Question

The ratio of the de Broglie wavelengths of a proton and an $\alpha$-particle will be 1 : 2 if their:

• A. velocities are in the ratio 1 : 8
• B. velocities are in the ratio 8 : 1
• C. kinetic energies are in the ratio 1 : 64
• D. kinetic energies are in the ratio 1 : 256

Answer 1

The correct option is B velocities are in the ratio 8 : 1

de Broglie equation is:
$\lambda$$=$$\frac{h}{mv}$
where m is the mass and v is the velocity
$\frac{{\lambda }_p}{{\lambda }_{\alpha }}$$=$$\frac{m_{\alpha }{v}_{\alpha }}{m_p{v}_p}$ $eq.\to 1$
Here,
${\lambda }_p$ = wavelength of the proton
${\lambda }_{\alpha }$ = wavelength of the alpha particle
$m_p$ = mass of the proton
$m_{\alpha }$ = mass of the alpha particle
$v_p$ = velocity of the proton
$v_{\alpha }$ = velocity of the alpha particle

We know,
$m_{\alpha }$ $=$ $4m_p$
given $\frac{{\lambda }_p}{{\lambda }_{\alpha }}$$=$$\frac{1}{2}$
Substituting the value of $m_{\alpha }$ in eq. 1, we get:

$\frac{{\lambda }_p}{{\lambda }_{\alpha }}$$=$$\frac{\left(4m_p\right)v_{\alpha }}{m_p{v}_p}$
$\frac{1}{2}$$=$$4$$\times$$\frac{v_{\alpha }}{v_p}$
$\frac{v_p}{v_{\alpha }}$$=$$\frac{8}{1}$