The ratio of de-Broglie wavelength of an α -particle and a proton accelerated from rest by the same potential is 1 m , the value of m is:

Question

TestHub · Accepted Answer

As we know, de Broglie wavelength in terms of energy of particle is given by, $λ = \frac{1}{\sqrt{2 m E}}$ .

Energy gained by proton, $E = q V$ .

Therefore,

$\frac{λ_{α}}{λ_{p}} = \frac{\frac{h}{\sqrt{2 m_{α} q_{α} V}}}{\frac{h}{\sqrt{2 m_{p} q_{p} V}}}$

$\Rightarrow \frac{λ_{α}}{λ_{p}} = \sqrt{\frac{1}{8}}$

$\Rightarrow m = 8$

Physics - Modern Physics Question with Solution | TestHub