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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 a particle is .......
(No need to use relativistic formula for the particle)

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answer is 40.

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Detailed Solution

$\begin{array}{l} \frac{1}{2} m v^{2} = \frac{h c}{λ_{p h}} \Rightarrow m v^{2} = \frac{2 h c}{λ_{p h}} \Rightarrow λ_{p h} = \frac{2 h c}{m v^{2}} \\ D e ‐ Broglie wavelength, λ_{p} = \frac{h}{p} = \frac{h}{m v} \\ ∴ \frac{λ_{p h}}{λ_{p}} = \frac{2 h c / m v^{2}}{h / m v} = \frac{2 c}{v} \\ ∵ v = c \times \frac{5}{100} = \frac{c}{20} \\ \Rightarrow \frac{c}{v} = 20 \\ ∴ \frac{λ_{p h}}{λ_{p}} = 2 (20) = 40 \end{array}$

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