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Q13.5(i):
The Q value of a nuclear reaction \(A + b \rightarrow C + d\) is defined by \(Q = [ m_{A} + m_{b} – m_{C} – m_{d} ]c^{2}\) 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) \(^{1}_{1}H+ ^{3}_{1}H \rightarrow ^{2}_{1}H+ ^{2}_{1}H\) 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\)

Solution :

Step 1: Identification of Reactants and Products

The nuclear reaction provided is: $^{1}_{1}H + ^{3}_{1}H \rightarrow ^{2}_{1}H + ^{2}_{1}H$.

The reactants are $A = ^{1}_{1}H$ and $b = ^{3}_{1}H$. The products are $C = ^{2}_{1}H$ and $d = ^{2}_{1}H$.

Note: The mass of $m(^{1}_{1}H)$ (hydrogen nucleus/proton) is required. Standard physical data provides $m(^{1}_{1}H) = 1.007825 \text{ u}$.

Step 2: Summation of Masses

According to the Q-value definition $Q = [m_{A} + m_{b} - m_{C} - m_{d}]c^{2}$, we first calculate the total initial and final mass.

Component Mass (u)
$m(^{1}_{1}H)$ $1.007825$
$m(^{3}_{1}H)$ $3.016049$
Total Reactant Mass ($\sum m_{initial}$) $4.023874$
$m(^{2}_{1}H)$ $2.014102$
$m(^{2}_{1}H)$ $2.014102$
Total Product Mass ($\sum m_{final}$) $4.028204$

Step 3: Calculating Mass Defect ($\Delta m$)

The mass defect is defined as the difference between the total initial mass and total final mass [Per the Mass-Energy Equivalence Principle]:

$\Delta m = \sum m_{initial} - \sum m_{final}$

$\Delta m = 4.023874 \text{ u} - 4.028204 \text{ u}$

$\Delta m = -0.004330 \text{ u}$

Step 4: Conversion to Energy (Q-value)

Using the conversion factor $1 \text{ u} = 931.5 \text{ MeV}/c^{2}$ [By Einstein's Mass-Energy Equivalence $E = \Delta m c^{2}$]:

$Q = (-0.004330 \text{ u}) \times (931.5 \text{ MeV/u})$

$Q = -4.0334 \text{ MeV}$

Step 5: Theoretical Analysis of the Reaction Type

Since the calculated $Q$ value is negative ($Q < 0$), the reaction consumes energy to proceed. In nuclear physics, such reactions are classified as endothermic [Per the First Law of Thermodynamics applied to nuclear binding energy].

Initial State: 4.023874 u Final State: 4.028204 u Q = -4.0334 MeV (Endothermic)

Final Solution: The Q-value of the reaction is $-4.0334 \text{ MeV}$. Because the Q-value is negative, the reaction is endothermic.


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