Question

The de-Broglie wavelength of a proton moving with a velocity $v$, is $\lambda$. If an alpha particle is moving with a velocity $\frac{v}{2}$ then, what would be its de-Broglie wavelength ?

• A. $\frac{\lambda }{2}$
• B. $\frac{\lambda }{4}$
• C. $2\lambda$
• D. $4\lambda$

Answer 1

The correct option is A $\frac{\lambda }{2}$

Let the masses of proton and alpha particles are $m_p\text{and}{m}_{\alpha }$ respectively. Then, $m_{\alpha }=4m_p$
The de-Broglie wavelength of proton is given by,
${\lambda }_p=\lambda =\frac{h}{p}=\frac{h}{m_{p}v}$

And, ${\lambda }_{\alpha }=\frac{h}{\left(4m_p\right)\left(\frac{v}{2}\right)}=\frac{h}{2m_{p}v}$
$=\frac{1}{2}\left(\frac{h}{m_{p}v}\right)=\frac{{\lambda }_p}{2}=\frac{\lambda }{2}$

Hence, $\left(A\right)$ is the correct answer.

Key concept: We need to express the unknown wavelength in terms of known variables like $\lambda ,m,v\text{or}h.$