MHT CET · Physics · Dual Nature of Matter

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 $\frac{λ}{3} \sqrt{\frac{M}{m}}$
B $\frac{λ}{3} . \frac{M}{m}$
C $\frac{λ}{3} \sqrt{\frac{m}{M}}$
D $\frac{λ}{3} . \frac{m}{M}$

Verified Solution

Answer & Solution

Correct Answer

(C) $\frac{λ}{3} \sqrt{\frac{m}{M}}$

Step-by-step Solution

Detailed explanation

When electron or any charged particle is accelerated through potential difference V, then kinetic energy gained is given by

E = e V

...... (i)

E = \frac{1}{2} m v_{0}^{2} = \frac{p^{2}}{2 m} = \frac{h^{2}}{2 m . λ^{2}}

...... (ii)

∴ e V = \frac{h^{2}}{2 m . λ^{2}} \Rightarrow λ = \frac{h}{\sqrt{2 m e V}}

....... (iii)
When proton of mas

M

is accelerated through a potential difference of

9 V

, then the de - Broglie wavelength obtained is

λ^{'} = \frac{h}{\sqrt{2 M e (9 V)}} = \frac{h}{3 \times \sqrt{2 M e V}} \times \frac{\sqrt{m}}{\sqrt{m}}

∴ λ^{'} = \frac{λ}{3} . \sqrt{\frac{m}{M}}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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