How much was the radius of the nucleus estima | vedclass

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(B) In the $\alpha -$ scattering experiment conducted by Rutherford,the distance of closest approach was used to estimate the upper limit of the nucle

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$10^{-14} \ m$

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(B) In the $\alpha -$ scattering experiment conducted by Rutherford,the distance of closest approach was used to estimate the upper limit of the nuclear size.For a gold nucleus $(Z = 79)$,the distance of closest approach $d$ is given by the formula $d = \frac{1}{4\pi\epsilon_0} \cdot \frac{2Ze^2}{K}$,where $K$ is the kinetic energy of the $\alpha -$ particle.Experimental results indicated that the nucleus is extremely small compared to the atom,with a radius estimated to be in the order of $10^{-14} \ m$ to $10^{-15} \ m$.Specifically,the radius of a nucleus is typically expressed as $R = R_0 A^{1/3}$,where $R_0 \approx 1.2 	imes 10^{-15} \ m$.Thus,the estimated order of magnitude for the radius of the nucleus from the scattering experiment is $10^{-14} \ m$.

How much was the radius of the nucleus estimated from the experiment of $\alpha -$ scattering?

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