The energy of a photon is equal to the kinetic energy of a proton. If λ₁ is the de-Broglie wavelength of a proton, λ₂ is the wavelength associated with the photon and if the energy of the photon is E then (λ₁/λ₂) is proportional to (a) E⁴ (b) √(E) (c) E² (d) E

Answer: (b). Proton: λ₁ = (h)/(√(2mE)). Photon: λ₂ = hc/E. (λ₁)/(λ₂) = (E)/(c√(2mE)) = (√(E))/(c√(2m)) So λ₁/λ₂ ∝ √(E): the answer is (b).

The energy of a photon is equal to the kinetic energy of a proton. If… | MHT CET 2026 (13 May, Shift 2)

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Matter Waves: de Broglie Wavelength

The energy of a photon is equal to the kinetic energy of a proton. If $\lambda_1$ is the de-Broglie wavelength of a proton, $\lambda_2$ is the wavelength associated with the photon and if the energy of the photon is $E$ then $(\lambda_1/\lambda_2)$ is proportional to

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Answer: (b)

Proton: $\lambda_1 = \dfrac{h}{\sqrt{2mE}}$ . Photon: $\lambda_2 = \dfrac{hc}{E}$ .
$\dfrac{\lambda_1}{\lambda_2} = \dfrac{E}{c\sqrt{2mE}} = \dfrac{\sqrt{E}}{c\sqrt{2m}}$
So $\lambda_1/\lambda_2 \propto \sqrt{E}$ : the answer is (b).

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