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Question: The activity of a radioactive sample is measured as \(9750 \,count/min\) at \(t = 0\) and \(975 \,co...

The activity of a radioactive sample is measured as 9750 count/min9750 \,count/min at t=0t = 0 and 975 count/min975 \,count/min at t=5mint = 5 min. The decay constant is approx.
A 0.922min⁡−10.922{\min ^{ - 1}}
B 0.691min⁡−10.691{\min ^{ - 1}}
C 0.461min⁡−10.461{\min ^{ - 1}}
D 0.230min⁡−10.230{\min ^{ - 1}}

Explanation

Solution

Hint
The radioactive decay constant is defined as the probability of a given unstable nucleus per unit time. It is denoted by λ\lambda, and it can be calculated by using the formula R=R0e−λtR = {R_0}{e^{ - \lambda t}}where, RR and R0R_0 are the activities at time t=0t = 0 and time t=tt = t respectively. The values of RR and R0R_0 are given in the question so by substituting these values and solving the equations, we get the value of decay constant.

Complete step by step answer
we know that the activity of the sample is given as,
R=R0e−λtR = {R_0}{e^{ - \lambda t}} where, λ\lambda is the decay constant, R0R_0 is the activity at the time t=0 and R is the activity at time tt.
⇒RR0=e−λt\Rightarrow \dfrac{R}{{{R_0}}} = {e^{ - \lambda t}}
Now, taking natural log both sides, we get
⇒ln⁡(RR0)=−λt\Rightarrow \ln \left( {\dfrac{R}{{{R_0}}}} \right) = - \lambda t
⇒ln⁡(R0R)=λt\Rightarrow \ln \left( {\dfrac{{{R_0}}}{R}} \right) = \lambda t
⇒λ=1tln⁡(R0R)\Rightarrow \lambda = \dfrac{1}{t}\ln \left( {\dfrac{{{R_0}}}{R}} \right)
Now, converting the natural log to base 10, we get
⇒λ=2.303tlog⁡(R0R)\Rightarrow \lambda = \dfrac{{2.303}}{t}\log \left( {\dfrac{{{R_0}}}{R}} \right) … (1)
As it is given that,
Initial activity of the sample is R0=9750 count/min{R_0} = 9750 \,count/min
Activity after t = 5min is R=975 count/minR = 975 \,count/min
Put these values in the equation (1), we get
⇒λ=2.3035log⁡(9750975)\Rightarrow \lambda = \dfrac{{2.303}}{5}\log \left( {\dfrac{{9750}}{{975}}} \right)
⇒λ=2.3035log⁡(10)\Rightarrow \lambda = \dfrac{{2.303}}{5}\log \left( {10} \right)
⇒λ=2.3035=0.461min⁡−1\Rightarrow \lambda = \dfrac{{2.303}}{5} = 0.461{\min ^{ - 1}}
Hence, (C) option is correct.

Note
The radioactive decay constant is defined as the probability of a given unstable nucleus per unit time. It is denoted by λ\lambda, while calculating the decay constant make sure that the base of log should be 1010 otherwise you will get the wrong answer.