Two protons must cross a Coulomb barrier of about 400 keV to fuse. In a gas of protons, the average kinetic energy of a proton equals this barrier height at a temperature of 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 4
The correct answer is Option (4) → \(3.1\times10^{9}\ \mathrm{K}\)
In a hot gas, the average kinetic energy of a particle is \( K=\frac{3}{2}kT \). We set this equal to the barrier height and solve for \( T \).
\( \begin{aligned} K &= 400\times10^{3}\times1.6\times10^{-19}\ \mathrm{J} \\ &= 6.4\times10^{-14}\ \mathrm{J} \end{aligned} \)
\( \begin{aligned} T &= \frac{2K}{3k} \\ &= \frac{2\times6.4\times10^{-14}}{3\times1.38\times10^{-23}} \\ &\approx 3.1\times10^{9}\ \mathrm{K} \end{aligned} \)
Therefore, protons of average energy would cross the barrier only at a temperature of about \( 3\times10^{9}\ \mathrm{K} \).
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.”