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An alpha particle of energy 5 MeV is directed towards a lead nucleus. Find the distance of closest approach
1. 40 fm
2. 47.1fm
3. 60 fm
4. 0.04 fm

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Correct Answer - Option 2 : 47.1fm

CONCEPT

  • The distance of the closest approach is the closest distance to the nucleus at which all the kinetic energy of the projected particle becomes potential energy and is given by
    \(\Rightarrow d = \frac{2Ze^{2}}{4\pi \epsilon_{0}K}\)
    Where d = Distance of closest approach, Z = Atomic number, K = Kinetic energy. e = charge 


CALCULATION : 

Given - K = 5 meV = 5 × 1.6 × 10-13J, Z = 82, and e = 1.6 × 10-19 C

  • The distance of closest approach is given by
    \(\Rightarrow d = \frac{2Ze^{2}}{4\pi \epsilon_{0}K}\)


Substituting the given values the above equation becomes 

\(\Rightarrow d = \frac{2\times82\times(1.6\times10^{-19})^{2}}{4\pi \epsilon_{0}\times 5\times1.6\times 10^{-13} }\)

\(\Rightarrow d = \frac{9\times 10^{9}\times 419.84\times 10^{-3 8}}{8\times 10^{-13}} = 47.1 \ \rm fermi\)

  • Hence, option 2 is the answer

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