Question

A proton and an alpha-particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths lambda_(p) and lambda_(a) related to each other?

Answer 1

Answer:

λ/λ'=$\sqrt{8}=2\sqrt{2}$

Explanation:

De broglie wavelength is given by : λ=$\frac{h}{mv}$

∴λ=$\frac{h}{\sqrt{2mqV} }$

Ratio of λ/λ'=$\frac{\sqrt{m'q'V} }{\sqrt{mqV} }$

where ' symbol denotes properties of alpha particle

∴λ/λ'= $\sqrt{\frac{(4m)(2q)}{{mq}}$ [∵ mass of alpha-particle is ≅4(mass of proton)]

λ/λ'=$\sqrt{8}=2\sqrt{2}$

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