Question

Phosphorus-32 has a half-life of 14.3 days. how many grams remain from a 10.0 gram sample after 30.0 days?

Answer 1

Answer: 2.23 grams

Explanation:

Radioactive decay follows first order kinetics.

Half-life of Phosphorus-32 = 14.3 days

$\lambda =\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{14.3}= 0.05days^{-1}$

$N=N_o\times e^{-\lambda t}$

N = amount left after time t= ?

$N_0$ = initial amount  = 10.0 g

$\lambda$ = rate constant= $0.05days^{-1}$

t= time  = 30 days

$N=10\times e^{- 0.05 days^{-1}\times 30days}$

$N=2.23g$

Answer 2

When the nuclear half-life of the radioactive isotope is showing the time needed for the isotope to be half of its initial value of mass.

so with each half-life, the isotope will be halved of its initial value as example:

after the first half-life, the isotope will lose 50 % of its initial value

and after the second half-life, the isotope will lose 25% of its initial value 

and after the third half-life, the isotope will lose 12.5 % of its initial value

and so on,

So here to get how many numbers of half-lives we will use this formula:

numbers of half-lives = total time passed / the half-life of the isotope

                                    = 30 days / 14 days

                                    =2 days

∴remainig mass = initial mass / 2^numbers of half-lives

                            = 10 g / 2^2

                            = 2.5 g