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An electron, an $$\alpha$$ particle, and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wave length?
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## QuestionPhysicsClass 12

An electron, an $$\alpha$$ particle, and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wave length?

For a particle, de Broglie wavelength,
$$\lambda =\frac h p$$
The relation between wavelength and kinetic energy of a particle is
$$K=\frac{p^2}{2m}$$
$$p=\frac h{\lambda }$$
$$K=\frac{\left(\frac h{\lambda }\right)^2}{2m}$$
$$\lambda =\frac h{\sqrt{2\mathit{mK}}}$$
From the above relation it can be concluded that for the same kinetic energy, de Broglie’s wavelength of a particle is inversely proportional to its mass.
Mass of a proton, $$m_p=1836m_e$$
Mass of an $$\alpha$$ particle, $$m_{\alpha }=4m_p$$
Therefore the masses are in the order, $$m_{\alpha }>m_p>m_e$$  .
Hence, the $$\alpha$$ particle has the shortest de Broglie’s  wavelength.