Question

In the nuclear reaction : \[ {}_{1}^{2}\text{H} + {}_{1}^{2}\text{H} \rightarrow {}_{2}^{\text{A}}\text{X} + {}_{0}^{1}\text{n} \] (a) Find the value of A.
(b) Calculate the amount of energy released in the reaction.
Given : \( m\left({}_{1}^{2}\text{H}\right) = 2.014102\text{ u} \), \( m\left({}_{2}^{\text{A}}\text{X}\right) = 3.016049\text{ u} \), \( m\left({}_{0}^{1}\text{n}\right) = 1.008665\text{ u} \), \( 1\text{ u} = 931.5\text{ MeV}/c^2 \)

Hint:

When computing values for mass defects in nuclear reactions, never round off numbers prematurely in intermediate steps! The differences occur past the third or fourth decimal place, so maintaining all six decimal places provided in the given values is essential to avoid significant rounding errors in your final energy calculation.

Answer 1

Concept: Nuclear reactions are governed by fundamental conservation laws that dictate how mass, charge, and energy are distributed between reactants and products.
• Law of Conservation of Nucleon Number (Mass Number): The total number of protons and neutrons (nucleons) must be equal before and after a nuclear reaction. In a standard balanced equation \({}_{Z_1}^{A_1}\text{P} + {}_{Z_2}^{A_2}\text{Q} \rightarrow {}_{Z_3}^{A_3}\text{R} + {}_{Z_4}^{A_4}\text{S}\), this requires: \[ A_1 + A_2 = A_3 + A_4 \]
• Law of Conservation of Atomic Number (Charge): The total electric charge of the reacting nuclei must equal the total electric charge of the resulting products: \[ Z_1 + Z_2 = Z_3 + Z_4 \]
• Mass-Energy Equivalence and Q-value: In nuclear reactions, the total rest mass of the products is typically less than the total rest mass of the initial reacting nuclei. This difference in mass, known as the mass defect (\(\Delta m\)), is converted directly into nuclear energy. The released energy (\(Q\)) is given by Albert Einstein's mass-energy relationship: \[ Q = \Delta m \cdot c^2 \] When the atomic mass units (u) are used, the energy equivalent of \(1\text{ u}\) is approximately \(931.5\text{ MeV}\). Thus, the energy released can be directly computed as: \[ Q = \Delta m \text{ (in u)} \times 931.5\text{ MeV} \] Part (a): Find the value of A.
Let us analyze the given nuclear fusion equation involving two deuterium nuclei: \[ {}_{1}^{2}\text{H} + {}_{1}^{2}\text{H} \rightarrow {}_{2}^{\text{A}}\text{X} + {}_{0}^{1}\text{n} \] To find the unknown mass number \(\text{A}\) of the product nucleus \(\text{X}\), we apply the Law of Conservation of Mass Number. Summing up the total mass numbers on the left-hand side (reactants): \[ \sum A_{\text{reactants}} = 2 + 2 = 4 \] Summing up the total mass numbers on the right-hand side (products): \[ \sum A_{\text{products}} = \text{A} + 1 \] Equating the two sides according to the conservation principle: \[ 4 = \text{A} + 1 \] Isolating \(\text{A}\) by subtracting 1 from both sides of the equation: \[ \text{A} = 4 - 1 \] \[ \text{A} = 3 \] Note on Verification: We can also verify this by checking the conservation of atomic number (\(\text{Z}\)): \[ \sum Z_{\text{reactants}} = 1 + 1 = 2 \] \[ \sum Z_{\text{products}} = 2 + 0 = 2 \] Since the atomic number matches perfectly, the product nucleus is an isotope of Helium, specifically Helium-3 (\({}_{2}^{3}\text{He}\)). Thus, the value of \(\text{A}\) is 3. Part (b): Calculate the amount of energy released in the reaction.

Step 1: Calculate the total mass of the reactants.
The reactants consist of two identical Deuterium nuclei (\({}_{1}^{2}\text{H}\)). The mass of one Deuterium atom is given as \(m\left({}_{1}^{2}\text{H}\right) = 2.014102\text{ u}\). \[ \text{Total initial mass } (m_{\text{reactants}}) = 2 \times m\left({}_{1}^{2}\text{H}\right) \] \[ m_{\text{reactants}} = 2 \times 2.014102\text{ u} \] Performing the multiplication step-by-step: \[ m_{\text{reactants}} = 4.028204\text{ u} \]

Step 2: Calculate the total mass of the products.
The products consist of one nucleus \({}_{2}^{3}\text{X}\) and one neutron (\({}_{0}^{1}\text{n}\)). We are given: \[ m\left({}_{2}^{3}\text{X}\right) = 3.016049\text{ u} \] \[ m\left({}_{0}^{1}\text{n}\right) = 1.008665\text{ u} \] The total final mass of the products is: \[ \text{Total final mass } (m_{\text{products}}) = m\left({}_{2}^{3}\text{X}\right) + m\left({}_{0}^{1}\text{n}\right) \] \[ m_{\text{products}} = 3.016049\text{ u} + 1.008665\text{ u} \] Adding the numbers carefully by aligning their decimal positions: {r@{}l} 3.016049 & u
+ 1.008665 & u
4.024714 & u
\[ m_{\text{products}} = 4.024714\text{ u} \]

Step 3: Calculate the mass defect (\(\Delta m\)).
The mass defect is the difference between the total mass of the reactants and the total mass of the products: \[ \Delta m = m_{\text{reactants}} - m_{\text{products}} \] Substituting our calculated values: \[ \Delta m = 4.028204\text{ u} - 4.024714\text{ u} \] Subtracting the decimals: {r@{}l} 4.028204 & u
- 4.024714 & u
0.003490 & u
\[ \Delta m = 0.003490\text{ u} = 0.00349\text{ u} \]

Step 4: Convert the mass defect into released energy.
We are given that \(1\text{ u} = 931.5\text{ MeV}/c^2\). Therefore, the energy released (\(E\)) in units of Mega-electron Volts (MeV) is: \[ E = \Delta m \times 931.5\text{ MeV} \] \[ E = 0.00349 \times 931.5\text{ MeV} \] Let us carry out the long multiplication to ensure absolute accuracy: \[ 349 \times 9315 = 3250935 \] Since there are 5 decimal places in total (\(3\) from \(0.00349\) and \(1\) from \(931.5\)), we position the decimal point: \[ E = 3.250935\text{ MeV} \] Rounding the energy value to three decimal places yields: \[ E \approx 3.251\text{ MeV} \] Thus, the total energy released during this nuclear fusion event is approximately \(3.251\text{ MeV}\).