Question

Determine the distance of closest approach (in fm) before the alpha particle reverses direction. Assume the lead nucleus remains stationary. Assume the alpha particles are initially very far from a stationary lead nucleus.

Answer 1

Answer:

Take E(alpha particle energy) = 5.5 MeV (5.5x106x1.6x10-19)

If the charge on the lead nucleus is +82e(atomic number of lead is 82) = +82x1.6x10-19 C and the charge on the alpha particle is +2e = 2x1.6x10-19 C

Using dc = (1/4πεo)qQ/Eα we have

dc = [9x10^9x(2x1.6x10-19x82x1.6x10-19)]/5.5x10-13 = 6.67x10^-13m. = 6.67 x 10^-13 x 10^15 = 6.67 x 10^2fm

Note: 1meter = 10^15fentometer

Explanation:

This is well inside the atom but some eight nuclear diameters from the centre of the lead nucleus.