Q. The temperature at which protons in proton gas would have enough energy to overcome Coulomb barrier of $4.14 \times 10^{- 14}$ J is (Boltzmann constant =1.38$\times 10^{- 23}\ {JK}^{- 1})\ $
A) $2 \times 10^{9}$ K
B) $10^{9}$ K
C) 6$\times 10^{9}$ K
D) $3 \times 10^{9}$ K
Explanation
(a) Given $\ \ k = 1.38 \times 10^{- 23}\ JK^{- 1}$
The energy of proton gas =$4.14 \times 10^{- 14}$ J
$E = \frac{3}{2}kT$
$4.14 \times 10^{- 14} = \frac{3}{2} \times 1.38 \times 10^{- 23} \times T$
$T = 2 \times 10^{9}$ K.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.