Concepts Of Physics MCQ Edition [Volume 2]
- Exam
- Concepts Of Physics MCQ Edition [Volume 2]
- Subject
- Physics
- Chapter
- The Nucleus
Concepts Of Physics MCQ Edition [Volume 2] · Physics · The Nucleus
Determine the Q -value for the fusion reaction: ^ 2 _ 1 H + ^ 3 _ 1...
Determine the -value for the fusion reaction:
^{2}_{1}\text{H} + ^{3}_{1}\text{H} \to ^{4}_{2}\text{He} + nm(^{2}_{1}\text{H}) = 2.014102 \text{ u}m(^{3}_{1}\text{H}) = 3.016049 \text{ u}m(^{3}_{2}\text{He}) = 3.016029 \text{ u}m(^{4}_{2}\text{He}) = 4.002603 \text{ u}.
^{2}_{1}\text{H} + ^{3}_{1}\text{H} \to ^{4}_{2}\text{He} + nm(^{2}_{1}\text{H}) = 2.014102 \text{ u}m(^{3}_{1}\text{H}) = 3.016049 \text{ u}m(^{3}_{2}\text{He}) = 3.016029 \text{ u}m(^{4}_{2}\text{He}) = 4.002603 \text{ u}.
Options
- A
- B
- C
- D
Correct answer
D.
Step-by-step solution
The given fusion reaction is:
^{2}_{1}\text{H} + ^{3}_{1}\text{H} \to ^{4}_{2}\text{He} + nm_{R} = m(^{2}_{1}\text{H}) + m(^{3}_{1}\text{H})m_{R} = 2.014102 \text{ u} + 3.016049 \text{ u} = 5.030151 \text{ u}m_{P} = m(^{4}_{2}\text{He}) + m_{n}m_{n} \approx 1.008665 \text{ u}m_{P} = 4.002603 \text{ u} + 1.008665 \text{ u} = 5.011268 \text{ u}\Delta m\Delta m = m_{R} - m_{P}\Delta m = 5.030151 \text{ u} - 5.011268 \text{ u} = 0.018883 \text{ u}Q1 \text{ u} \approx 931.5 \text{ MeV}Q = \Delta m \times 931.5 \text{ MeV/u}Q = 0.018883 \times 931.5 \approx 17.59 \text{ MeV}17.57 \text{ MeV}.
^{2}_{1}\text{H} + ^{3}_{1}\text{H} \to ^{4}_{2}\text{He} + nm_{R} = m(^{2}_{1}\text{H}) + m(^{3}_{1}\text{H})m_{R} = 2.014102 \text{ u} + 3.016049 \text{ u} = 5.030151 \text{ u}m_{P} = m(^{4}_{2}\text{He}) + m_{n}m_{n} \approx 1.008665 \text{ u}m_{P} = 4.002603 \text{ u} + 1.008665 \text{ u} = 5.011268 \text{ u}\Delta m\Delta m = m_{R} - m_{P}\Delta m = 5.030151 \text{ u} - 5.011268 \text{ u} = 0.018883 \text{ u}Q1 \text{ u} \approx 931.5 \text{ MeV}Q = \Delta m \times 931.5 \text{ MeV/u}Q = 0.018883 \times 931.5 \approx 17.59 \text{ MeV}17.57 \text{ MeV}.