A particle A of mass 'm' and charge 'q' is accelerated by a potential difference of 50 V. Another particle B of mass ' 4 m' and charge 'q' is accelerated by a potential difference of 2500 V. The ratio of de-Broglie wavelengths \lambda _A \over \lambda _B is close to :

Dual Nature of Radiation PYQ — JEE Main 2019 · 12 Jan · Shift 1 · Q63 with Solution

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Correct answer: C

For a particle accelerated through a potential difference $V$ :

$qV = \frac{p^2}{2m}$

So,

$p = \sqrt{2mqV}$

Hence the de-Broglie wavelength is

$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}$

Therefore,

$\lambda \propto \frac{1}{\sqrt{mV}}$

(since $h$ , $q$ are same for both particles).

For particle $A$ :

So,

$\lambda_A \propto \frac{1}{\sqrt{m \cdot 50}}$

For particle $B$ :

So,

$\lambda_B \propto \frac{1}{\sqrt{4m \cdot 2500}}$

$\frac{\lambda_A}{\lambda_B} = \sqrt{\frac{4m \cdot 2500}{m \cdot 50}}$

Cancel $m$ :

$\frac{\lambda_A}{\lambda_B} = \sqrt{\frac{4 \cdot 2500}{50}}$

$= \sqrt{4 \cdot 50}$

$= \sqrt{200}$

$= 10\sqrt{2} \approx 14.14$

The value is closest to:

$\boxed{14.14}$

So the correct option is C.