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CUET · PHYSICS · PYQ PAPER 2023

The binding energy per nucleon is 1.1 MeV for \({ }_1^2 H\) and 7.0 MeV for \({ }_2^4 He\). The energy released in the given fusion reaction will be:
\({ }_1^2 H+{ }_1^2 H \rightarrow{ }_2^4 He\)

  1. A 1.1 MeV
  2. B 11.9 MeV
  3. C 23.6 MeV
  4. D 28.0 MeV
Verified Solution

Answer & Solution

Correct Answer

(C) 23.6 MeV

Step-by-step Solution

Detailed explanation

\( Q = E_{B,products} - E_{B,reactants} \) \( Q = (7.0 \text{ MeV/nucleon} \times 4) - 2 \times (1.1 \text{ MeV/nucleon} \times 2) \) \( Q = 28.0 \text{ MeV} - 4.4 \text{ MeV} \) \( Q = 23.6 \text{ MeV} \)