# An alpha particle of energy 5 MeV is directed towards a lead nucleus. Find the distance of closest approach

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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