A) Electron
B) Alpha particle
C) Proton
D) Neutron
Correct Answer: B
Solution :
de-Broglie relation is given by \[\lambda =\frac{h}{mv}\] This can also be written as \[\lambda =\frac{h}{\sqrt{2mv}}\] So, \[\lambda \propto \frac{1}{\sqrt{m}}\] Because, an alpha particle having the highest mass among the others, so alpha particle have the least wavelength.
You need to login to perform this action.
You will be redirected in
3 sec
