Answer :
The value of the decay constant will be approximately 503.28 s⁻¹ and the half-life is 5 seconds when the activity of a sample of a radioisotope at some time is 10.1mCi and 0.32 hours later it is 7.60mCi. The number of nuclei in the sample, when the activity was 10.1 mCi, will be 742,647. The activity of the sample 2.50 h after it was 10.1mCi is 6.54 mCi.
(a) The value of the decay constant can be calculated using the following formula:
⇒ [tex]\lambda=\frac{ln(A1/A2)}{t2-t1} = \frac{ln(10.1/7.6)}{0.32-0}[/tex]
⇒ λ ≈ 0.1398 h⁻¹
Converting the value in seconds:
⇒ λ ≈ 0.1398 × 3600 ≈ 503.28 s⁻¹
(b) The half-life can be calculated as follows:
[tex]t1/2=\frac{ln(2)}{503.28} = 0.00138 h[/tex] ≈ 5s
(c) The number of nuclei can be found by rearranging the radioactivity formula:
[tex]N=\frac{A}{\lambda} = \frac{10.1 \times 3.7 \times 10^7}{503.28}[/tex]
⇒ N ≈ 742646.97 nuclei
(d) Using the exponential decay equation to find the activity of the sample after 2.50 h
[tex]A(t)= A_0 \times e^{- \lambda t[/tex]
⇒ [tex]A(t)= 10.1 \times e^{-0.00283 \times 2.50 \times 3600}[/tex]
⇒ A(t) ≈ 6.54 mCi
Therefore, the value of the decay constant will be approximately 503.28 s⁻¹ and the half-life is 5 seconds. The number of nuclei in the sample, when the activity was 10.1 mCi, will be 742,647. The activity of the sample 2.50 h after it was 10.1mCi is 6.54 mCi.
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