Jason Yuhas

2021-12-28

Actinium-226 has a half-life of 29 hours. If 100 mg of actinium-226 disintegrates over a period of 58 hours, how many mg of actinium-226 will remain? Use $A\left(t\right)=C{e}^{kt}$.

Matthew Rodriguez

Step 1
The given data is:
Half life
Initial 100 mg
To find how many actinium-226 will remain after 58 hours.
Step 2
The given equation is:
$A\left(t\right)=C{e}^{kt}$ (1)
At $t=0,A\left(0\right)=100mg$
$A\left(0\right)=100$
$C{e}^{k\left(0\right)}=100$
$C=100$
Substitute the value of C in equation (1),
$A\left(t\right)=100{e}^{kt}$ (2)
Step 3
At half life
$A\left(29\right)=100{e}^{29k}$
$50=100{e}^{29k}$
${e}^{29k}=\frac{1}{2}$
$29k=\mathrm{ln}\left(0.5\right)$
$k=-0.0239$
Substitute the value of k in equation (2),
$A\left(t\right)=100{e}^{-0.0239t}$
Step 4
Now find actinium-226 remain after 58 hours is,
$A\left(58\right)=100{e}^{-0.0239×58}$
$=25mg$
Thus, the actinium remains after 58 hours is 25mg.

Jillian Edgerton

no of half lives (n)

Amount left
$=\frac{100mg}{{2}^{2}}=\frac{100mg}{4}$

karton

Solution:
Using formula:
$A={A}_{0}\left(\frac{1}{2}{\right)}^{\frac{y}{n}}$
A is final amount after time "t".
h is the half-life
Given: ${A}_{0}=100mg$
h=29 hours
t=58 hours
A=?

$=100mg\left(\frac{1}{2}{\right)}^{2}$
$=100mg×\frac{1}{4}$
A=25mg
25mg of actinium-226 will remain after 58 hours

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