A particle A of mass ' m ' and charge ' q ' is accelerated by

Question:

A particle $\mathrm{A}$ of mass ' $\mathrm{m}$ ' and charge ' $\mathrm{q}$ ' is accelerated by a potential difference of $50 \mathrm{v}$ Another particle B of mass ' $4 \mathrm{~m}$ ' and charge ' $q$ ' is accelerated by a potential differnce of

$2500 \mathrm{~V}$. The ratio of de-Broglie wavelength $\frac{\lambda_{A}}{\lambda_{B}}$ is

  1. (1) $10.00$

  2. (2) $0.07$

  3. (3) $14.14$

  4. (4) $4.47$


Correct Option: 3,

Solution:

(3) de Broglie wavelength $(\lambda)$ is given by $\mathrm{K}=\mathrm{qV}$

$\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}(\because \mathrm{p}=\sqrt{2 \mathrm{mK}})$

Substituting the values we get

$\therefore \frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{\sqrt{2 \mathrm{~m}_{\mathrm{B}} \mathrm{q}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}}}{\sqrt{2 \mathrm{~m}_{\mathrm{A}} \mathrm{q}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}}}=\sqrt{\frac{4 \mathrm{~m} \cdot \mathrm{q} \cdot 2500}{\text { m.q. } 50}}$

$=2 \sqrt{50}=2 \times 7.07=14.14$

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