Question

The energy released if hydrogen atoms are combined to form \(^{4}_{2}\text{He}\) is _____ MeV. (Take binding energies per nucleon of \(^{2}_{1}\text{H}\) and \(^{4}_{2}\text{He}\) as \(1.1\,\text{MeV}\) and \(7.2\,\text{MeV}\), respectively).

• A. \(6.1\)
• B. \(24.4\)
• C. \(26.6\)
• D. \(5\)

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the energy released during the nuclear fusion process where hydrogen isotopes (specifically Deuterium, given the binding energy data) combine to form a Helium nucleus.
Step 2: Key Formula or Approach:
Energy released in a nuclear reaction (\(Q\)-value) is the difference between the total binding energy of the products and the total binding energy of the reactants:
\[ Q = \text{B.E.}(\text{Products}) - \text{B.E.}(\text{Reactants}) \]
Total B.E. = (Binding Energy per nucleon) \(\times\) (Mass number \(A\)).
Step 3: Detailed Explanation:
The reaction can be represented as:
\[ 2 \left( {}_{1}^{2}\text{H} \right) \rightarrow {}_{2}^{4}\text{He} + \text{Energy} \]
1. Binding Energy of the products (Helium, \(A = 4\)):
\[ \text{B.E.}_{\text{He}} = 4 \times 7.2 = 28.8 \text{ MeV} \]
2. Binding Energy of the reactants (Two Deuterium nuclei, each with \(A = 2\)):
\[ \text{B.E.}_{\text{reactants}} = 2 \times (2 \times 1.1) = 2 \times 2.2 = 4.4 \text{ MeV} \]
3. Calculation of energy released (\(Q\)):
\[ Q = 28.8 - 4.4 = 24.4 \text{ MeV} \]
Step 4: Final Answer:
The total energy released in the formation of one Helium nucleus is 24.4 MeV.