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  4. In Geiger-Marsden experiment, in a head-on collision between an α-particle and a gold nucleus, the minimum separation is 4 × 10-14 m. The energy of α-particle is (Take, atomic number of gold =79 )

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Q. In Geiger-Marsden experiment, in a head-on collision between an $\alpha$-particle and a gold nucleus, the minimum separation is $4 \times 10^{-14} m$. The energy of $\alpha$-particle is (Take, atomic number of gold $=79$ )

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Solution:

At distance of minimum separation, $KE = PE$.
$\Rightarrow KE =\frac{k Ze \cdot 2 e}{r}$
or $KE =\frac{9 \times 10^{9} \times 79 \times 2 \times\left(1.6 \times 10^{-19}\right)^{2}}{4 \times 10^{-14} J }$
$=\frac{9 \times 10^{9} \times 79 \times 2\left(1.6 \times 10^{-19}\right)^{2}}{4 \times 10^{-14} \times 1.6 \times 10^{-19}} eV$
$=5.688\, MeV \left(\because 1\, MeV =1.6 \times 10^{-13} J \right)$