Question

A proton of energy $8\text{eV}$ is moving in a circular path in a uniform magnetic field. The energy of an $\alpha$ - particle moving in the same magnetic field and along the same path will be

• A. $2\text{eV}$
• B. $4\text{eV}$
• C. $6\text{eV}$
• D. $8\text{eV}$

Answer 1

The correct option is D $8\text{eV}$

We know that,
$q_{\alpha }=2q_p$, $m_{\alpha }=4m_p$

$KE=\frac{1}{2}m{v}^2=\frac{P^2}{2m}.......\left(1\right)$

Force due to magnetic field,

$F_m=qBv=\frac{m{v}^2}{r}$

$Bq=\frac{mv}{r}=\frac{P}{r}$

$P=Bqr..........\left(2\right)$

Using equation $\left(1\right)$ and $\left(2\right)$

$KE=\frac{B^2{q}^2{r}^2}{2m}$

$KE\propto \frac{q^2}{m}$ [ $\because$ $B$ and $r$ are constants]

$\frac{K{E}_{\alpha }}{K{E}_p}={\left(\frac{q_{\alpha }}{q_p}\right)}^2\times \left(\frac{m_p}{m_{\alpha }}\right)$

$={\left(\frac{2}{1}\right)}^2\times \left(\frac{1}{4}\right)=\frac{4}{4}$

$K{E}_{\alpha }=K{E}_p=8eV$