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

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

Answer 2

Answer:

37.50g of ⁹¹Nb will remain after 2040 years.

Explanation:

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

$Ln\frac{N}{N_0} = -Kt$ (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:

$t_{1/2} = \frac{ln2}{K}$

680 years = ln 2 / K

K = 1.019x10⁻³ years⁻¹

Replacing these values in (1):

$Ln\frac{N}{3.297 moles ^{91}Nb} = -1.019x10^{-3}years^{-1}*2040years$

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.