An alpha particle of charge +2e and mass 41.66 × 10^-27) kg, and an ¹⁶O nucleus of charge +8e and mass 161.66 × 10^-27) kg have been accelerated from rest through the same electric potential. They are then injected into a uniform magnetic field B, where both move at right angles to the field. The ratio of the radius of the path of the alpha particle to the radius of the path of the nucleus ¹⁶O is

Question

Quizplus · Accepted Answer

The radius of path of a charged particle in a magnetic field is given by r = mv/qB, where m is the mass of the particle, v is its velocity, q is its charge and B is the magnetic field strength.

Both the alpha particle and the nucleus have the same electric potential difference and hence the same kinetic energy, which is given by K = eV, where e is the elementary charge and V is the electric potential difference.

The velocity of the particles can be found from the conservation of energy principle as 1/2 mv^2 = eV, or v = sqrt(2eV/m).

The ratio of the radii of path of the alpha particle and the nucleus is given by r_alpha/r_nucleus = (m_alpha/m_nucleus) * (q_nucleus/q_alpha).

Substituting the given values, we get r_alpha/r_nucleus = (41.66/161.66) * (8/2) = 1/2.

Therefore, the correct answer is D.

An Alpha Particle of Charge +2e and Mass 41 α\alphaα /rO = 1/1

An Alpha Particle of Charge +2e and Mass 41 $\alpha$ /r_O = 1/1