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Dual Nature of Radiation question

2023 · 12 Apr · Shift 1 · Q52

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Dual Nature of Radiation
2023 · 12 Apr · Shift 1 · Q52

Dual Nature of Radiation question

2023 · 12 Apr · Shift 1 · Q52

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1

A proton and an

\alpha

-particle are accelerated from rest by

2 \mathrm{~V}

and

4 \mathrm{~V}

potentials, respectively. The ratio of their de-Broglie wavelength is :

A
4 : 1
B
2 : 1
C
8 : 1
D
16 : 1

View written solutionFree

Correct answer: A

Use the de-Broglie wavelength formula

For a particle accelerated through a potential difference $V$ from rest,

\lambda = \frac{h}{p}

and the kinetic energy gained is

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

So,

p = \sqrt{2mqV}

Hence,

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

Therefore,

\lambda \propto \frac{1}{\sqrt{mqV}}

For the proton

For proton:

mass $= m_p$
charge $= e$
potential $= 2\,\text{V}$

So,

\lambda_p \propto \frac{1}{\sqrt{m_p \cdot e \cdot 2}}

For the $\alpha$ -particle

For $\alpha$ -particle:

mass $= 4m_p$
charge $= 2e$
potential $= 4\,\text{V}$

So,

\lambda_\alpha \propto \frac{1}{\sqrt{(4m_p)(2e)(4)}}

Find the ratio

\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{(4m_p)(2e)(4)}{m_p \cdot e \cdot 2}}

Cancel common factors:

\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{32m_pe}{2m_pe}} = \sqrt{16} = 4

Thus,

\lambda_p : \lambda_\alpha = 4 : 1

Check options

A: $4:1$ ✅
B: $2:1$ ❌
C: $8:1$ ❌
D: $16:1$ ❌

So the correct option is A.

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