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At the center of the Sun the density of the hydrogen is 35% from the gas density,

${\mathit{蚁}}_{\mathbf{H}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{35}\mathit{蚁}$

The gas density is the number density of H multiplied by the mass of H.

${\mathit{蚁}}_{\mathbf{H}}\mathbf{=}{\mathit{n}}_{\mathbf{H}}{\mathit{m}}_{\mathbf{H}}$

Combine above two equations.

$$\begin{gathered} {\mathit{n}}_{\mathbf{H}}{\mathit{m}}_{\mathbf{H}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{35}\mathit{蚁} \\ {\mathit{n}}_{\mathbf{H}}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{35}\mathbf{蚁}}{{\mathbf{m}}_{\mathbf{H}}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right) \end{gathered}$$

Here,$\mathit{蚁}$ is density, and${\mathit{m}}_{\mathbf{H}}$ is mass of hydrogen atom.

Substitute all the known values in equation (1).

$$\begin{gathered} n_H=\frac{0.35\left(1.5脳{10}^{5}kg/m^3\right)}{1.67脳{10}^{-27}kg} \\ =3.1脳{10}^{31}protons/m^3 \end{gathered}$$

Therefore, the number of density of protons is$3.1脳{10}^{31}protons/m^3$ .

From the ideal gas law,

$$\begin{gathered} PV=NkT \\ \frac{N}{V}=\frac{P}{kT}...........\left(2\right) \end{gathered}$$

Substitute all the known values in equation (2).

$$\begin{gathered} \frac{N}{V}=\frac{1.01脳{10}^{5}Pa}{\left(1.38脳{10}^{-23}J/K\right)\left(273K\right)} \\ =2.68脳{10}^{25}protons/m^3 \end{gathered}$$

Find the ratio of proton density to the density of particles as follows.

$$\begin{gathered} \frac{n_H}{N/V}=\frac{3.1脳{10}^{31}protons/m^3}{2.68脳{10}^{25}protons/m^3} \\ =1.2脳{10}^6 \end{gathered}$$

Therefore, the required ratio is$1.2脳{10}^6$ .