Question

The wavelength of a photon is equal to the de Broglie wavelength of a particle moving with a speed of 75% of the speed of light in vacuum. The ratio of the kinetic energy of the particle and the energy of the photon is

• A. 1 : 1
• B. 3 : 8
• C. 1 : 3
• D. 2 : 3

Hint:

Use de Broglie relation $\lambda = \dfrac{h}{p}$ and compare particle kinetic energy with photon energy $\dfrac{hc}{\lambda}$.

Answer 1

The Correct Option is B

Let $\lambda$ be the common wavelength. For the photon: $E = \dfrac{hc}{\lambda}$
For the particle: de Broglie wavelength $\lambda = \dfrac{h}{p} \Rightarrow p = \dfrac{h}{\lambda}$
Kinetic energy of the particle: $K = \dfrac{p^2}{2m} = \dfrac{h^2}{2m\lambda^2}$
So, ratio: $\dfrac{K}{E} = \dfrac{h^2}{2m\lambda^2} \cdot \dfrac{\lambda}{hc} = \dfrac{h}{2mc\lambda}$
Now substitute $v = 0.75c$, so $p = mv = 0.75mc \Rightarrow \lambda = \dfrac{h}{0.75mc}$
Put $\lambda$ back in expression: $\dfrac{K}{E} = \dfrac{(0.75mc)^2}{2m} \cdot \dfrac{1}{hc/\lambda} = \dfrac{0.5625mc^2}{2} \cdot \dfrac{\lambda}{hc}$
Simplify and substitute to get: $\dfrac{3}{8}$ ⇒ Ratio = 3 : 8