Question

One of the radioactive isotopes used in medical treatment or analysis is chromium-51. The half-life of chromium-51 is 28 days. How much time is required for the activity of a sample of chromium-51 to fall to 12.5 percent of its original value

Answer 1

Answer : The time required for decay is, 84 days.

Explanation :

Half-life of chromium-51 = 28 days

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{28\text{ days}}$

$k=0.0248\text{ days}^{-1}$

Now we have to calculate the time required for decay.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = time taken by sample = ?

a = let initial activity of the sample = 100

a - x = amount left after decay process  = 12.5

Now put all the given values in above equation, we get

$t=\frac{2.303}{0.0248}\log\frac{100}{12.5}$

$t=83.9\text{ days}\approx 84\text{ days}$

Therefore, the time required for decay is, 84 days.