Question

Beryllium-11 is a radioactive isotope of the alkaline metal Beryllium. It decays at a rate of 4.9% every second. Assuming you started with 100%, how much would be left after 45 seconds?\

Answer 1

Answer:

11.02 % of an isotope will be left after 45 seconds.

Explanation:

$N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

where,

$N_o$ = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

$t_{\frac{1}{2}}$ = half life of the isotope

$\lambda$ = rate constant

We have :

Mass of Beryllium-11 radioactive isotope= $N_o=100$

Mass of Beryllium-11 radioactive isotope after 45 seconds = $N=?$

t = 45 s

$\lambda$ = rate constant  = $4.9 \% s^{-1}=0.049 s^{-1}$

$N=N_o\times e^{-(\lambda \times t}$

Now put all the given values in this formula, we get

$N=100\times e^{-0.049 s^{-1}\ties 45 s}$

$N=11.02 g$

Percentage of isotope left :

$\frac{N}{N_o}\times 100=\frac{11.02 g}{100 g}\times 100=11.02\%$

11.02 % of an isotope will be left after 45 seconds.