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_{(1)/(2)}}=(0.693)/(14.3)= 0.05days^(-1)`

`N=N_o* 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* e^{- 0.05 days^(-1)* 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