$\begin{array}{1 1}(a)\;\text{An electron}\\(b)\;\text{A proton}\\(c)\;\text{An }\alpha-\text{particle}\\(d)\;\text{All have same wavelength}\end{array}$

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- Ratio of the mass of proton and $\alpha$ particle is 1 : 4.
- Mass of electron compared to proton and alpha particle is negligible.
- De-Broglie wavelength $\lambda=\large\frac{h}{mu}$

Since, de-Broglie wavelength $\lambda=\large\frac{h}{mu}$

$\therefore \lambda \propto \frac{1}{m}$

Hence, heavier the particle, lesser the wavelength.

$\therefore$ The particle posessing smallest de-Broglie wavelength is $\alpha$-particle

Hence (c) is the correct answer.

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