1. √(E/2m) 2. c√(2mE) 3. c√(2m/E) 4. 1/c√(E/2m)

Answer: 3. c√(2m/E) Key idea: λ=h/p , with different momentum-energy relations for photon and electron. Photon: E=pc⇒λ_ph=hc/E . Electron: E=p²/2m⇒λ_e=h/√(2mE) . fracλ_phλ_e=hc/E·√(2mE)/h=c√(2m/E)

Topic: Matter waves, de Broglie relation· Dual Nature of Matter and Radiation

A photon and an electron (mass m) have the same energy E. The ratio ≤ft(λ_photon/λ_electron) of their de Broglie wavelengths is: (c is the speed of light)

1. $\sqrt{E/2m}$

2. $c\sqrt{2mE}$

3. $c\sqrt{\frac{2m}{E}}$

4. $\frac{1}{c}\sqrt{E/2m}$

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Answer: 3. $c\sqrt{\frac{2m}{E}}$

Key idea: $\lambda=h/p$, with different momentum-energy relations for photon and electron.

Photon: $E=pc\Rightarrow\lambda_{ph}=\dfrac{hc}{E}$.
Electron: $E=\dfrac{p^2}{2m}\Rightarrow\lambda_e=\dfrac{h}{\sqrt{2mE}}$.

$$\frac{\lambda_{ph}}{\lambda_e}=\frac{hc}{E}\cdot\frac{\sqrt{2mE}}{h}=c\sqrt{\frac{2m}{E}}$$

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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A photon and an electron (mass m) have the same energy E. The ratio ≤ft(λ_photon/λ_electron) of their de Broglie wavelengths is: (c is the speed of light)

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