Dual Nature of Radiation and Matter - Result Question 9
9. The wavelength $\lambda_e$ of an electron and $\lambda_p$ of a photon are of same energy $E$ are related by [2013]
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======= ####9. The wavelength $\lambda_e$ of an electron and $\lambda_p$ of a photon are of same energy $E$ are related by [2013]
c3eec34ec6b1fad69db54a20ad4b2dca40c2aa54 (a) $\lambda_p \propto \lambda_e$
(b) $\lambda_p \propto \sqrt{\lambda_e}$
(c) $\lambda_p \propto \frac{1}{\sqrt{\lambda_e}}$
(d) $\lambda_p \propto \lambda_e^{2}$
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Answer:
Correct Answer: 9. (d)
Solution:
- (d) $As P=\frac{E}{c}$
$\lambda_p=\frac{hc}{E}$
$\lambda_e=\frac{h}{\sqrt{2 mE}} \Rightarrow \lambda_e^{2}=\frac{h^{2}}{2 mE}$
From equations (i) and (ii)
$\lambda_p \propto \lambda_e^{2}$
de-Broglie wavelength, $\lambda=\frac{h}{p}$
Here, $h=$ plank’s constant
$p=$ momentum
Momentum, $p=\sqrt{2 m k}$
$\therefore \lambda=\frac{h}{\sqrt{2 m E}} \quad$ (Here, $E=$ kinetic energy)
$\therefore E=\frac{h^{2}}{2 m \lambda^{2}} \quad \Rightarrow \lambda^{2}=\frac{h^{2}}{2 m E}$
