The ratio of de Broglie wavelength of a deuteron with kinetic energy E to that of an alpha particle with kinetic energy 2E, is n : 1. The value of n is . (Assume mass of proton = mass of neutron)
Answer
2
Solution
= h 2mK
Deuteron: mass = 2m, K = E; alpha particle: mass = 4m, K = 2E (m = mass of a proton = mass of a neutron)
d = m K md Kd = 4m 2E2m E = 4 = 2
n = 2
Answer: 2