Consider a nucleus { }_{30}^{60} X. Its appro | vedclass

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(D) The density of a nucleus is given by the formula: $ho = \frac{	ext{Mass}}{	ext{Volume}}$.Mass of the nucleus $M = A 	imes m_p$,where $A$ is t

Vedclass · Accepted Answer

$2.2 	imes 10^{17} 	ext{ kg m}^{-3}$

Vedclass · Answer

(D) The density of a nucleus is given by the formula: $ho = \frac{	ext{Mass}}{	ext{Volume}}$.Mass of the nucleus $M = A 	imes m_p$,where $A$ is the mass number and $m_p \approx 1.6 	imes 10^{-27} 	ext{ kg}$.Volume of the nucleus $V = \frac{4}{3} \pi R^3$,where $R = R_0 A^{1/3}$.Substituting $V$: $V = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A$.Density $ho = \frac{A 	imes m_p}{\frac{4}{3} \pi R_0^3 A} = \frac{3 m_p}{4 \pi R_0^3}$.Substituting the values: $ho = \frac{3 	imes 1.6 	imes 10^{-27}}{4 	imes 3.14 	imes (1.2 	imes 10^{-15})^3}$.$ho = \frac{4.8 	imes 10^{-27}}{12.56 	imes 1.728 	imes 10^{-45}} \approx 2.2 	imes 10^{17} 	ext{ kg m}^{-3}$.

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

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