Two positively charged particles m1 and m2 have been accelerated across the same potential difference of 200 kV as shown below. [figure] [Given mass of m1 = 1 amu and m2 = 4 amu] The de Broglie wavelength of m1 will be x times of m2. The value of x is (nearest integer)
Answer
2
Solution
A particle of charge q accelerated through potential difference V gains kinetic energy K = qV.
= hp = h 2mK = h 2mqV
Both particles carry the same charge and are accelerated through the same V
1 m
1 2 = m2m1 = 41 = 2
x = 2
Answer: 2