Question

half life of Au 198 is 2.7 days what will be the activity of 1mg Au 198​

Answer 1

Answer:

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Answer 2

Given info : the half life of Au 198 is 2.7 days.

To find : the activity of 1 mg of Au 198 is ...

solution : we know, half life of radioactive decay is given by, $T_{1/2}=\frac{ln2}{\lambda}$

here, $T_{1/2}$ = 2.7 days

so radioactive decay constant for Au 198 is, λ = ln2/2.7  day⁻¹ = 0.2567 day⁻¹

activity of any radioactive decay is given as, A = λN

where, N is no of nuclei present in the given amount of a substance.

here, mass of Au = 1 mg

atomic mas of Au = 198 g

so the no of nuclei = given mass/atomic mass × Avogadro's no

                               = 10⁻³g/198g × 6.023 × 10²³ = 3.04 × 10¹⁸

now activity, A = λN

= 0.2567 × 3.04 × 10¹⁸ disintegrations per day

= 0.78 × 10¹⁸ disintegrations per day.

= 0.78 × 10¹⁸/(24 × 3600) disintegrations per second

= 9 × 10¹² disintegrations per sec.

therefore the activity of 1 mg Au is 9 ×  10¹² disintegrations per second.