A proton accelerated through a potential V has de Broglie wavelength $\lambda$. Then the de Broglie

Question

A proton accelerated through a potential V has de Broglie wavelength $\lambda$. Then the de Broglie wavelength of an $\alpha$-particle, when accelerated through the same potential $V$ is\

Professor PC Thomas Classes · Answer

The de Broglie wavelength of a proton when accelerated through potential V is $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}$.\ Now the de Broglie wavelength of $\alpha$ - particle, when accelerated through V is\ $\lambda^{\prime}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}^{\prime} \mathrm{qV}}} ; \mathrm{m}^{\prime} 4 \mathrm{~m} \mathrm{q}^{\prime}=2 \mathrm{q}$\ $\lambda^{\prime}=\frac{\mathrm{h}}{\sqrt{2 	imes 4 \mathrm{~m} 	imes 2 \mathrm{qV}}}=\frac{\lambda}{2 \sqrt{2}}$\