Question

Explain and solve the following physics question.

An alpha particle with kinetic energy 14.0 MeV makes a collision with lead nucleus, but it is not "aimed" at the center of the lead nucleus, and has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude L=p0b, where p0 is the magnitude of the initial momentum of the alpha particle and b=1.20×10⁻¹² m . (Assume that the lead nucleus remains stationary and that it may be treated as a point charge. The atomic number of lead is 82. The alpha particle is a helium nucleus, with atomic number 2.)

1. What is the distance of closest approach? Find r. (meters)

2. Repeat for b =1.30×10⁻¹³ m. Find r. (meters)

3. Repeat for b =1.30×10⁻¹⁴ m. (meters)

Answer 1

The trajectory of the Alpha particle is a hyperbola as it is repelled by the positive charge on Lead nucleus.

For an axis through the Lead nucleus perpendicular to the figure, angular momentum is conserved.

Let the initial velocity of Alpha particle be when it is far away from Lead nucleus with the velocity aimed along a line with a separation of   from the nucleus.

Let the velocity of Alpha particle be when it is closest to the Lead nucleus at a distance of .

Conserving the angular momentum,

----------- (1)

Conserving the total energy of Alpha particle,

(when the charges are faraway electric potential energy is zero)

Substitute,

Solving the above quadratic equation,

Only the positive sign gives a valid root.

Initial kinetic energy of Alpha particle

1. For

Distance of closest approach

2) For

Distance of closest approach

3)

For

Distance of closest approach