Question

The activity of a radioactive substance is $ 4700 $ counts per minute. Five minutes later the activity is $ 2700 $ counts per minute. Find the (a) decay constant (b) half life of the radioactive substance.

Answer 1

Hint:By using the decay equation for the first order decay we can substitute the counts per minute at the beginning and the counts per minute at the time five minutes later and calculate the decay constant. Then using the decay constant we can find the half life by the formula, $ {t_{1/2}} = \dfrac{{\ln 2}}{\lambda } $ .

Formula Used: In this solution, we will be using the following formula,
 $ N = {N_o}{e^{ - \lambda t}} $
Where $ {N_o} $ is the initial activity and $ N $ is the final activity,
 $ \lambda $ is the radioactive decay constant and $ t $ is the time interval
and $ {t_{1/2}} = \dfrac{{\ln 2}}{\lambda } $
where $ {t_{1/2}} $ is the half life of the substance.

Complete Step by Step Solution
In this problem we are provided that the activity of the radioactive substance was initially $ 4700 $ counts per minute. So we have,
 $ {N_o} = 4700 $
Again after 5 minutes the activity was reduced to $ 2700 $ counts per minute. So we get,
 $ N = 2700 $ and $ t = 5\min $
So we can write the first order decay equation as,
 $ N = {N_o}{e^{ - \lambda t}} $
So by substituting the values in this equation we can find out the decay constant. So after substituting the values we get,
 $ 2700 = 4700{e^{\left( { - \lambda \times 5} \right)}} $
On keeping the exponential on one side,
 $ \dfrac{{2700}}{{4700}} = {e^{\left( { - \lambda \times 5} \right)}} $
Taking ln on both the sides of the equation we get,
 $ - 5\lambda = \ln \left( {\dfrac{{27}}{{47}}} \right) $
On doing the calculation, we get
 $ - 5\lambda = - 0.55 $
Therefore, from here we get the value of the decay constant as,
 $ \lambda = 0.11{\min ^{ - 1}} $
Now using this value of the decay constant, we can find the half life of the radioactive substance using the formula,
 $ {t_{1/2}} = \dfrac{{\ln 2}}{\lambda } $
So on substituting the values we have,
 $ {t_{1/2}} = \dfrac{{0.693}}{{0.11}}\min $
On calculating this gives us,
 $ {t_{1/2}} = 6.25\min $

Hence the decay constant of the radioactive substance is $ 0.11{\min ^{ - 1}} $ and the half life is $ 6.25\min $ .

Note The decay constant of a nuclide is the probability that the nuclide will decay by that given mechanism. It is related to the half-life of that substance by the formula $ {t_{1/2}} = \dfrac{{\ln 2}}{\lambda } $ , where the half life is the time taken by the concentration of the substance to become exactly half of the initial concentration.