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According to classical physics, $10^{-15}\ m$ is distance of closest approach $(d_c)$ for fusion to occur between two protons. A more accurate and quantum approach says that ${d_c} = \frac{{{\lambda _p}}}{{\sqrt 2 }}$ where $'\lambda _p'$ is de-broglie's wavelength of proton when they were far apart. Using quantum approach, find equation of temperature at centre of star. [Given: $M_p$ is mass of proton, $k$ is boltzman constant]
$\frac{{{e^4}{M_p}}}{{24{\pi ^2}\varepsilon _0^2k{h^2}}}$
$\frac{{{e^4}{M_p}}}{{12{\pi ^2}\varepsilon _0^2k{h^2}}}$
$\frac{{{e^2}{M_p}}}{{24{\pi ^2}\varepsilon _0^2k{h^2}}}$
$\frac{{{e^4}{M_p}}}{{6{\pi ^2}\varepsilon _0^2k{h^2}}}$
Solution
${\lambda _P} = \frac{h}{{{M_P}{V_{rms}}}}$ and ${d_C} = \frac{{{\lambda _p}}}{{\sqrt 2 }}$
$\frac{3}{2}{\rm{KT}} = \frac{1}{2}{{\rm{M}}_{\rm{P}}}{\rm{V}}_{rms}^2 \Rightarrow {{\rm{V}}_{rms}} = \sqrt {\frac{{3{\rm{KT}}}}{{{{\rm{M}}_{\rm{p}}}}}} $
and for fusion,
$2\left( {\frac{1}{2}{M_P}{V_{rm{s^2}}}} \right) = \frac{{{q^2}}}{{4\pi {\varepsilon _0}{d_C}}} \Rightarrow {T_C} = \frac{{{q^2}}}{{12\pi {\varepsilon _0}{d_C}K}}$
$ \Rightarrow {T_C} = \frac{{{q^2}}}{{12\pi {\varepsilon _0}\frac{{\lambda p}}{{\sqrt 2 }}K}}\quad {\lambda _p} = \frac{h}{{{M_p}{V_{rms}}}}$
$\Rightarrow \mathrm{T}_{\mathrm{C}}=\frac{\mathrm{q}^{2}}{\frac{12 \pi \varepsilon_{0} \mathrm{k}}{\sqrt{2}} \times \frac{\mathrm{h}}{\mathrm{M}_{\mathrm{P}} \mathrm{V}_{\mathrm{ma}}}}=\frac{\mathrm{q}^{2} \mathrm{M}_{\mathrm{P}}(\sqrt{\frac{3 \mathrm{KT}_{\mathrm{C}}}{\mathrm{M}_{\mathrm{p}}}}) \sqrt{2}}{12 \pi \varepsilon_{0} \mathrm{Kh}}$
squaring
$\Rightarrow \mathrm{T}_{\mathrm{C}}^{2} \frac{\mathrm{q}^{4} \mathrm{M}_{\mathrm{P}} \times \frac{3 \mathrm{KT}_{\mathrm{C}}}{\mathrm{M}_{\mathrm{P}}} \times 2}{144 \pi^{2} \varepsilon_{0}^{2} \mathrm{K}^{2} \mathrm{h}^{2}} \Rightarrow \mathrm{T}_{\mathrm{C}}=\frac{\mathrm{q}^{4} \mathrm{M}_{\mathrm{P}}}{24 \pi^{2} \varepsilon_{0} \mathrm{Kh}^{2}}$
