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=(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=(2.303)/(k)\log(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=(2.303)/(0.0248)\log(100)/(12.5)`

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

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