Ncert Exercises & Solutions

13.15 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) \({ }_{1}^{1} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \rightarrow{ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H}\)

(ii) \({ }_{6}^{12} \mathrm{C}+{ }_{6}^{12} \mathrm{C} \rightarrow{ }_{10}^{20} \mathrm{Ne}+{ }_{2}^{4} \mathrm{He} \)

Atomic masses are given to be

m () = 2.014102 u

m () = 3.016049 u

m () = 12.000000 u

m () = 19.992439 u

Hint: The Q-value of a reaction is given by \(Q=\Delta mc^2\).
(i)
Step 1: Find the Q-value for the first reaction.
The given nuclear reaction is:

It is given that: \({ }_{1}^{1} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \rightarrow{ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H}\)

Atomic mass of

m ({}_{1}^{1}H)

=1.007825 u

Atomic mass of

m ({}_{1}^{3}H)

=3.016049 u

Atomic mass of

m ({}_{1}^{2}H)

=2.014102 u

According to the question, the Q-value of the reaction can be written as:

Q = [m ({}_{1}^{1}H) + m ({}_{1}^{3}H) - 2 m ({}_{1}^{2}H)] c^{2}

= [1.00725 + 3.016049 - 2 \times 2.014102] c^{2}

Q = (- 0.004905 c^{2}) u

B u t 1 u = 931.5 M e V / c^{2}

∴ Q = - 0.004905 \times 931.5 = - 4 .5690 M e V

The negative Q-value of the reaction shows that the reaction is endothermic.
(ii)
Step 2: Find the Q-value for the second reaction.

The given nuclear reaction is:

It is given that: \({ }_{6}^{12} \mathrm{C}+{ }_{6}^{12} \mathrm{C} \rightarrow{ }_{10}^{20} \mathrm{Ne}+{ }_{2}^{4} \mathrm{He} \)

Atomic mass of

m ({}_{6}^{12}C) = 12.0 u

Atomic mass of

m ({}_{10}^{20}N e) = 19.992439 u

Atomic mass of

m ({}_{2}^{4}H e) = 4.002603 u

The Q-value of this reaction is given as:

Q = [2 m ({}_{6}^{12}C) - m ({}_{10}^{20}N e) - m ({}_{2}^{4}H e)] c^{2}

= [2 \times 12.0 - 19.992439 - 4.00203] c^{2}

= (0.005531 c^{2}) u

= 0.005531 \times 931.5 = 5.1521 M e V

The positive Q-value of the reaction shows that the reaction is exothermic.

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