Question

${\lambda }_e,{\lambda }_p$ and ${\lambda }_{\alpha }$ are the de-Broglie wavelength of electron, proton and $\alpha$ -particle. If all are accelerated by same potential, then

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

Answer 1

Answer: (E)

${\lambda }_e=\frac{12.27}{\sqrt{V}}\mathring{A}$
${\lambda }_p=\frac{0.201}{\sqrt{V}}\mathring{A}$
${\lambda }_{\alpha }=\frac{0.101}{\sqrt{V}}\mathring{A}$