Kinetic energy of a proton is equal to energy ' E ' of a photon. Let ' λ₁ ' be the de-Broglie wavelength of proton and ' λ₂ ' is the wavelength of photon. If (λ₁)/(λ₂) ∝ Eⁿ, then the value of ' n ' is (a) 1/2 (b) 1/4 (c) 2 (d) 4

Answer: (a). Proton: λ₁ = (h)/(√(2mE)). Photon: λ₂ = hc/E. (λ₁)/(λ₂) = (√(E))/(c√(2m)) ∝ E1/2 So n = 1/2: the answer is (a).

Kinetic energy of a proton is equal to energy ' E ' of a photon. Let… | MHT CET 2021 (22 Sep, Shift 1)

Q104MHT CET 2021 · 22 SEP · SHIFT 1Medium

Matter Waves: de Broglie Wavelength

Kinetic energy of a proton is equal to energy ' $E$ ' of a photon. Let ' $\lambda_1$ ' be the de-Broglie wavelength of proton and ' $\lambda_2$ ' is the wavelength of photon. If $\frac{\lambda_1}{\lambda_2} \propto E^{n}$ , then the value of ' $n$ ' is

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Answer: (a)

Proton: $\lambda_1 = \dfrac{h}{\sqrt{2mE}}$ . Photon: $\lambda_2 = \dfrac{hc}{E}$ .
$\dfrac{\lambda_1}{\lambda_2} = \dfrac{\sqrt{E}}{c\sqrt{2m}} \propto E^{1/2}$
So $n = \frac{1}{2}$ : the answer is (a).

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