Answer:
de Broglie wavelength of a particle of mass m and kinetic energy K is \[\lambda =\frac{h}{\sqrt{2mK}}\] When the kinetic energy is \[K/\text{4}\], \[\lambda '=\frac{h}{\sqrt{2mK/4}}=\frac{2h}{\sqrt{2mK}}=2\lambda .\]
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