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

Answer

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.
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