Question:
An electron of mass \(m\) and a photon have the same energy E. The ratio of de-Broglie wavelengths associated with them \(\left(\frac{\lambda _{e}}{\lambda _{p}}\right)\) is {\(c\) = velocity of light}
An electron of mass \(m\) and a photon have the same energy E. The ratio of de-Broglie wavelengths associated with them \(\left(\frac{\lambda _{e}}{\lambda _{p}}\right)\) is {\(c\) = velocity of light}
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Use the de Broglie wavelength lambda = h/p for the electron with momentum sqrt(2mE) and for the photon with momentum E/c.
Updated On: Nov 12, 2026
- \(\sqrt{\frac{E}{2m}}\)
- \(c\sqrt{2mE}\)
- \(\frac{1}{c}\sqrt{\frac{2m}{E}}\)
- \(\frac{1}{c}\sqrt{\frac{E}{2m}}\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
The de Broglie wavelength is \(\lambda = \dfrac{h}{p}\), where p is the momentum.
Step 2: Electron
With energy E, \(p = \sqrt{2mE}\), so \(\lambda_e = \dfrac{h}{\sqrt{2mE}}\).
Step 3: Photon
For a photon of energy E, \(p = \dfrac{E}{c}\), so \(\lambda_p = \dfrac{hc}{E}\).
Step 4: Ratio
\(\dfrac{\lambda_e}{\lambda_p} = \dfrac{h}{\sqrt{2mE}}\cdot\dfrac{E}{hc} = \dfrac{1}{c}\sqrt{\dfrac{E}{2m}}\), which is option (D).
Final Answer:
The ratio is (1/c) times the square root of E/2m. \[ \boxed{\text{(D) }\dfrac{1}{c}\sqrt{\dfrac{E}{2m}}} \]
The de Broglie wavelength is \(\lambda = \dfrac{h}{p}\), where p is the momentum.
Step 2: Electron
With energy E, \(p = \sqrt{2mE}\), so \(\lambda_e = \dfrac{h}{\sqrt{2mE}}\).
Step 3: Photon
For a photon of energy E, \(p = \dfrac{E}{c}\), so \(\lambda_p = \dfrac{hc}{E}\).
Step 4: Ratio
\(\dfrac{\lambda_e}{\lambda_p} = \dfrac{h}{\sqrt{2mE}}\cdot\dfrac{E}{hc} = \dfrac{1}{c}\sqrt{\dfrac{E}{2m}}\), which is option (D).
Final Answer:
The ratio is (1/c) times the square root of E/2m. \[ \boxed{\text{(D) }\dfrac{1}{c}\sqrt{\dfrac{E}{2m}}} \]
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