In the nuclear fusion reaction - 21H- 31H - 42He-n - given that the repulsive potential energy between the two nuclei is - 7-7 - 10-14J- the temperature to which the gases must be heated to initiate the reaction is nearly Boltzmann-s constant k-1-38 x 10 -23J

The correct option is B 10^9K The repulsive potential energy =32kT 7.7× 10^ 14=32× 1.38×× 10^ 23× T → T = 3.7× 10^9 K ∼ 10^9 K ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Physics

Nuclear Fusion

In the nuclea...

Question

In the nuclear fusion reaction : $_{1}^{2} H +_{1}^{3} H \to$ $_{2}^{4} H e + n$ , given that the repulsive potential energy between the two nuclei is $\sim 7.7 \times 10^{- 14}$ J, the temperature to which the gases must be heated to initiate the reaction is nearly
(Boltzmann's constant $k =$ 1.38 x 10 $^{- 23}$ J)

10^{7} K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10^{5} K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10^{3} K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10^{9} K

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is B $10^{9} K$
The repulsive potential energy $= \frac{3}{2} k T$

$7.7 \times 10^{- 14} = \frac{3}{2} \times 1.38 \times \times 10^{- 23} \times T$

$\to T = 3.7 \times 10^{9} K \sim 10^{9} K$

Suggest Corrections

Similar questions

Q. Consider the DT reaction (deuteriumtritium fusion).

_{1}^{2} H +_{1}^{3} H \to_{2}^{4} H e + n

(a) Calculate the energy released in MeV in this reaction from the data:

m (_{1}^{2} H) = 2.014102 u

m (_{1}^{3} H) = 3.016049 u

(b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction? (Hint: Kinetic energy required for one fusion event =average thermal kinetic energy available with the interacting particles = 2(3kT/2); k = Boltzmans constant, T = absolute temperature.)

Q. Calculate
Given Avogadro's number

= 6.02 \times 10^{23}

and Boltzmann's constant

= 1.38 \times 10^{- 23} J / (m o l - K)

Q. Consider the D–T reaction (deuterium–tritium fusion)2 3 41 1 2 H H He n + → +(a) Calculate the energy released in MeV in this reaction from thedata:m(21H )=2.014102 um(31H ) =3.016049 u(b) Consider the radius of both deuterium and tritium to beapproximately 2.0 fm. What is the kinetic energy needed toovercome the coulomb repulsion between the two nuclei? To whattemperature must the gas be heated to initiate the reaction?(Hint: Kinetic energy required for one fusion event =averagethermal kinetic energy available with the interacting particles= 2(3kT/2); k = Boltzman’s constant, T = absolute temperature.)

Consider the D−T reaction (deuterium−tritium fusion)

(a) Calculate the energy released in MeV in this reaction from the data:

= 2.014102 u

= 3.016049 u

(b)Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction? (Hint: Kinetic energy required for one fusion event =average thermal kinetic energy available with the interacting particles = 2(3kT/2); k = Boltzman’s constant, T = absolute temperature.)

Q. The average translational kinetic energy of air molecules is

0.040 e V

(

1 e V = 1.6 \times 10^{- 19} J

). Calculate the temperature of the air. Boltzman's constant

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

Join BYJU'S Learning Program

In the nuclear fusion reaction : 21H+31H→ 42He+n , given that the repulsive potential energy between the two nuclei is ∼7.7×10−14J, the temperature to which the gases must be heated to initiate the reaction is nearly (Boltzmann's constant k=1.38 x 10−23J)

The correct option is B 109KThe repulsive potential energy =32kT7.7×10−14=32×1.38××10−23×T→ T=3.7×109K∼109K

The correct option is B $10^{9} K$
The repulsive potential energy $= \frac{3}{2} k T$

$7.7 \times 10^{- 14} = \frac{3}{2} \times 1.38 \times \times 10^{- 23} \times T$

$\to T = 3.7 \times 10^{9} K \sim 10^{9} K$