Determine the resulting nucleus in the reaction
Chapter 31, Problem 31EA(choose chapter or problem)
Determine the resulting nucleus in the reaction \(n+{ }_{56}^{137} B a \rightarrow ?+\gamma\).
Energy and momentum are also conserved in nuclear reactions, and can be used to determine whether or not a given reaction can occur. For example, if the total mass of the final products is less than the total mass of the initial particles, this decrease in mass
(recall \(\Delta E=\Delta m c^{2}\)) is converted to kinetic energy \((K E)\) of the outgoing particles. But if the total mass of the products is greater than the total mass of the initial reactants, the reaction requires energy. The reaction will then not occur unless the bombarding particle has sufficient kinetic energy. Consider a nuclear reaction of the general form
\(a+X \rightarrow Y+b\) (31-1)
where particle a is a moving projectile particle (or small nucleus) that strikes nucleus X, producing nucleus Y and particle b (typically, \(p, n, \alpha, \gamma\)). We define the reaction energy, or Q-value, in terms of the masses involved, as
\(Q=\left(M_{a}+M_{x}-M_{b}-M_{y}\right)c^{2}\) (31-2a)
For a \(\gamma\) ray, \(M=0\). If energy is released by the reaction, \(Q>0 \). If energy is required, \(Q<0\)
Because energy is conserved, Q has to be equal to the change in kinetic energy (final minus initial):
\(Q=K E_{b}+K E_{Y}-K E_{a}-K E_{x}\) (31-2b)
If X is a target nucleus at rest (or nearly so) struck by incoming particle a, then \(K E_{x}=0\). For \(Q>0\), the reaction is said to be exothermic or exoergic; energy is released in the reaction, so the total kinetic energy is greater after the reaction than before. If Q is negative, the reaction is said to be endothermic or endoergic: an energy input is required to make the reaction happen. The energy input comes from the kinetic energy of the initial colliding particles (a and X).
Equation Transcription:
Text Transcription:
n+_56^137 B a \rightarrow ?+\gamma
\Delta E=\Delta m c^{2}
(K E)
a+X \rightarrow Y+b
p, n, \alpha, \gamma
Q=(Ma+Mx-Mb-My)c2
\gamma
M=0
Q>0
Q<0
Q=KEb+KEY-KEa-KEx
KEx=0
Q>0
Unfortunately, we don't have that question answered yet. But you can get it answered in just 5 hours by Logging in or Becoming a subscriber.
Becoming a subscriber
Or look for another answer