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Sagot :
Step 1
All radiation decay follows first order kinetics as follows:
[tex]\text{A = A}_0xe^{-\lambda t}[/tex]λ = decay constant
t = time taken
A0 = initially present mass
A = mass present after t time
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Step 2
Information provided:
12.3 y = half-life time = t 1/2
A0 = 48.0 mg
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Step 3
Procedure:
λ is calculated as follows:
[tex]\begin{gathered} t_{\frac{1}{2}\text{ }}=\text{ }\frac{ln2}{\lambda} \\ \lambda\text{ = }\frac{ln\text{ 2}}{12.3\text{ years}}=\text{ 0.056 1/y} \end{gathered}[/tex]Now,
From step 1:
[tex]\begin{gathered} A\text{ = 48.0 mg x }e^{-0.056\frac{1}{years}x\text{ 49.2 years}} \\ A\text{ = 3.05 mg} \end{gathered}[/tex]For t = 98.4 years => A = 0.194 mg
Answer:
What mass of the nuclide will remain after 49.2 y? 3.05 mg
And then after 98.4 y? 0.194 mg
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