The isotope Mg-28 has a half life of 21 hours. If a sample initially contains exactly 10000atoms of Mg-28, approximately how many of these atoms will remain after one week?

You can find your constant of proportionality, k, by using

T=\frac{-1}{k}ln(2)

Since it is given that T=21 hours, solve for k.

k will be negative because it is decaying.

Then, sub that into:

A=10000e^{kt}, set t=168 (there are 168 hours in a week) to find how many atoms remain.

thank you so much :)

I have another problem,

iodine-131,half life=8.0 days, initially emits 9.95 x 10^(18) beta particles per day. How long will it take for the activity to drop 6.22 x 10^(17) beta particles per day?

it has beta particle thingy so I don't know what to do anymore... hope you can help me with this. Thanks :)

It's the same thing, and use the same formula.

I personally prefer the formula:

\frac{N}{N_o} = \(\frac12\)^{t/h}

Where N is the number of 'current' atoms, No the initial number, t the time since the start of the decay and h the half life.

It is also valid for activity;

\frac{A}{A_o} = \(\frac12\)^{t/h}

Here,
Ao = 9.95 x 10^18
A = 6.22 x 10^17
h = 8 days
t = ? Days