Question

A proton and an electron are accelerated by the same potential difference, let lambda_(e) and lambda_(p) denote the de-Broglie wavelengths of the electron and the proton respectively

Answer 1

Answer:

λ/λ'=44.72=$\sqrt{2000}$

Explanation:

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

$\frac{1}{2}mv^{2}=qV$

$mv=\sqrt{ 2qVm }$

∴λ=$\frac{h}{\sqrt{2mqV} }$

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

where ' symbol denotes properties of proton

∴λ/λ'=$\sqrt{ \frac{2000meV}{meV} }$  [∵ mass of proton is ≅2000(mass of $e^{-1}$)]

λ/λ'=44.72

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