   Chapter 10, Problem 62P ### University Physics Volume 3

17th Edition
William Moebs + 1 other
ISBN: 9781938168185

#### Solutions ### University Physics Volume 3

17th Edition
William Moebs + 1 other
ISBN: 9781938168185
Textbook Problem

# Two fusion reactions mentioned in the text are n + 3 H e → 4 H e + γ and n + 1 H → 2 H + γ . Both reactions release energy, but the second also creates more fuel. Con film that the energies produced in the reactions are 20.58 and 2.22 MeV, respectively. Comment on which product nuclide is most tightly bound, 4He or 2H.

To determine

The total energy released in reaction 1 is 20.58MeV and the total energy released in reaction 2 is 2.22MeV, and also compare which nucleus is more tightly bound.

Explanation

Given:

The fusion reaction 1 is given by,

n+3He+3He4He+γ

The fusion reaction 1 is given by,

n+1H2H+γ

Formula used:

The total energy released in reaction 1 is given by,

E1=Δm1c2

Here, Δm1 is the mass defect, and c is velocity of light.

The mass defect in reaction 1 is given by,

Δm1=n+m3Hem4Hemγ

Here, n is the mass of the neutron, m3He is the mass of the nucleus of 3He, m4He is the mass of the nucleus of 4He, and mγ is the mass of photon emitted.

The energy released during the reaction 2 is given by,

E2=Δm2c2

Here, Δm2 is the mass defect.

The mass defect in reaction 2 is given by,

Δm2=n+m1Hm2Hmγ

Here, m1H is the mass of the nucleus of 1H, m2H is the mass of the nucleus of 2H.

Calculation:

The mass defect in reaction 1 is calculated by,

Δm1=n+m3Hem4Hemγ=1.008645u+3.016049u4.002603u0u=4.024694u4.002603u=0.022091u

The total energy released in reaction 1 is calculated by,

E1=Δm1c2=(0

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