An ' $\alpha$ ' particle of energy $10 \mathrm{eV}$ is moving in a circular path in uniform magnetic field. The energy of proton moving in the same path and same magnetic field will be [mass of ' $\alpha$ ' particle $=4$ times mass of proton $]$

(C) From the formula mentioned above, momentum of particle moving in a magnetic field $\mathrm{mv}=\mathrm{p}=\mathrm{qBr}$ Therefore, Kinetic Energy of that particle can be written as $\mathrm{KE}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}=$ $\frac{q^{2} B^{2} r^{2}}{2 m}$ In the same magnetic field for the same path, $\mathrm{KE} \propto \frac{\mathrm{q}^{2}}{\mathrm{~m}}$ This ratio is same for the alpha particle and the proton. $\left(\frac{(2 \mathrm{e})^{2}}{4 \mathrm{amu}}=\frac{4 \mathrm{e}^{2}}{4 \mathrm{amu}}=\right.$ $\frac{\mathrm{e}^{2}}{\mathrm{amu}}$; 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 $8 \mathrm{eV}$ too.