Question

A lithium nucleus (3 protons, 4 neutrons) is accelerated through a potential difference of 100,000 V over a distance of 20 cm before entering a uniform magnetic field of magnitude 0.5 T pointing perpendicular to the path of the nucleus. What is the radius of the path followed by the nucleus? (1) 24 cm (2) 14 cm (3) 9.1 cm (4) 5.3 cm (5) 7.5 cm O 1 O 2 O 3 O 5

Answer 1

S. The radius of the path followed by the lithium nucleus is 9.1 cm.

According to the Lorentz force equation, the force acting on a particle with charge q, mass m, and velocity v in a uniform magnetic field B is given by F = qv x B.

The radius of the path followed by the nucleus can be calculated by equating the centripetal force to the Lorentz force.

F_c = F_L

mv^2/r = qv x B

Substituting the given values,

(4*1.67 * 10^-27) * (2.99 * 10^8)^2 / r = (3.2 * 10^-19) * (2.99 * 10^8) * 0.5

Solving for r,

r = (4*1.67 * 10^-27) * (2.99 * 10^8)^2 / [(3.2 * 10^-19) * (2.99 * 10^8) * 0.5]

r = 9.1 cm

Hence, the correct answer is (3) 9.1 cm.

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