Question

An $\alpha$-particle, a proton and an electron have the same kinetic energy. Which one of the following is correct in case of their de-Broglie wavelength:

• A. ${\lambda }_{\alpha }>{\lambda }_p>{\lambda }_e$
• B. ${\lambda }_{\alpha }<{\lambda }_p<{\lambda }_e$
• C. ${\lambda }_{\alpha }={\lambda }_p={\lambda }_e$
• D. ${\lambda }_{\alpha }>{\lambda }_p<{\lambda }_e$

Answer 1

Answer: (B)

${\lambda }_D=\frac{h}{p}=\frac{h}{\sqrt{2mK}}$
$\therefore \lambda \propto \frac{1}{\sqrt{m}}$
$\because m_{\alpha }>m_p>m_e$
${\lambda }_c>{\lambda }_p>{\lambda }_{\alpha }$