Dual Nature of Radiation and Matter - Result Question 6
6. An electron of mass $m$ and a photon have same energy E. The ratio of de-Broglie wavelengths associated with them is :
[2016]
(a) $\frac{1}{c}(\frac{E}{2 m})^{\frac{1}{2}}$
(b) $(\frac{E}{2 m})^{\frac{1}{2}}$
(c) $c(2 mE)^{\frac{1}{2}}$
(d) $\frac{1}{xc}(\frac{2 m}{E})^{\frac{1}{2}}$
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Answer:
Correct Answer: 6. (a)
Solution:
- (a) For electron De-Broglie wavelength,
$\lambda_e=\frac{h}{\sqrt{2 mE}}$
For photon $E=pc$
$\Rightarrow$ De-Broglie wavelength, $\lambda _{Ph}=\frac{hc}{E}$
$\therefore \quad \frac{\lambda_e}{\lambda _{Ph}}=\frac{h}{\sqrt{2 mE}} \times \frac{E}{hc}=(\frac{E}{2 m})^{1 / 2} \frac{1}{c}$
