MHT CET · Physics · Dual Nature of Matter
An electron of mass ' \(m\) ' and a photon have same energy \(E\). The ratio of the de-Broglie wavelengths associated with them is (c = velocity of light in air)
- A \(\left[\frac{\mathrm{E}}{2 \mathrm{M}}\right]^{\frac{1}{2}}\)
- B \(\frac{1}{\mathrm{c}}\left[\frac{\mathrm{E}}{2 \mathrm{~m}}\right]^{\frac{1}{2}}\)
- C \(\mathrm{c}(2 \mathrm{mE})^{\frac{1}{2}}\)
- D \(\frac{1}{\mathrm{c}}\left[\frac{2 \mathrm{~m}}{\mathrm{E}}\right]^{\frac{1}{2}}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{\mathrm{c}}\left[\frac{\mathrm{E}}{2 \mathrm{~m}}\right]^{\frac{1}{2}}\)
Step-by-step Solution
Detailed explanation
De-broglie wavelength
For an electron, \(\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}\)
For photon, \(\mathrm{E}=\mathrm{pc} \Rightarrow \lambda_{\text {Photon }}=\frac{\mathrm{hc}}{\mathrm{E}}\)
\(\Rightarrow \frac{\lambda_{\mathrm{e}}}{\lambda_{\text {photon }}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}} \times \frac{\mathrm{E}}{\mathrm{hc}}=\left(\frac{\mathrm{E}}{2 \mathrm{~m}}\right)^{1 / 2} \frac{1}{\mathrm{c}}\)
For an electron, \(\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}\)
For photon, \(\mathrm{E}=\mathrm{pc} \Rightarrow \lambda_{\text {Photon }}=\frac{\mathrm{hc}}{\mathrm{E}}\)
\(\Rightarrow \frac{\lambda_{\mathrm{e}}}{\lambda_{\text {photon }}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}} \times \frac{\mathrm{E}}{\mathrm{hc}}=\left(\frac{\mathrm{E}}{2 \mathrm{~m}}\right)^{1 / 2} \frac{1}{\mathrm{c}}\)
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