The Q value of a nuclear reaction A + b → C + d is defined by

Solution:

(i) The given nuclear reaction is:

${ }_{1} \mathrm{H}^{1}+{ }_{1}^{3} \mathrm{H} \rightarrow{ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H}$

It is given that:

Atomic mass $m\left({ }_{1}^{1} \mathrm{H}\right)=1.007825 \mathrm{u}$

Atomic mass $m\left({ }_{1}^{3} \mathrm{H}\right)=3.016049 \mathrm{u}$

Atomic mass $m\left({ }_{1}^{2} \mathrm{H}\right)=2.014102 \mathrm{u}$

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

$Q=\left[m\left({ }_{1}^{1} \mathrm{H}\right)+m\left({ }_{1}^{3} \mathrm{H}\right)-2 m\left({ }_{1}^{2} \mathrm{H}\right)\right] c^{2}$

$=[1.007825+3.016049-2 \times 2.014102] c^{2}$

$Q=\left(-0.00433 c^{2}\right) \mathrm{u}$

But $1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^{2}$

$\therefore Q=-0.00433 \times 931.5=-4.0334 \mathrm{MeV}$

The negativeQ-value of the reaction shows that the reaction is endothermic.

 

(ii) The given nuclear reaction is:

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

It is given that:

Atomic mass of $m\left({ }_{6}^{12} \mathrm{C}\right)=12.0 \mathrm{u}$

Atomic mass of $m\left({ }_{10}^{20} \mathrm{Ne}\right)=19.992439 \mathrm{u}$

Atomic mass of $m\left({ }_{2}^{4} \mathrm{He}\right)=4.002603 \mathrm{u}$

The Q-value of this reaction is given as:

$Q=\left[2 m\left({ }_{6}^{12} \mathrm{C}\right)-m\left({ }_{10}^{20} \mathrm{Ne}\right)-m\left({ }_{2}^{4} \mathrm{He}\right)\right] c^{2}$

$=[2 \times 12.0-19.992439-4.002603] c^{2}$

$=\left(0.004958 c^{2}\right) \mathrm{u}$

$=0.004958 \times 931.5=4.618377 \mathrm{MeV}$

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