In the nuclear fusion- 12H-13H- 24He-n given that the repulsive potential energy between the two nuclie is 7-7- 10-14J- the temperature at which the gases must be heated to initiate the reaction is nearly -Boltzmann-s constant k-1-38- 10-23J-K-

The correct option is D 10^9K Energy #160; #160; E ≈ kT So, #160; #160; #160; 7.7× 10^ 14≈ 1.38× 10^ 23× T T≈ #160; 5.6× 10^9 ...

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Standard X

Physics

Nuclear Fusion

In the nuclea...

Question

In the nuclear fusion, $_{1}^{2} H +_{1}^{3} H \to_{2}^{4} H e + n$ given that the repulsive potential energy between the two nuclie is $7.7 \times 10^{- 14} J$ , the temperature at which the gases must be heated to initiate the reaction is nearly [Boltzmann's constant $k = 1.38 \times 10^{- 23} J / K$ ]-

10^{7} K

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10^{5} K

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10^{3} K

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10^{9} K

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Solution

The correct option is D $10^{9} K$
Energy $E \approx k T$
So, $7.7 \times 10^{- 14} \approx 1.38 \times 10^{- 23} \times T$
$⟹ T \approx 5.6 \times 10^{9} K$
Correct answer is option D.

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Similar questions

Q. In the nuclear fusion reaction,

_{1} H^{2} +_{1} H^{3} \to_{2} {He}^{4} +_{0} n^{1}

The repulsive potential energy between the two fusing nuclei is

7.7 \times 10^{- 14} J

. The temperature to which the gas must be heated is nearly

(Boltzmen constant K = 1.38 \times 10^{- 23} {JK}^{- 1})

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. At what temperature does the average transnational kinetic energy of a molecule in a gas become equal to kinetic energy of an electron accelerated from rest through a potential difference of

1

volt? (

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

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In the nuclear fusion, 21H+31H→42He+n given that the repulsive potential energy between the two nuclie is 7.7×10−14J, the temperature at which the gases must be heated to initiate the reaction is nearly [Boltzmann's constant k=1.38×10−23J/K]-

The correct option is D 109KEnergy E≈kTSo, 7.7×10−14≈1.38×10−23×T⟹ T≈5.6×109 KCorrect answer is option D.

The correct option is D $10^{9} K$
Energy $E \approx k T$
So, $7.7 \times 10^{- 14} \approx 1.38 \times 10^{- 23} \times T$
$⟹ T \approx 5.6 \times 10^{9} K$
Correct answer is option D.