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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 $10^{-14} \ m$.

List-$I$	List-$II$
$(A)$ Strong nuclear force	$(i)$ $10^{-2}$
$(B)$ Weak nuclear force	(ii) $1$
$(C)$ Electromagnetic force	(iii) $10^{-39}$
$(D)$ Gravitational force	(iv) $10^{-13}$

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

Similar Questions

State the dimension of the nucleus from the Rutherford experiment.

Consider the following nuclear reactions:
$I$. ${ }_{7}^{14} N +{ }_{2}^{4} He \longrightarrow{ }_{8}^{17} O + X$
$II$. ${ }_{4}^{9} Be +{ }_{2}^{4} He \longrightarrow{ }_{6}^{12} C + Y$
Then,

Consider a nucleus ${ }_{30}^{60} X$. Its approximate density is (Take $1 \text{ amu} = 1.6 \times 10^{-27} \text{ kg}$,$R_0 = 1.2 \times 10^{-15} \text{ m}$)

Match the following fundamental forces of nature with their relative strength:
List-$I$ List-$II$
$(A)$ Strong nuclear force $(i)$ $10^{-2}$
$(B)$ Weak nuclear force (ii) $1$
$(C)$ Electromagnetic force (iii) $10^{-39}$
$(D)$ Gravitational force (iv) $10^{-13}$

Explain isotopes,isobars,isotones,and isomers by giving examples.

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