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NCERT Class XII Chapter
Nuclei
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Question : 15 of 31
Marks: +1, -0
The Q value of a nuclear reaction A + b → C + d is defined by Q = [mA+mb−mc−md]c2[m_A + m_b - m_c - m_d]c^2, 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) 11H+13H{}_1^1\mathrm{H} + {}_1^3\mathrm{H} → 12H+12H{}_1^2\mathrm{H} + {}_1^2\mathrm{H}
(ii) 612C+612C{}_6^{12}\mathrm{C} + {}_6^{12}\mathrm{C} → 1020Ne+24He{}_{10}^{20}\mathrm{Ne} + {}_2^4\mathrm{He}
Atomic masses are given to be
m (11H)({}_1^1\mathrm{H}) = 1.007825 u, m (12H)({}_1^2\mathrm{H}) = 2.014102 u, m (13H)({}_1^3\mathrm{H}) = 3.016049 u,
m (612C)({}_6^{12}\mathrm{C}) = 12.000000 u , m (1020Ne)({}_{10}^{20}\mathrm{Ne}) = 19.992439u
Solution:  
(i) Let us find the Q value in given first equation
11H+13H{}_1^1\mathrm{H} + {}_1^3\mathrm{H} → 12H+12H{}_1^2\mathrm{H} + {}_1^2\mathrm{H}
Q =
[m(11H)+m(13H)−2m(12H)]c2[m({}_1^1\mathrm{H}) + m({}_1^3\mathrm{H}) - 2m({}_1^2\mathrm{H})]c^2
= [1.007825 + 3.016049 – 2 × 2.014102] × (931 MeV)
Q = [4.023874 – 4.028204] 931.5 MeV = – 4.033 MeV
Negative Q value shows that reaction is endothermic.
(ii) Q value in the given second equation
612C+612C{}_6^{12}\mathrm{C} + {}_6^{12}\mathrm{C} → 1020Ne+24He{}_{10}^{20}\mathrm{Ne} + {}_2^4\mathrm{He}
Q =
[2m(612C)−m(1020Ne)−m(24He)]c2[2m({}_6^{12}\mathrm{C}) - m({}_{10}^{20}\mathrm{Ne}) - m({}_2^4\mathrm{He})]c^2
Q = [2 × 12.0000 – 19.992439 – 4.002603] × 931.5 MeV
Q = 0.004958 × 931.5 MeV = 4.618 MeV
Positive Q shows that the reaction is exothermic.
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