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.]

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Detailed Solution

Let m be the mass of particle.

$\frac{m v^{2}}{2} = \frac{h c}{λ_{photon}}, where symbols have their usual meanings.$

$\begin{array}{l} \frac{p^{2}}{2 m} = \frac{h c}{λ_{photon}} \\ and p = \frac{h}{λ_{particle}} \Rightarrow \frac{h^{2}}{2 m λ_{particle}^{2}} = \frac{h c}{λ_{photon}} \\ \Rightarrow \frac{λ_{photon}}{λ_{particle}} = \frac{2 m c}{h} \times λ_{particle} = \frac{2 m c}{h} \times \frac{h}{m v} \\ = \frac{2 c}{0.05 c} = 40 \end{array}$