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  4. Which of these particles (having the same kinetic energy) has the shortest de-Broglie wave length?

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Q. Which of these particles (having the same kinetic energy) has the shortest de-Broglie wave length?

Delhi UMET/DPMTDelhi UMET/DPMT 2010

Solution:

de-Broglie relation is given by $ \lambda =\frac{h}{mv} $
This can also be written as $ \lambda =\frac{h}{\sqrt{2mv}} $
So, $ \lambda \propto \frac{1}{\sqrt{m}} $
Because, an alpha particle having the highest mass among the others,
so alpha particle have the least wavelength.