13.21 From the relation R = <math><mo><math><msub><mrow><mrow><mi mathvariant="bold-italic">R</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><msup><mrow><mrow><mi mathvariant="bold-italic">A</mi></mrow></mrow><mrow><mrow><mn>1</mn><mo>/</mo><mn>3</mn></mrow></mrow></msup></math></mo></math> , where<math><msub><mrow><mrow><mi mathvariant="bold-italic">R</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></math> is a constant and A is the mass number of a nucleus, shows that the nuclear matter density is nearly constant (i.e. independent of A).

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Physics Ncert Solutions Class 12th Physics Class 12th Nuclei

13.21 From the relation R = $R_{0} A^{1 / 3}$
, where $R_{0}$ is a constant and A is the mass number of a nucleus, shows that the nuclear matter density is nearly constant (i.e. independent of A).

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Payal Gupta | Contributor-Level 10

4 months ago

13.21 We have the expression for nuclear radius as:

R = $R_{0} A^{1 / 3}$

Where $R_{0}$ = constant

A = mass number of nucleus

Let m be the average mass of the nucleus, hence mass of the nucleus = mA

Nuclear matter density $ρ$ can be written as

$ρ = \frac{M a s s o f t h e n u c l e u s}{V o l u m e o f t h e n u c l e u s}$ = $\frac{m A}{\frac{4}{3} π R^{3}}$ = $\frac{3 m A}{4 π {(R_{0} A^{\frac{1}{3}})}^{3}}$ = $\frac{3 m A}{4 π R_{0}^{3} A}$ = $\frac{3 m}{4 π R_{0}^{3}}$

Hence, the nuclear mass density is independent of A. It is nearly constant

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13.21 From the relation R = $R_{0} A^{1 / 3}$
, where $R_{0}$ is a constant and A is the mass number of a nucleus, shows that the nuclear matter density is nearly constant (i.e. independent of A).

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13.21 From the relation R = R0A1/3 , whereR0 is a constant and A is the mass number of a nucleus, shows that the nuclear matter density is nearly constant (i.e. independent of A).

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13.21 From the relation R = $R_{0} A^{1 / 3}$
, where $R_{0}$ is a constant and A is the mass number of a nucleus, shows that the nuclear matter density is nearly constant (i.e. independent of A).