Explanation Consider that the energies are small enough that the classical mechanics is applicable. Write the equation for 1-D momentum conservation. p p = p n + p N = ( m n + m N ) v (I) Here, p p is momentum of proton, p n is the momentum of neutron, p N is the momentum of nitrogen, m n is the mass of neutron, m N is the mass of nitrogen and v is the speed. Write the equation for kinetic energy. K p = p p 2 2 m p Substitute equation I in the above equation. K p = [ ( m n + m N ) v ] 2 2 m p = ( m n + m N ) m p 1 2 ( m n + m N ) v 2 = ( m n + m N ) m p ( K n + K N ) (II) Here, K p is the kinetic energy of proton, K n is the kinetic energy of neutron and K N is the kinetic energy of nitrogen. Write the equation for energy conservation. K p + Q = K n + K N (III) Here, Q is the reaction energy. Substitute equation II in III and rewrite to find kinetic energy of proton. K p + Q = m p ( m n + m N ) K p K p = − Q m n + m N m n + m N − m p (IV) Write the equation for Q value

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Chapter 42, Problem 16P

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Prove that the bombarding proton energy must be 3.23 MeV for the given reaction to occur.

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Chapter 42, Problem 15P

Chapter 42, Problem 17P

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