In a Geiger Marsden experiment, an alpha particle of energy 6 meV hits a nucleus of z = 72. The distance of closest approach nearly is

35 fm

For the distance of closest we know

$\frac{1}{4\pi \in_0}\times \frac{Ze\times 2e}{r_0}=K.E$

$⇒\frac{9\times 10^9\times 72 \times (1.6\times 10^{-19})^2}{r_0}$

$=6\times 10^{-3}\times 1.6\times 10^{-19}$

$⇒r_0=\frac{9\times 10^9 \times 72 \times 1.6 \times 10^{-19}}{6\times 10^{-3}}$

$⇒r_0=\frac{9\times 72 \times 1.6}{6}\times 10^{-7}$

$⇒r_0=17.28\times 10^{-7} m$

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