Question

The half-life of an isotope is 2.0 hours. How much of a 50.0 g sample is left after 6.0 hours?

Answer 1

The half-life of a radioactive isotope is 2.0 hours, i.e, $t_{1/2}$ = 2.0 hours.

Therefore, the radiactive decay constant is $λ = \frac{0.693}{t_{1/2}} = \frac{0.693}{2.0 \;hours}$

Let N₀ = 50.0 g the initial amount of the radioisotope and N_t is the amount remaining after t = 6.0 hours.

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

$ln (\frac{N_t }{50.0}) = -\frac{0.693}{2.0 \;hours} \times 6.0 \; hours = -2.079 \\ \frac{N_t }{50.0} = e^{-2.079} \\ \frac{N_t }{50.0} = 0.12505 \\ N_t = 6.2525 \; g$

The amount of the radioactive isotope left is 6.2525 g (ans).