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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

Potassium-40 can decay in three modes: β^ - -emission, β^ +...

Potassium-40 can decay in three modes: β−\beta^{-}-emission, β+\beta^{+}-emission or electron capture. The atomic masses of ^{40}_{18}\text{Ar},1940K, ^{40}_{19}\text{K} and ^{40}_{20}\text{Ca}areare39.9624 \text{ u},,39.9640 \text{ u}andand39.9626 \text{ u}respectively.Computetherespectively. Compute theQ-values for each of the three decay modes respectively.

Options

  1. A1.3034 MeV1.3034 \text{ MeV}, 1.490 MeV1.490 \text{ MeV}, 1.490 MeV1.490 \text{ MeV}
  2. B1.3034 MeV1.3034 \text{ MeV}, 0.4676 MeV0.4676 \text{ MeV}, 1.490 MeV1.490 \text{ MeV}
  3. C1.3034 MeV1.3034 \text{ MeV}, 0.9786 MeV0.9786 \text{ MeV}, 1.490 MeV1.490 \text{ MeV}
  4. D1.3034 MeV1.3034 \text{ MeV}, 1.490 MeV1.490 \text{ MeV}, 0.4676 MeV0.4676 \text{ MeV}

Correct answer

B. 1.3034 MeV1.3034 \text{ MeV}, 0.4676 MeV0.4676 \text{ MeV}, 1.490 MeV1.490 \text{ MeV}

Step-by-step solution

The QQ-value for β−\beta^{-}-emission is given by the difference in atomic masses of the parent and daughter nuclei:

Qβ−=[M(1940K)−M(2040Ca)]c2Q_{\beta^{-}} = [M(^{40}_{19}\text{K}) - M(^{40}_{20}\text{Ca})] c^2

Using 1 u=931 MeV/c21 \text{ u} = 931 \text{ MeV}/c^2, we get:

Qβ−=(39.9640−39.9626)×931=0.0014×931=1.3034 MeVQ_{\beta^{-}} = (39.9640 - 39.9626) \times 931 = 0.0014 \times 931 = 1.3034 \text{ MeV}

The QQ-value for β+\beta^{+}-emission accounts for the mass of two electrons (one from the positron emitted and one from the difference in atomic electron count):

Qβ+=[M(1940K)−M(1840Ar)−2me]c2Q_{\beta^{+}} = [M(^{40}_{19}\text{K}) - M(^{40}_{18}\text{Ar}) - 2m_e] c^2

Using mec2=0.511 MeVm_e c^2 = 0.511 \text{ MeV}, we get:

Qβ+=(39.9640−39.9624)×931−2(0.511)=0.0016×931−1.022=1.4896−1.022=0.4676 MeVQ_{\beta^{+}} = (39.9640 - 39.9624) \times 931 - 2(0.511) = 0.0016 \times 931 - 1.022 = 1.4896 - 1.022 = 0.4676 \text{ MeV}

The QQ-value for electron capture is given by:

QEC=[M(1940K)−M(1840Ar)]c2Q_{EC} = [M(^{40}_{19}\text{K}) - M(^{40}_{18}\text{Ar})] c^2

QEC=(39.9640−39.9624)×931=0.0016×931=1.4896 MeV≈1.490 MeVQ_{EC} = (39.9640 - 39.9624) \times 931 = 0.0016 \times 931 = 1.4896 \text{ MeV} \approx 1.490 \text{ MeV}
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