In Fig. 43-10, the equation for n(K), the number density per unit energy for particles, n(K)$\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{13}\mathit{n}\frac{{\mathbf{K}}^{\mathbf{1}\mathbf{/}\mathbf{2}}}{{\mathbf{\left(}\mathbf{k}\mathbf{T}\mathbf{\right)}}^{\mathbf{3}\mathbf{/}\mathbf{2}}}{\mathit{e}}^{\mathbf{-}\frac{\mathbf{K}}{\mathbf{k}\mathbf{T}}}$iswhere nis the total particle number density. At the center of the Sun, the temperature is$\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathit{K}$and the average proton energy${\mathit{K}}_{\mathbf{a}\mathbf{v}\mathbf{g}}$is 1.94 keV. Find the ratio of the proton number density at 5.00keVto the number density at the average proton energy.

$n\left(K\right)=1.3n\frac{K^{{\displaystyle \frac{1}{2}}}}{{\left(kT\right)}^{{\displaystyle \frac{3}{2}}}{e}^{\frac{K}{T}}}$

1. The temperature at the center of the sun, role="math" localid="1661752758764" $T_{sun}=1.50\times {10}^{7}K$
2. The average proton energy, ${\mathrm{K}}_{\mathrm{avg}}=1.94\mathrm{keV}$
3. Given proton energy, K = 5.00 keV

Formula:

The equation for number density per unit energy for particles,

$\mathbf{n}\mathbf{\left(}\mathbf{K}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{3}\mathbf{n}\frac{{\mathbf{K}}^{{\displaystyle \frac{1}{2}}}}{{\mathbf{\left(}\mathbf{kT}\mathbf{\right)}}^{{\displaystyle \frac{3}{2}}}{\mathbf{e}}^{{\displaystyle \frac{K}{T}}}}$ …… (i)

From the expression for given, it can be said using equation (i) that,

$n\left(K\right)\alpha K^{\frac{1}{2}}{e}^{-\frac{K}{kT}}$ ….. (a)

Thus, with the value of Boltzmann’s constant as:

$\mathrm{k}=8.62\times {10}^{-5}\frac{\mathrm{eV}}{\mathrm{K}}\mathrm{or}8.62\times {10}^{-8}\frac{\mathrm{keV}}{\mathrm{K}}$

The ratio of theproton number density atto the number density at the average proton energy can be calculated using the above relation (a) and the given data as follows:

$$\begin{gathered} \frac{\mathrm{n}\left(\mathrm{K}\right)}{\mathrm{n}\left({\mathrm{K}}_{\mathrm{avg}}\right)}={\left(\frac{5.00\mathrm{keV}}{1.94\mathrm{keV}}\right)}^{\frac{1}{2}}\exp \left(-\frac{5.00\mathrm{keV}-1.94\mathrm{keV}}{\left(8.62\times {10}^{-8}{\displaystyle \frac{\mathrm{keV}}{\mathrm{K}}}\right)\left(1.50\times {10}^7\mathrm{K}\right)}\right) \\ =0.151 \end{gathered}$$

Hence, the required ratio is 0.151 .