Question

A certain radioactive material is decaying at a rate proportional to the amount present.

If a sample of 50 grams of the material was present initially and after 2 hours the sample lost 10% of its mass, find:

(a) An expression for the mass of the material remaining at any time t.

(b) The mass of the material after 4 hours.

(c) The time at which the material has decayed to one half of its initial mass.

Answer 1

Solution;

(a)

The law of radioactivity;

$N=N_oe^{-\lambda t}$

At $t=0,N_o=50$

Hence we have;

$N=50e^{-\lambda t}$

Amount remain after 2 hours;

$50-0.1(50)=45g$

Hence,we can equate;

$45=50e^{-2\lambda}$

$\frac{9}{10}=e^{-2\lambda}$

Therefore mass at any time at any time t is;

$N=50(\frac{9}{10})^{\frac t2}$

(b)

t=4hours

$N=50(\frac{9}{10})^2=40.5g$

(c)

One half of the initial mass is 25g;

Equate;

$25=50(\frac{9}{10})^\frac t2$

$2ln(0.5)=tln(9)$

$t=13.16 hours$