Ask Question
11 July, 09:26

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.

+2
Answers (1)
  1. 11 July, 13:12
    0
    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.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Determine the distance of closest approach (in fm) before the alpha particle reverses direction. Assume the lead nucleus remains ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers