A 1 MeV proton is sent against a gold leaf (Z = 79). Calculate the distance of closest approach for head on collision.

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Solution

The condition for the closest approach
$= \frac{1}{2} m v^{2} = \frac{1}{4 π ϵ_{\circ}} \frac{Z e^{2}}{d} = e \times 10^{6}$
Since $1 M e V = 10^{6} \times 1.6 \times 10^{- 19} = e \times 10^{6}$ J
$\Rightarrow d = \frac{9 \times 10^{9} \times 79 \times 1.6 \times 10^{- 19}}{10^{6}}$
$= 1.136 \times 10^{- 13}$ m

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A 1 MeV proton is sent against a gold leaf (Z = 79). Calculate the distance of closest approach for head on collision.

The condition for the closest approach=12mv2=14πϵ∘Ze2d=e×106Since 1MeV=106×1.6×10−19=e×106J⇒d=9×109×79×1.6×10−19106=1.136×10−13m