From the relation R = R₀A^1/3, where R₀ is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e., independent of A).

Solution

Radius of a nucleus of mass number A is given by,

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

where,

R_{0} = 1.2 \times 10^{- 15} m .

This implies that the volume of the nucleus, which is proportional to R³ is proportional to A.

Volume of nucleus =

\frac{4}{3} {πr}^{3}

= \frac{4}{3} π (R_{0} A^{1 / 3})^{3} = \frac{4}{3} π R_{0}^{3} A

Density of nucleus =

\frac{mass of nucleus}{volume of nucleus}

= \frac{mA}{\frac{4}{3} {πR}_{0}^{3} A}

= \frac{3 m}{4 π R_{0}^{3}}

The above derived equation shows that density of nucleus is constant, independent of A for all nuclei. Density of nuclear matter is approximately 2.3 × 10⁷ kg/m³ which is very large as compared to ordinary matter.

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From the relation R = R₀A^1/3, where R₀ is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e., independent of A).

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Wave Optics

From the relation R = R0A1/3, where R0 is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e., independent of A).

Some More Questions From Wave Optics Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

From the relation R = R₀A^1/3, where R₀ is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e., independent of A).