Explanation:
de-Broglie wavelength,
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\lambda {\rm{ = }}\frac{h}{{mv}} = \frac{h}{{\sqrt {2mE} }}\)
\(E\) is same for all, so \(\lambda \propto \frac{1}{{\sqrt m }}\)
Hence, de-Broglie wavelength will be
maximum for particle with lesser mass.
Mass of the given particels in
increasing order are given as
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{m_e} < {m_p} < {m_n} < {m_\alpha }\)
Thus, de-Broglie wavelength will be maximum for electron.