A proton of energy 12 eV is moving in a circular path in uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will be

Solution:

When a charged particle moves in a uniform magnetic field perpendicular to its velocity, the magnetic force provides the necessary centripetal force for circular motion.

The magnetic force is given by $F_B = qvB$.

The centripetal force is given by $F_c = \frac{mv^2}{r}$.

Equating these forces: $qvB = \frac{mv^2}{r}$

From this, the radius of the circular path is $r = \frac{mv}{qB}$.

The kinetic energy of the particle is $K = \frac{1}{2}mv^2$.

From the radius equation, we can express velocity $v = \frac{qBr}{m}$.

Substitute $v$ into the kinetic energy equation:

$K = \frac{1}{2}m \left(\frac{qBr}{m}\right)^2 = \frac{1}{2}m \frac{q^2B^2r^2}{m^2} = \frac{q^2B^2r^2}{2m}$.

Given for a proton:

Charge of proton, $q_p = e$

Mass of proton, $m_p = m$

Energy of proton, $K_p = 12 \text{ eV}$

So, for the proton: $K_p = \frac{e^2B^2r^2}{2m} = 12 \text{ eV}$.

For an alpha particle:

Charge of alpha particle, $q_\alpha = 2e$

Mass of alpha particle, $m_\alpha = 4m$ (since an alpha particle consists of 2 protons and 2 neutrons)

It moves in the same magnetic field ($B$) and along the same path (same radius $r$).

The energy of the alpha particle, $K_\alpha$, will be:

$K_\alpha = \frac{q_\alpha^2B^2r^2}{2m_\alpha} = \frac{(2e)^2B^2r^2}{2(4m)} = \frac{4e^2B^2r^2}{8m} = \frac{e^2B^2r^2}{2m}$.

Comparing $K_\alpha$ with $K_p$, we see that:

$K_\alpha = K_p$.

Therefore, the energy of the alpha particle will be $12 \text{ eV}$.