In a fusion reaction for two protons to overcome a Coulomb potential barrier of height of 400 keV, the minimum temperature required is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 3.0\times10^{9} \) K
In thermonuclear fusion the protons get their energy from thermal motion. Taking the average thermal kinetic energy of a proton, \( \frac{3}{2}kT \), equal to the barrier height gives the temperature needed.
\( \displaystyle \frac{3}{2}kT = 400\,\mathrm{keV} = 4\times10^{5}\,\mathrm{eV} \)
\( \displaystyle T = \frac{2 \times 4\times10^{5}\,\mathrm{eV}}{3 \times 8.62\times10^{-5}\,\mathrm{eV\,K^{-1}}} \)
\( \displaystyle T \approx 3.1\times10^{9}\,\mathrm{K} \approx 3.0\times10^{9}\,\mathrm{K} \)
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 – “… 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:
\( \displaystyle (3/2)kT = K \simeq 400\ \mathrm{keV} \)
which gives T ~ 3 × 109 K.”