Dual Nature of Radiation and Matter - Result Question 9

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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]

3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/dual-nature-of-radiation-and-matter/dual-nature-of-radiation-and-matter—result-question-9.md (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:

  1. (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)

$\therefore E=\frac{h^{2}}{2 m \lambda^{2}} \quad \Rightarrow \lambda^{2}=\frac{h^{2}}{2 m E}$