The temperature at which protons in proton gas would have enough energy to overcome Coulomb barrier of 4.14× 10-14J is (Boltzmann constant ) =1.38 × 10-23 JK-1 )

Question

The temperature at which protons in proton gas would have enough energy to overcome Coulomb barrier of 4.14× 10-14J is (Boltzmann constant ) =1.38 × 10-23 JK-1 ) (A)  2× 109K  (B)  109K  (C)  6× 109K  (D)  3× 109K

Tardigrade Admin · Accepted Answer

Given: k=1.38× 10-23JK-1 The energy of proton gas =4.14× 10-14J ∴ E=(3/2)kT or 4.14× 10-14=(3/2)× 1.38× 10-23× T T=2× 109K

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 )
$= 1.38 \times 10^{- 23} J K^{- 1}$ )

$2 \times 10^{9} K$

$10^{9} K$

$6 \times 10^{9} K$

$3 \times 10^{9} K$

$4.5 \times 10^{9} K$

Given: $k = 1.38 \times 10^{- 23} J K^{- 1}$
The energy of proton gas
$= 4.14 \times 10^{- 14} J$
$∴$ $E = \frac{3}{2} k T$
or $4.14 \times 10^{- 14} = \frac{3}{2} \times 1.38 \times 10^{- 23} \times T$
$T = 2 \times 10^{9} K$

Q. The temperature at which protons in proton gas would have enough energy to overcome Coulomb barrier of 4.14×10−14J is (Boltzmann constant ) =1.38×10−23JK−1 )

2×109K

109K

6×109K

3×109K

4.5×109K

Given: k=1.38×10−23JK−1 The energy of proton gas =4.14×10−14J ∴ E=23​kT or 4.14×10−14=23​×1.38×10−23×T T=2×109K