Question

The Q value of a nuclear reaction A + b → C + d is defined by
Q = [m_A+m_b–m_C–m_d]c² 
where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.
(i) \(^{1}_{ 1} H + ^{3} _{1}H → ^{2}_{1} H + ^{2}_{1} H\)
(ii) \(^{12}_{6} C + ^{12}_{6} C → ^{12}_{10} Ne + ^{4}_{2} He\)
Atomic masses are given to be
\(m (^{2}_{ 1} H)\) = 2.014102 u
\(m (^{3}_{1} H) \)= 3.016049 u
\(m (^{12}_{ 6} C) \)= 12.000000 u
\(m (^{20}_{10} Ne)\) = 19.992439 u

Answer 1

(i) The given nuclear reaction is:
\(^{1} _{1} H + ^{3}_{1} H → ^{2}_{1} H + ^{2} _{1} H\)
It is given that:
Atomic mass \( m (^{1}_{1} H) \)= 1.0078 25 u
Atomic mass \(m (^{3}_{1} H)\) = 3.016049 u
Atomic mass \(m (^{2}_{1} H)\) = 2.014102 u
According to the question, the Q-value of the reaction can be written as:
Q = [\( m (^{1}_{1} H) \) + \(m (^{3}_{1} H)\)- \(m (^{2}_{1} H)\)]c²
Q = [1.0078 25 + 3.016049 -2 x 2.014102]c²
Q = (-0.00433 c²)u
But 1u = 931.5 \(\frac{Mev}{c^{2}}\)
Q = -0.00433 x 931.5 = -4.0334 MeV
The negative Q-value of the reaction shows that the reaction is endothermic.

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(ii) The given nuclear reaction is:
\(^{12}_{6} C + ^{12}_ {6} C → ^{12}_{10} Ne + ^{4 }_{2} He\)
It is given that:
Atomic mass of m \((^{12}_{6} C) = 12.000000 u\)
Atomic mass of \(m (^{20}_{10} Ne)\) = 19.992439 u
Atomic mass of \(m (^{4}_{2} He)\) = 4.002603 u
The Q-value of this reaction is given as:
Q = [2m\((^{12}_{6}C)\) - \(m(^{20}_{10}Ne)\) - \(m(^{4}_{2}He)]c^2\)
Q = [2 x 12.0 -19.992439 - 4.002603]c²
Q = (0.004958 c²⁾ u
Q = 0.004958 x 931.5
Q = 4.618377 MeV
The positive Q-value of the reaction shows that the reaction is exothermic.