Question

A 1,200g block of phosphorus-32, which has a half life of 14.3 days, is stored for 100.1 days. At the end of this period, how much phosphorus-32 remains? Round to the nearest tenth.

Answer 1

Answer:

The amount of phosphorus-32 left after 100.1 days is 9.3 g.

Step-by-step explanation:

Given:

Initial amount of Phosphorus-32 is, $N_0=1200\ g$

Time period of decay is, $t=100.1\ days$

Half life of the block is, $t_{1/2}=14.3\ days$

Now, final amount left is, $N=?$

We know that, the decay equation for a radioactive material is given as:

$N=N_0e^{-kt}\\k\to decay\ constant$

The value of the decay constant is given as:

$k=\frac{\ln 2}{t_{1/2}}\\\\k=\frac{0.693}{14.3}=0.0485$

Now, plug in all the given values and calculate 'N'. This gives,

$N=(1200)e^{(-0.0485\times 100.1)}\\\\N=9.349\approx 9.3\ g$

Therefore, the amount of phosphorus-32 left after 100.1 days is 9.3 g