A proton, an electron and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as :
- p > e >
- < p < e
- e > > p
- p < e <
Answer
(2) < p < e
Solution
= hp = h 2mK
Since K is the same for all three, 1 m.
As me < mp < m , we get < p < e
Answer: (2)