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  4. In a head on collision between an α -particle and gold nucleus, the closest distance of approach is 41.3 fm. Calculate the energy of α -particle (Z of gold =79)

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Q. In a head on collision between an $\alpha$ -particle and gold nucleus, the closest distance of approach is $41.3 fm$. Calculate the energy of $\alpha$ -particle $(Z$ of gold $=79)$

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Solution:

$E =\frac{( Ze )(2 e )}{4 \pi \varepsilon_{0} r }=\frac{9 \times 10^{9} \times 79 \times 2\left(1.6 \times 10^{-19}\right)^{2}}{41.3 \times 10^{-15}}$
$=\frac{9 \times 79 \times 3.2 \times 1.6 \times 10^{-14}}{41.3} J$
$=8.814 \times 10^{-13}$ Joule $=\frac{8.814 \times 10^{-13}}{1.6 \times 10^{-13}}$
$E =5.51 MeV$