Question

Plot a graph showing variation of de-Broglie wavelength λ versus `1/sqrt V`, where V is accelerating potential for a particle of mass m and charge q. Obtain the slope of this graph.

Answer 1

The de-Broglie wavelength of a particle accelerated through potential V is:

λ = `h/p`

Since the kinetic energy gained is:

`1/2 mv^2 = qV`

mv = `sqrt(2 mq V)`

Substituting into the de-Broglie expression:

λ = `h/(sqrt(2 mqV))`

Rewrite as:

λ = `h/(sqrt(2 mq)) * 1/sqrtV`

This shows λ is directly proportional to `1/sqrtV`.