Explanation Given data: Sample of 3 H is 2 mg. Formula used: Write the expression for activity of radioactive sample. R = N τ = 0.693 N t 1 2 (1) Here, R is the activity, N is the number of samples, τ is the time constant, and t 1 2 is the half-life. Consider r as the decay rate such that, r = 1 τ = 0.693 t 1 2 (2) From Appendix D, the half-life of 3 H is 12.33 y and atomic mass is 3.016 g mol . Consider the conversion factor from years to seconds. 1 y = 3.15 × 10 7 s Convert 12.33 y into seconds by using conversion factor. 12 .33 y = 12.33 × 3.15 × 10 7 s Substitute 12.33 × 3.15 × 10 7 s for t 1 2 in equation (2), r = 0.693 12.33 × 3.15 × 10 7 s = 1.784 × 10 − 9 s - 1 Find the number of atoms in the sample. Number of atoms in the sample = 2 × 10 − 3 g = 2 × 10 − 3 ( 1 mole 3.016 ) { ∵ 1 mole = 6.023 × 10 23 ∵ 3.016 g = 1 mole } = 2 × 10 − 3 ( 6

Student Workbook for College Physics: A Strategic Approach

3rd Edition

ISBN: 9780321907240

Author: Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field

Publisher: PEARSON

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Radioactive decay

The emission of energy to produce ionizing radiation is known as radioactive decay. Alpha, beta particles, and gamma rays are examples of ionizing radiation that could be released. Radioactive decay happens in radionuclides, which are imbalanced atoms. T…

Artificial Radioactivity

The radioactivity can be simply referred to as particle emission from nuclei due to the nuclear instability. There are different types of radiation such as alpha, beta and gamma radiation. Along with these there are different types of decay as well.

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Chapter 30, Problem 58GP

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The activity in Bq and in Ci of a 2 mg sample of 3H .

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Chapter 30, Problem 57GP

Chapter 30, Problem 59GP

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The activity in Bq and in Ci of a 2 mg sample of 3 H .

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The activity in Bq and in Ci of a 2 mg sample of 3H .

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Chapter 30 Solutions