An ' ' particle of energy 10 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 ' ' particle =4 times mass of proton ]

Step-by-step solution

(C) 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 = p ² 2 ~m = q² B² r² 2 m In the same magnetic field for the same path, KE q ² ~m This ratio is same for the alpha particle and the proton. ( (2 e )² 4 amu = 4 e ² 4 amu = . 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 8 eV too.