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Chapter 6: Q36E (page 1520)

Calculate the energy (in MeV) released in the triple-alpha process localid="1667794947974" 34He鈫�C12.

Short Answer

Expert verified

The energy released in the triple-alpha process 34He鈫�C13 is 7.274MeV.

Step by step solution

01

Define the reaction energy

The reaction energy is determined by using the expression:

Q=[MA+MB-MC-MD]脳c2

Where, MAand MBare the mass of reactants, MQand MDare the mass of products.

c is the speed of light that and c2=931.5MeV/u.

02

Determine the reaction energy

The given reaction is:

34He鈫�C12

The atomic mass of He is 4.002503u and the atomic mass of C is 12.0u.

Now, Q=3MH4胃-MC12

Substituting the values, and we get,

Q=[3(4.002603-12.0)](931.5)=(0.007809)(931.5)=7.274MeV

Hence, the energy released in the triple-alpha process 34He鈫�C12is 7.274MeV.

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