# (solved)Question 2.28 of NCERT Class XI Physics Chapter 2

The unit of length convenient on the nuclear scale is a fermi : 1 f = 10^-15 m. Nuclear sizes obey roughly the following empirical relation : r = r0 A^1/3 where r is the radius of the nucleus, A its mass number, and ro is a constant equal to about, 1.2 f. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.
(Rev. 24-Aug-2023)

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### Question 2.28 NCERT Class XI Physics

The unit of length convenient on the nuclear scale is a fermi : 1 f = 10^-15 m. Nuclear sizes obey roughly the following empirical relation : r = r0 A^1/3 where r is the radius of the nucleus, A its mass number, and ro is a constant equal to about, 1.2 f. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.

### Physical Concept

Take nucleus to be a sphere of radius r to get its volume.

Mass of the nucleus is mass of 1 proton x A - the atomic mass number

The above will be used to obtain the density m/V.

### Video Explanation

Please watch this youtube video for a quick explanation of the solution:

### Solution

Let $\displaystyle m_p$ = mass of proton/neutron

hence, mass of nucleus = $\displaystyle m_p \times A \text{ kg}$

\begin{aligned} \text{density } & = \frac{\text{mass of nucleus}}{\text{volume of nucleus}}\\\\ &= \frac{m_p \times A}{4\pi/3 \times r^3}\\\\ &= \frac{m_p \times A}{4\pi/3 \times (r_0 A^{1/3})^3}\\\\ &= \frac{m_p \times \bcancel{A}}{4\pi/3 \times (r_0^3 \times \bcancel{A})}\\\\ &= \frac{m_p}{4\pi/3 \times r_0^3}\\\\ &= \frac{1.67 \times 10^{-27}}{4\pi/3 \times (1.2 \times 10^{-15})^3}\\\\ &=2.3 \times 10^{17} \text{ kg m}^{-3}\\ &\text{(rounded to 2 SF due to r}_0 \text{)}\\\\ &= \text{constant} \:\underline{QED}\\ \end{aligned}

\begin{aligned} &\underline{\text{Part II}}\\ &\text{from above density of }\\ &\text{sodium nucleus } \\ &=2.3 \times 10^{17} \text{kg m}^{-3}\:\underline{Ans} \end{aligned}

NOTE: The answer in the NCERT book 0.3 x 10^18 seems wrong. Could be a misprint 0.3 instead of 2.3?