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.