Question

Iobium-91 has a half-life of 680 years. After 2,040 years, how much niobium-91 will remain from a 300.0-g sample?

Answer 1

`A=A_(0)*( (1)/(2))^{ (t)/(h)} \\ \\A - final\ amount \\ \\A_(0) - initial\ amount \\ \\ t - time \\ \\ h - half\ life \\ \\ A=300.0 g*( (1)/(2))^{ (2040)/(680)} \\ A = 37.50 g`

Answer 2

37.50g of ⁹¹Nb will remain after 2040 years.

Step-by-step explanation:

The rate of decay of a radioactive isotope obeys the following formula:

`Ln(N)/(N_0) = -Kt` (1)

Where N is moles of atoms in time t, N₀ is initial moles of atoms, K is decay constant and t is time.

300.0g of niobium are:

300.0g × (1mol / 91g) = 3.297 moles of ⁹¹Nb

It is possible to obtain decay constant from half-life, thus:

`t_(1/2) = (ln2)/(K)`

680 years = ln 2 / K

K = 1.019x10⁻³ years⁻¹

Replacing these values in (1):

`Ln(N)/(3.297 moles ^(91)Nb) = -1.019x10^(-3)years^(-1)*2040years`

N / 3.297 moles of ⁹¹Nb = -0.125

N = 0.4121 moles of ⁹¹Nb after 2040 years. In mass:

0.4121 moles of ⁹¹Nb × (91g / mol) = 37.50g of ⁹¹Nb will remain after 2040 years.