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Assuming the velocity to be same,which sub-atomic particle possesses smallest de-Brogile wavelength

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

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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.
answered Jan 10, 2014 by sreemathi.v
edited Mar 19, 2014 by mosymeow_1

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