A proton and an alpha particle, both initially at rest, are (suitablly) accelerated so as to have the same kinetic energy. What is the ratio of their de-Broglie wavelengths?

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Solution

Given, a proton and an alpha particle, both initially at rest are accelerated such that they have the same kinetic energy.

De-broglie wavelength is given by,

λ = \frac{h}{p} = \frac{h}{\sqrt{2 m E}}

(for same K.E)

Therefore, the ratio of their de-broglie wavelength is,

\frac{λ_{p}}{λ_{α}} = \sqrt{\frac{m_{α}}{m_{p}}} = \sqrt{\frac{4 m}{m}} = 2 .

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Wave Optics

A proton and an alpha particle, both initially at rest, are (suitablly) accelerated so as to have the same kinetic energy. What is the ratio of their de-Broglie wavelengths?

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