Question

What should be the order of arrangement of de-Broglie wavelength of electron $\left({\lambda }_e\right)$, an $\alpha$-particle $\left({\lambda }_{\alpha }\right)$ and proton $\left({\lambda }_p\right)$ given that all have the same kinetic energy ?

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

Answer 1

Answer: (C)

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mE}}\propto \frac{1}{\sqrt{m}}$
$m_a>m_p>m_e$
$SO{\lambda }_e>{\lambda }_p>{\lambda }_a$