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Chapter 11, Problem 6C
To determine

To verify:

The Q-values for fusion of two deuterium atoms and helium - 3 nucleus.

Expert Solution & Answer
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Explanation of Solution

Given info:

Nuclear Reaction

= H12+H23eH24e +p11+18.3 MeV

Mass of Helium- 4 ( H24e)

= 4.00260 u

Mass of Helium- 3

= 3.01603 u

Mass of nucleus

= 2.01410 u

Mass of deuterium

= 2.01410 u

Mass of neutron

= 1.00896 u

Mass of proton

= 1.00875 u

Speed of light in vacuum

= 3×108 m/s.

Formula used:

H12+H12H23e +n01+Q1

The mass defect relation and its Q value from Einstein’s Mass- energy relation is given by,

Δm1 = (2mH (mHe+ mn) (1)Q1 = Δm1c2(2)

For the second equation,

H12+H23eH24e +P11+Q2

The mass defect relation and its Q value from Einstein’s Mass- energy relation is given by,

Δm2 = (mH+mH23e (mH24e+ mP) (3)Q2 = Δm2c2(4).

Calculation:

Substituting the given values in equation 1 and 2, we get

Δ m1 = (2×2.01410 u)  (3.01603 u + 1.00869 u)Δ m1 = 0.00348 uΔ m1=0.00348 u × ( 1.66× 10 27 kg 1 u)Δ m1 = 5.7768×1030kgQ1 = 5.7768×1030kg × (3×108m/s)2 Q1 = 5.2×1013JQ1 = 5.2×1013×( 1 eV 1.6× 10 19  J)=3.3×106eVQ1 = 3.3×106eV ×( 1 MeV 1× 10 3  eV)

Q1 = 3.3 MeV

For second equation substituting the values in 3 and 4,

Δ m2 = (2.01410 u + 3.01603 u)  (4.00260 u + 1.00785 u)Δ m2 = 0.01968 uΔ m2=0.01968 u × ( 1.66× 10 27 kg 1 u)Δ m2 = 3.26688×1029kgQ2 = 3.26688×1029kg × (3×108m/s)2 Q2 = 2.941×1012JQ2 = 2.941×1012×( 1 eV 1.6× 10 19  J)=18.3×106eVQ2 = 18.3×106eV ×( 1 MeV 1× 10 3  eV)

Q2 = 18.3 MeV.

Conclusion:

Thus, the values for fusion of two deuterium atoms and helium - 3 nucleus is verified.

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