An electron of mass m has de-Broglie wavelength when accelerated … - Vidyadip

MCQ

An electron of mass $m$ has de-Broglie wavelength $\lambda$ when accelerated through potential difference $V$. When proton of mass $M$ is accelerated through potential difference $9 V$, the de-Broglie wavelength associated with it will be (Assume that wavelength is determined at low voltage)

A
$\frac{\lambda}{3} \sqrt{\frac{M}{m}}$
B
$\frac{\lambda}{3} \cdot \frac{M}{m}$
✓
$\frac{\lambda}{3} \sqrt{\frac{m}{M}}$
D
$\frac{\lambda}{3} \cdot \frac{m}{M}$

✓

Answer

Correct option: C.

$\frac{\lambda}{3} \sqrt{\frac{m}{M}}$

(c) : de-Broglie wavelength associated with an electron of mass $m$ when accelerated through potential difference $V$ is
$
\lambda=\frac{h}{\sqrt{2 m e V}}
$. . . . . .(i)
and for proton is
$
\lambda_p=\frac{h}{\sqrt{2 M e(9 V)}}
$. . . . . .(ii)
From eqns. (i) and (ii),
$
\frac{\lambda_p}{\lambda}=\sqrt{\frac{m}{9 M}}=\frac{1}{3} \sqrt{\frac{m}{M}} \text { or } \lambda_p=\frac{\lambda}{3} \sqrt{\frac{m}{M}}
$

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