COMEDK · Physics · 29. Nuclear Physics
Find the binding energy of the tritium nucleus:
[Given: mass of \(1 \mathrm{H}^3=3.01605 \mathrm{~u} ; \mathrm{~m}_{\mathrm{p}}=1.00782 \mathrm{~u} ; \mathrm{~m}_{\mathrm{n}}=1.00866 \mathrm{~u}\).]
- A 8.5 MeV
- B 8.5 J
- C 0.00909 eV
- D 0.00909 MeV
Answer & Solution
Correct Answer
(A) 8.5 MeV
Step-by-step Solution
Detailed explanation
The tritium nucleus \(^{3}_{1}H\) consists of 1 proton and 2 neutrons. Mass of constituents = \(m_p + 2m_n = 1.00782 + 2 \times 1.00866 = 1.00782 + 2.01732 = 3.02514 \text{ u}\). Mass defect \(\Delta m = (m_p + 2m_n) - m_{nucleus} = 3.02514 - 3.01605 = 0.00909 \text{ u}\).…
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