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Q.
Which of these particles (having the same kinetic energy) has the largest de-Broglie wavelength?
Delhi UMET/DPMTDelhi UMET/DPMT 2009Dual Nature of Radiation and Matter
Solution:
de-Broglie wavelength $ = \lambda = \frac{ h}{ mv} = \frac{ h}{ \sqrt{ 2 m E }} $
$E$ is same for all, so $ \lambda \propto \frac{ 1}{ \sqrt m } $
Hence, de-Broglie waveiength will be maximum for particle with lesser mass. Mass of the given particles in increasing order are given as
$ m_e < m_p < m_n < m_\alpha $
Thus, de-Broglie wavelength will be maximum for electron.