Physics - Modern Physics Question with Solution | TestHub

(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 $\mathrm{KE}=\frac{{\mathrm{p}}^2}{2\mathrm{m}}=$ $\frac{q^2{B}^2{r}^2}{2m}$ 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. $(\frac{(2\mathrm{e}{)}^2}{4\mathrm{amu}}=\frac{4{\mathrm{e}}^2}{4\mathrm{amu}}=$ $\frac{{\mathrm{e}}^2}{\mathrm{amu}}$; 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.