The Coulomb barrier height for two interacting protons is 400 keV. What will be the temperature at which protons in a proton gas will be able to overcome the Coulomb barrier?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 3 × 109 K
Two protons can come close enough to fuse only if their thermal kinetic energy lets them climb the Coulomb barrier. The average kinetic energy of a particle in a gas at temperature T is \( \frac{3}{2}kT \) (Boltzmann constant \( k = 1.38\times10^{-23} \) J/K); in the standard estimate this is set equal to the barrier height.
\( \displaystyle \frac{3}{2}kT = 400\,\mathrm{keV} \)
\( 400\,\mathrm{keV} = 400\times10^{3} \times 1.6\times10^{-19}\,\mathrm{J} = 6.4\times10^{-14}\,\mathrm{J} \)
\( \displaystyle T = \frac{2 \times 6.4\times10^{-14}\,\mathrm{J}}{3 \times 1.38\times10^{-23}\,\mathrm{J/K}} \)
\( T \approx 3.1\times10^{9}\,\mathrm{K} \approx 3\times10^{9}\,\mathrm{K} \)
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Nuclear Fusion – Energy Generation in Stars – “… They, therefore, must have enough energy to overcome this coulomb barrier. The height of the barrier depends on the charges and radii of the two interacting nuclei. It can be shown, for example, that the barrier height for two protons is ~ 400 keV, and is higher for nuclei with higher charges. We can estimate the temperature at which two protons in a proton gas would (averagely) have enough energy to overcome the coulomb barrier:
\( \displaystyle (3/2)k T = K \simeq 400\ \mathrm{keV}, \text{ which gives } T \sim 3 \times 10^{9}\ \mathrm{K} \)”