Looking for classes? Ksquare Career Institute, Bengaluru →
A proton and an $\alpha$-particle are accelerated from rest to the same energy. The de Broglie wavelengths $\lambda_p$ and $\lambda_\alpha$ are in the ratio,
Detailed Solution
$\lambda=\frac{h}{\sqrt{2mK}}$; with equal K, $\frac{\lambda_p}{\lambda_\alpha}=\sqrt{\frac{m_\alpha}{m_p}}=\sqrt{\frac{4m}{m}}=2:1$.