Question

What will be the ratio of de-Broglie wavelengths of proton and a-particle of the same energy?

• A. $2:1$
• B. $1:2$
• C. $4:1$
• D. $1:4$

Answer 1

Answer: (A)

de-Broglie wavelength, $\lambda =\frac{h}{p}$
$=\frac{h}{\sqrt{2mKE}}$
For particles of same energy, $\lambda \propto \frac{1}{\sqrt{m}}$
$\therefore \frac{{\lambda }_p}{{\lambda }_{\alpha }}=\sqrt{\frac{m_{\alpha }}{m_p}}$
$=\sqrt{\frac{4m_p}{m_p}}=\frac{2}{1}$