Question

An $\alpha$ - particle moving horizontally makes a head on collision elastically with a proton (at rest). What are the ratio of de - Broglie’s wavelength’s associated with $\alpha$ - particle and proton just after collision?

• A. 2:1
• B. 4:3
• C. 1:2
• D. 2:3

Answer 1

The correct option is D 2:3

Let v_i be initial speed of $\alpha$ -particle and $v_{\alpha }$ and $v_p$ are final speeds of $\alpha$ -particle and proton.
Then, $4m_p.v_i=4m_p{v}_{\alpha }+m_p{v}_p\Rightarrow$ momentum is converved.
Also, coefficient of restitution = 1 $\Rightarrow$ Collision being elastic.
$\Rightarrow v_i-0=v_p-v_{\alpha }$
Eliminating $v_i$ from there, we get
$4v_p-4v_{\alpha }=4v_{\alpha }+v_p$
$\Rightarrow 3v_p=8v_{\alpha }$
$or{v}_{\alpha }=\frac{3}{8}{v}_p$
Now, de - Broglie wavelengths of proton and $\alpha$ - particle are
${\lambda }_p=\frac{h}{m_p{v}_p}$
and ${\lambda }_{\alpha }=\frac{h}{m_{\alpha }{v}_{\alpha }}$
$\Rightarrow \frac{{\lambda }_{\alpha }}{{\lambda }_p}=\frac{h}{m_{\alpha }{v}_{\alpha }}\times \frac{m_p{v}_p}{h}$
$=(\frac{m_p}{m_{\alpha }}.)\left(\frac{v_p}{v_{\alpha }}\right)=\frac{1}{4}\times \frac{8}{3}=\frac{2}{3}$