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de Broglie Wavelength

Alpha particle and a proton are accelerated from rest by the same potential. Find the ratio of their de- broglie wavelength

(Asked by Amritha)

Answer:

Charge of alpha particle = 2e

Mass of alpha particle = 4 u

Charge of proton = e

mass of proton = u

The energy acquired by proton when accelerated through a pd of V,

E=eV

The momentum acquired by proton= ${\sqrt{2ueV}}$

The de Broglie wavelength is given by $\lambda =\frac{h}{mv}$

Therefore, de Broglie wavelength of Proton, $\lambda _{proton}=\frac{h}{\sqrt{2ueV}}$

Similarly, $\lambda _{alpha}=\frac{h}{\sqrt{2\times 4u\times 2e\times V}}$

$\frac{\lambda _{alpha}}{\lambda _{proton}}=\frac{\sqrt{2ueV}}{\sqrt{2\times 4u\times 2e\times V}}=\frac{1}{2\sqrt{2}}$

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