According to classical physics, 10^{-15} m is | vedclass

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Question

${\lambda _P} = \frac{h}{{{M_P}{V_{rms}}}}$ and ${d_C} = \frac{{{\lambda _p}}}{{\sqrt 2 }}$

$\frac{3}{2}{\rm{KT}} = \frac{1}{2}{{\rm{M}}_{\rm{P}}}{

Vedclass · Accepted Answer

$\frac{{{e^4}{M_p}}}{{6{\pi ^2}\varepsilon _0^2k{h^2}}}$

Vedclass · Answer

${\lambda _P} = \frac{h}{{{M_P}{V_{rms}}}}$ and ${d_C} = \frac{{{\lambda _p}}}{{\sqrt 2 }}$

$\frac{3}{2}{m{KT}} = \frac{1}{2}{{m{M}}_{m{P}}}{m{V}}_{rms}^2 \Rightarrow {{m{V}}_{rms}} = \sqrt {\frac{{3{m{KT}}}}{{{{m{M}}_{m{p}}}}}} $

and for fusion,

$2\left( {\frac{1}{2}{M_P}{V_{rm{s^2}}}} ight) = \frac{{{q^2}}}{{4\pi {\varepsilon _0}{d_C}}} \Rightarrow {T_C} = \frac{{{q^2}}}{{12\pi {\varepsilon _0}{d_C}K}}$

$ \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}}}}$

$\Rightarrow \mathrm{T}_{\mathrm{C}}=\frac{\mathrm{q}^{2}}{\frac{12 \pi \varepsilon_{0} \mathrm{k}}{\sqrt{2}} 	imes \frac{\mathrm{h}}{\mathrm{M}_{\mathrm{P}} \mathrm{V}_{\mathrm{ma}}}}=\frac{\mathrm{q}^{2} \mathrm{M}_{\mathrm{P}}(\sqrt{\frac{3 \mathrm{KT}_{\mathrm{C}}}{\mathrm{M}_{\mathrm{p}}}}) \sqrt{2}}{12 \pi \varepsilon_{0} \mathrm{Kh}}$

squaring

$\Rightarrow \mathrm{T}_{\mathrm{C}}^{2} \frac{\mathrm{q}^{4} \mathrm{M}_{\mathrm{P}} 	imes \frac{3 \mathrm{KT}_{\mathrm{C}}}{\mathrm{M}_{\mathrm{P}}} 	imes 2}{144 \pi^{2} \varepsilon_{0}^{2} \mathrm{K}^{2} \mathrm{h}^{2}} \Rightarrow \mathrm{T}_{\mathrm{C}}=\frac{\mathrm{q}^{4} \mathrm{M}_{\mathrm{P}}}{24 \pi^{2} \varepsilon_{0} \mathrm{Kh}^{2}}$

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