Question

The half-life of an isotope is 10.0 minutes. If 25.0 grams are left after 60.0 minutes, how many grams were in the original sample?

Answer 1

The half-life of a radioactive isotope is 10.0 min, i.e, $t_{1/2}$ = 10.0 min.

Therefore, the radioactive decay constant is $λ = \frac{0.693}{t_{1/2}} = \frac{0.693}{10.0 \;min}$

Let N₀ is the initial amount of the radio isotope and N_t = 25.0 g is the amount remaining after t = 60.0 min.

Use the radioactive decay law.

$N(t) = N_0 \times e^{-λt} \\ ln (\frac{N_t }{N_0}) = -λ \times t$

Plug in values and get

$ln (\frac{25.0 }{N_0}) = -\frac{0.693}{10.0 \; min} \times 60.0 \; min = -4.158 \\ \frac{25.0 }{N_0} = e^{-4.158} \\ \frac{25.0 }{N_0} = 0.015638 \\ N_0 = 1598.67 \; g$

The amount of the radioactive isotope originally present is 1598.67 g (ans).