Proton, Deuteron and alpha particle of the same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton, deuteron and alpha particle are respectively rp, rd and rα. Which one of the following relations is correct?

Engineering

Physics

Magnetic Field

Moton of Charged Particle in Uniform Magnetic Field

Lorentz Force

Question

Proton, Deuteron and alpha particle of the same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton, deuteron and alpha particle are respectively r_p, r_d and r_α. Which one of the following relations is correct?

r_α = r_p < r_d

r_α > r_d > r_p

r_α = r_d > r_p

r_α = r_p = r_d

$R = \frac{mv}{qB} = \frac{\sqrt{2 mk}}{qB}$

$R_{p} : R_{d} : R_{α} = \frac{\sqrt{m}}{e} : \frac{\sqrt{2 m}}{e} : \frac{\sqrt{4 m}}{2 e}$

$= 1 : \sqrt{2} : 1$

When charged particles move perpendicular to a uniform magnetic field, they experience a force that acts as a centripetal force, causing them to move in circular paths. The radius of this circular path is given by the formula:

$r = \frac{m v}{q B}$

where $m$ is the mass of the particle, $v$ is its speed, $q$ is its charge, and $B$ is the magnetic field strength.

Since all particles have the same kinetic energy (K), we can express their speed in terms of K. The kinetic energy is $K = \frac{1}{2} m v^{2}$ . Solving for $m v$ :

$m v = \sqrt{2 m K}$

Substituting this into the radius formula:

$r = \frac{\sqrt{2 m K}}{q B}$

Since K and B are the same for all particles, the radius depends on the ratio $\frac{\sqrt{m}}{q}$ .

Let's analyze the particles:

Proton (p): mass = m, charge = +e
Deuteron (d): a nucleus of deuterium, containing 1 proton and 1 neutron. mass = 2m, charge = +e
Alpha particle (α): a helium nucleus, containing 2 protons and 2 neutrons. mass = 4m, charge = +2e

Now, let's calculate the ratio $\frac{\sqrt{m}}{q}$ for each:

For proton: $\frac{\sqrt{m}}{e}$

For deuteron: $\frac{\sqrt{2 m}}{e} = \sqrt{2} \times \frac{\sqrt{m}}{e}$

For alpha particle: $\frac{\sqrt{4 m}}{2 e} = \frac{2 \sqrt{m}}{2 e} = \frac{\sqrt{m}}{e}$

Comparing the ratios:

Proton and alpha particle have the same ratio: $\frac{\sqrt{m}}{e}$

Deuteron has a larger ratio: $\sqrt{2} \times \frac{\sqrt{m}}{e}$

Since the radius is proportional to this ratio, we find:

$r_{α} = r_{p} < r_{d}$

Therefore, the correct relation is $r_{α} = r_{p} < r_{d}$ .

Key Formulae

Radius of Circular Path: $r = \frac{m v}{q B}$

Kinetic Energy: $K = \frac{1}{2} m v^{2}$

Radius in terms of Kinetic Energy: $r = \frac{\sqrt{2 m K}}{q B}$

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