Question:
A photon and an electron have equal energy ' \(E\) '. The ratio of wavelength of photon to wavelength of electron is proportional to
A photon and an electron have equal energy ' \(E\) '. The ratio of wavelength of photon to wavelength of electron is proportional to
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Write each wavelength in terms of the common energy E and compare how they depend on E.
Updated On: Nov 13, 2026
\(\frac{1}{\mathrm{E}}\)
\(\sqrt{E}\)
E
\(\frac{1}{\sqrt{\mathrm{E}}}\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
For a photon, \(E=\dfrac{hc}{\lambda_{p}}\). For an electron, \(E=\dfrac{p^{2}}{2m}\).
Step 2: Wavelengths
\(\lambda_{p}=\dfrac{hc}{E}\) and \(\lambda_{e}=\dfrac{h}{\sqrt{2mE}}\).
Step 3: Ratio
\(\dfrac{\lambda_{p}}{\lambda_{e}}=\dfrac{hc}{E}\cdot\dfrac{\sqrt{2mE}}{h}=c\sqrt{2m}\,\dfrac{1}{\sqrt{E}}\).
Step 4: Result
So the ratio is proportional to \(\dfrac{1}{\sqrt{E}}\).
Final Answer:
The ratio of photon wavelength to electron wavelength is proportional to \(1/\sqrt{E}\). \[ \boxed{\text{(D) }\dfrac{1}{\sqrt{\mathrm{E}}}} \]
For a photon, \(E=\dfrac{hc}{\lambda_{p}}\). For an electron, \(E=\dfrac{p^{2}}{2m}\).
Step 2: Wavelengths
\(\lambda_{p}=\dfrac{hc}{E}\) and \(\lambda_{e}=\dfrac{h}{\sqrt{2mE}}\).
Step 3: Ratio
\(\dfrac{\lambda_{p}}{\lambda_{e}}=\dfrac{hc}{E}\cdot\dfrac{\sqrt{2mE}}{h}=c\sqrt{2m}\,\dfrac{1}{\sqrt{E}}\).
Step 4: Result
So the ratio is proportional to \(\dfrac{1}{\sqrt{E}}\).
Final Answer:
The ratio of photon wavelength to electron wavelength is proportional to \(1/\sqrt{E}\). \[ \boxed{\text{(D) }\dfrac{1}{\sqrt{\mathrm{E}}}} \]
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