Question

A proton of energy $8eV$ 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. $4eV$
• B. $2eV$
• C. $8eV$
• D. $6eV$

Answer 1

Answer: (C)

$8eV$

From the formula mentioned above, momentum of particle moving in a magnetic field $mv=p=qBr$
Therefore, Kinetic Energy of that particle can be written as $KE=\frac{p^2}{2m}=\frac{q^2{B}^2{r}^2}{2m}$
In the same magnetic field for the same path, $KE\propto \frac{q^2}{m}$
This ratio is same for the alpha particle and the proton. ($\frac{(2e{)}^2}{4amu}=\frac{4e^2}{4amu}=\frac{e^2}{amu}$ ; Here $amu$ is the atomic mass unit)
So, in such conditions, both will have the same energy. Hence, energy of the alpha particle will be $8eV$ too.