Question

Calculate the energy released (or energy input required) for the reaction

69-31Ga + n → α + 66-29Cu.

Q=

Answer 1

Energy released in above reaction will be given by:

Q = dm*c^2

Q = (m_Ga + m_n - m_alpha - mCu)*c^2

mn = mass of neutron = 1.008665 u

m_alpha = mass of alpha particle = 4.002602 u

m-Ga = mass of Galium = 69.723 u

m_Cu = mass of Copper = 63.546 u

(See that Check these values with your reference book for accurate answer, if any of these values are different than let me know in comments)

Using these values:

Q = (69.723 u + 1.008665 u - 4.002602 u - 63.546 u)*(931.5 MeV/u)

Q = 3.183063 u*(931.5 MeV/u)

Q = +2965.02318 MeV = Energy release from the above reaction

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