Question

The kinetic energy of a particle is equal to the energy of a photon. The particle moves at $5\%$ of the speed of light. The ratio of the photon wavelength to the de Broglie wavelength of the particle is [No need to use relativistic formula for the particle.]

• A. 40
• B. 4
• C. 2
• D. 80

Answer 1

Answer: (A)

Let $m$ be the mass of particle.
$\frac{m{v}^2}{2}=\frac{hc}{{\lambda }_{\text{photon }}}$,
where symbols have their usual meanings.
$\frac{p^2}{2m}=\frac{hc}{{\lambda }_{\text{photon }}}$
and $p=\frac{h}{{\lambda }_{\text{particle }}}$
$\Rightarrow \frac{h^2}{2m{\lambda }_{\text{particle }}^2}=\frac{hc}{{\lambda }_{\text{photon }}}$
$\Rightarrow \frac{{\lambda }_{\text{photon }}}{{\lambda }_{\text{particle }}}=\frac{2mc}{h}\times {\lambda }_{\text{particle }}$
$=\frac{2mc}{h}\times \frac{h}{mv}$
$=\frac{2c}{0.05c}=40$