# Initial stage you have 128.0 mg radioactive sample and after 24.0 days 2.00 mg of sample remain Calculate the half life of the sample?

### 4 Answers

- pisgahchemistLv 72 years ago
Half-life... t½

Radioactive decay follows first-order kinetics. The integrated rate equation is:

A = Ao e^(-kt) ..... A are the amounts (or activities), t is the elapsed time and k is the decay constant which is related to the half-life by

k = ln2 / t½ ..... or ...... t½ = ln2 / k

A = Ao e^(-kt) can be written as:

ln A = -kt + ln Ao

or

ln (A/Ao) = -kt

k = ln(A/Ao) / -t ......... solve for k

k = ln(2.00 mg / 128 mg) / -24.0 days

k = 0.1733 days⁻¹

t½ = ln2 / 0.1733 days⁻¹ = 4.00 days

- Dr WLv 72 years ago
use this equation

.. A(t) = A(o) * (1/2)^(time / half life)

rearranging

.. A(t) / A(o) = (1/2)^(t / h.l.)

.. ln[ A(t) / A(o) ] = (t / h.l.) * ln(1/2)

.. ln[ A(t) / A(o) ] = (t / h.l.) * (ln(1) - ln(2)).. .. . .recall that ln(a/b) = ln(a) - ln(b)

.. ln[ A(t) / A(o) ] = (t / h.l.) * (-ln(2)).. .. . .recall that ln(1) = 0

.. ln[ A(o) / A(t) ] = (t / h.l.) * ln(2).. .. . .recall that ln(a/b) = - ln(b/a)

and finally, we get

.. h.l = t * ln(2) / ln(Ao/At)

solving

.. h.l. = 24.0days * ln(2) / ln(64.0) = 4.0 days

- Roger the MoleLv 72 years ago
(2.00 mg / 128.0 mg) = (1/2)^(24.0 days / x days)

Do the division on the left:

1/64 = (1/2)^(24.0 days / x days)

Convert the fraction to powers of 2:

(1/2)^6 = (1/2)^(24.0 days / x days)

So:

24.0 days / x days = 6

6x days = 24 days

x = 4.0 days

- Anonymous2 years ago
After 1 half life you have an activity of 64, after 2 half lives 32 etc.

Divide 24 days by the number of half lives to reach 2 activity.

(I'm not sure the mass change is the correct form of the question though, it should be REM, becquerels, curies or grays).