Question

an alpha particle is accelerated through a potential difference of V volts from rest. the de broglie's wavelength associated with it is??

Answer 1

Dear Student


Mass of $\alpha$- particle = 4u

Charge of $\alpha$-particle = 2e

Therefore,

the kinetic energy of the $\alpha$-particle when accelerated through a potential of V volts will be,

E = eV


The de broglie's wavelength is given by the equation:

$\lambda$ = $\frac{h}{mv}$ ( or $\frac{h}{p}$)

So, the momentum acquired by the $\alpha$-particle, p = $\sqrt{2mE}$ = $\sqrt{2meV}$

Substituting the values, we get, momentum = $\sqrt{2x4ux2exV}$

Thus, $\lambda$ = $\frac{h}{\sqrt{2x4ux2exV}}$




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