An electron and an alpha particle are accelerated by the same potential difference. Let e and denote the de-Broglie wavelengths of the electron and the alpha particle, respectively, then:
- e >
- e = 4
- e =
- e <
Answer
(1) e >
Solution
= h 2mqV
e = m q me qe
m 4mp 7344 me and q = 2e
e = 7344 2 121
e >
Answer: (1)