An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vaccuum)
Show Hint
- \( {1 \over c}{\left( {{{2E} \over m}} \right)^{{1 \over 2}}} \)
- \( {1 \over c}{\left( {{E \over {2m}}} \right)^{{1 \over 2}}} \)
- \( {\left( {{E \over {2m}}} \right)^{{1 \over 2}}} \)
- \( c{\left( {2mE} \right)^{{1 \over 2}}} \)
The Correct Option is B
Solution and Explanation
Step 2: \(\dfrac{\lambda_e}{\lambda_{ph}}=\dfrac{h}{\sqrt{2mE}}\cdot\dfrac{E}{hc}=\dfrac{E}{c\sqrt{2mE}}\).
Step 3: \(=\dfrac1c\sqrt{\dfrac{E}{2m}}=\dfrac1c\left(\dfrac{E}{2m}\right)^{1/2}\).
Answer: (B) \(\dfrac1c\left(\dfrac{E}{2m}\right)^{1/2}\)
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