Question

³⁷S decays by emitting a β^- particle. This decay process has a half-life of 5.05 min. Answer the following questions about this process and report all answers to three significant figures.



1. Determine the initial activity (in dis/min, do not input units) of a 6.90 mg sample of this isotope. Report your answer to three significant figures in scientific notation.

2. Determine the time (in min) that it takes for the activity of ³⁷S to decrease to 6.47×10⁶ dis/min.

Answer 1

Given that the half life of ³⁷S is 5.05 min

Decay constant = ln 2/ half life

= 0.693 / 5.05 min

   = 0.137 min^-1

Number of nuclei present initially= 6.9*1g/1000 mg) * (1 mol / 37 g)*(6.022*10²³ nuclei/1 mol)

= 1.123*10²⁰ nuclei

Initial activity = Number of nuclei * Decay constant

= 1.123*10²⁰ nuclei * 0.137 min^-1

= 1.54*10¹⁹ dis/min

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We know that,

-28.5 = -0.137 t

t = 208.03min