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## Q. 16.5

Determine the radius of a neutron star with a mass of 2 solar masses.

Strategy We use Equation (16.16) to determine the volume, and from that we can determine the radius. One solar mass $=1.99 \times 10^{30}$ kg.

$V^{1 / 3}=\frac{6.5 \hbar^{2}}{N^{1 / 3} m^{3} G}$ (16.16)

## Verified Solution

First, we find the number of neutrons N.

$N=\frac{2 M_{\text {sun }}}{m_{\text {neutron }}}=\frac{2\left(1.99 \times 10^{30} kg \right)}{1.675 \times 10^{-27} kg }=2.4 \times 10^{57} \text { neutrons }$

We now determine the cube root of the volume to be $V^{1 / 3}$

\begin{aligned}&=\frac{6.5\left(1.06 \times 10^{-34} J \cdot s \right)^{2}}{\left(2.4 \times 10^{57}\right)^{1 / 3}\left(1.675 \times 10^{-27} kg \right)^{3}\left(6.67 \times 10^{-11} J \cdot m / kg ^{2}\right)} \\&=1.7 \times 10^{4} m\end{aligned}

The radius R is calculated to be, from $V=\frac{4}{3} \pi R^{3}$,

$R=\left(\frac{3}{4 \pi}\right)^{1 / 3} V^{1 / 3}=\left(\frac{3}{4 \pi}\right)^{1 / 3}\left(1.7 \times 10^{4} m \right)=11 km$

The radius of a neutron star twice as massive as our sun is only 11 km! It is interesting to compare the density of a neutron star with that of a typical nucleus and with a nucleon (see Problem 11).