A) 1:2
B) 1:1
C) \[1:\sqrt{2}\]
D) 4 : 1
Correct Answer: A
Solution :
de - Broglie wavelength, \[\lambda =\frac{h}{\sqrt{2m{{E}_{k}}}}\] \[\frac{{{\lambda }_{\alpha }}}{{{\lambda }_{p}}}=\sqrt{\frac{{{m}_{p}}}{{{m}_{\alpha }}}}\] \[=\sqrt{\frac{1}{4}}=\frac{1}{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
