1. Identify the fusion reaction

Two deuterons fuse to form one helium nucleus:

${}^2\text{H} + {}^2\text{H} \rightarrow {}^4\text{He} + \text{energy}$

A deuteron has mass number $2$, so two deuterons together contain $4$ nucleons. The helium nucleus formed is ${}^4\text{He}$, which also has $4$ nucleons.

2. Use binding energy data

Given:

• Binding energy per nucleon of deuteron $= 1.1\,\text{MeV}$
• Binding energy per nucleon of helium $= 7.0\,\text{MeV}$

So,

• Total binding energy of one deuteron: $BE_d = 2 \times 1.1 = 2.2\,\text{MeV}$

• Total binding energy of two deuterons: $BE_{\text{initial}} = 2 \times 2.2 = 4.4\,\text{MeV}$

• Total binding energy of helium nucleus: $BE_{\text{final}} = 4 \times 7.0 = 28.0\,\text{MeV}$

3. Energy released in fusion

Energy released is the increase in total binding energy:

$Q = BE_{\text{final}} - BE_{\text{initial}}$

$Q = 28.0 - 4.4 = 23.6\,\text{MeV}$

4. Check options

• A: $30.2\,\text{MeV}$ ❌
• B: $32.4\,\text{MeV}$ ❌
• C: $23.6\,\text{MeV}$ ✅
• D: $25.8\,\text{MeV}$ ❌

Therefore, the correct answer is Option C.