A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in a uniform magnetic field. What should be the energy of an α-particle to describe a circle of the same radius in the same field?

For a particle of mass m and charge Q in a magnetic field, the radius of the path is given by,

R = mv/QB

where v is the velocity of the particle and B is the magnetic field strength.

The kinetic energy of the particle is given by KE = 1/2 mv².

From the radius equation, we have v = QBR/m.

Substituting this into the kinetic energy equation, we get:

KE = 1/2 m (QBR/m)² = Q²B²R²/2m

For the proton (p), we have KEp = Qp²B²R²/2mp = 1 MeV

For the alpha particle (α), we have KEα = Qα²B²R²/2mα

Since the radius R and magnetic field B are the same for both particles, we can write the ratio of their kinetic energies as:

KEα/KEp = (Qα²/2mα) / (Qp²/2mp) = (Qα²/Qp²) * (mp/mα)

An alpha particle has a charge Qα = 2Qp and a mass mα = 4mp. Substituting these values:

KEα/KEp = ( (2Qp)²/Qp² ) * (mp/4mp) = 4 * (1/4) = 1

Therefore, KEα = KEp = 1 MeV

The energy of the α-particle should be 1 MeV to describe a circle of the same radius in the same field.