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

Answer: 1. 1/c(E/2m)^(1/2) Key idea: use λ=h/p for the electron and λ=hc/E for the photon. Electron: p=√(2mE) , so λ_e=h/√(2mE) . Photon: λ_ph=hc/E . dfracλ_eλ_ph=E/c√(2mE)=1/c√(E/2m) .

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

An electron of mass m and a photon have same energy E. The ratio of de-Broglie wavelengths associated with them is : (c being velocity of light)

1. \(\frac{1}{c}\left(\frac{E}{2m}\right)^{\frac{1}{2}}\)

2. \(\left(\frac{E}{2m}\right)^{\frac{1}{2}}\)

3. \(c(2mE)^{\frac{1}{2}}\)

4. \(\frac{1}{c}\left(\frac{2m}{E}\right)^{\frac{1}{2}}\)

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Answer: 1. \(\frac{1}{c}\left(\frac{E}{2m}\right)^{\frac{1}{2}}\)

Key idea: use \(\lambda=h/p\) for the electron and \(\lambda=hc/E\) for the photon.

Electron: \(p=\sqrt{2mE}\), so \(\lambda_e=\dfrac{h}{\sqrt{2mE}}\).
Photon: \(\lambda_{ph}=\dfrac{hc}{E}\).
\(\dfrac{\lambda_e}{\lambda_{ph}}=\dfrac{E}{c\sqrt{2mE}}=\dfrac1c\sqrt{\dfrac{E}{2m}}\).

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

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An electron of mass m and a photon have same energy E. The ratio of de-Broglie wavelengths associated with them is : (c being velocity of light)

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