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

Beginner2021-12-29Added 32 answers

Step 1

The given data is:

Half life $t\frac{1}{2}=29\text{}hours$

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 $t=29\text{}hours,A\left(29\right)=\frac{100}{2}=50mg$

$A\left(29\right)=100{e}^{29k}$

$50=100{e}^{29k}$

$e}^{29k}=\frac{1}{2$

$29k=\mathrm{ln}\left(0.5\right)$

$k=-0.0239$

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,

$A\left(58\right)=100{e}^{-0.0239\times 58}$

$=25mg$

Thus, the actinium remains after 58 hours is 25mg.

Jillian Edgerton

Beginner2021-12-30Added 34 answers

no of half lives (n)

Amount left

karton

Expert2022-01-04Added 613 answers

Solution:

Using formula:

A is final amount after time "t".

h is the half-life

Given:

h=29 hours

t=58 hours

A=?

A=25mg

25mg of actinium-226 will remain after 58 hours

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