to perform this action.
You will be redirected in
3 sec