The temperature of the interior of the sun is about \(1.5\times10^{7}\ \mathrm{K}\). The average kinetic energy of a proton at this temperature is about
(Take \(k=1.38\times10^{-23}\ \mathrm{J\,K^{-1}}\) and \(1\ \mathrm{eV}=1.6\times10^{-19}\ \mathrm{J}\).)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1.9 keV
In a hot gas, the average kinetic energy of a particle at temperature \( T \) is \( K=\frac{3}{2}kT \).
\( \begin{aligned} K &= \frac{3}{2}kT \\ &= 1.5\times1.38\times10^{-23}\times1.5\times10^{7} \\ &\approx 3.1\times10^{-16}\ \mathrm{J} \end{aligned} \)
In electron volts,
\( \begin{aligned} K &= \frac{3.1\times10^{-16}}{1.6\times10^{-19}}\ \mathrm{eV} \\ &\approx 1.9\times10^{3}\ \mathrm{eV} \\ &= 1.9\ \mathrm{keV} \end{aligned} \)
Therefore, an average proton in the sun has about 1.9 keV of kinetic energy. This is far below the barrier of about 400 keV for two protons.
NCERT (2026-27):
Nuclear fusion – energy generation in stars – “… We can estimate the temperature at which two protons in a proton gas would (averagely) have enough energy to overcome the coulomb barrier: (3/2)k T = K ≃ 400 keV, which gives T ~ 3 × 109 K.”