Question

A count rate meter is used to measure the activity of a given sample. At one instant, the meter shows 4750 counts per minute. Five minutes later, it shows 2700 counts per minute. Find:
(a) the decay constant and
(b) the half-life of the sample. (${log}_{10}1.760 = 0.2455$)

Answer 1

Hint: The radioactive decay constant is defined as the probability of a given unstable nucleus decaying per unit time. It is denoted by $\lambda$.

Complete step by step answer:
(a) We know that the activity of the sample is given as,
$ R\quad =\quad { R }_{ 0 }{ e }^{ -\lambda t }$
where $\lambda$ is the decay constant, ${R}_{0}$ is the activity at time 0 and R is the activity at time t.
$\implies \cfrac { R }{ { R }_{ 0 } } \quad =\quad { e }^{ -\lambda t }$
Now, taking natural log both sides we get,
$ ln(\cfrac { R }{ { R }_{ 0 } } )\quad =\quad -\lambda t$
$\implies ln(\cfrac { { R }_{ 0 } }{ { R } } )\quad =\quad \lambda t$
$\implies \lambda \quad =\quad \cfrac { 1 }{ t } ln(\cfrac { { R }_{ 0 } }{ { R } } )$
Now converting the natural log to base 10 we get,
$ \lambda \quad =\quad \cfrac { 2.303 }{ t } log(\cfrac { { R }_{ 0 } }{ { R } } )$ -----(1)
Now, it is given in the question that t=5 mins, Activity at time 0, ${R}_{0}$ is 4750 count/minute and Activity at time t, R is 2700 count/minute. Substituting the value in equation (1) we get,
$ \lambda \quad =\quad \cfrac { 2.303 }{ 5 } log(\cfrac { 4750 }{ { 2700 } } )$
$\implies \lambda \quad =\quad \cfrac { 2.303 }{ 5 } log(1.76)$
$\implies \lambda \quad =\quad \cfrac { 2.303 }{ 5 } \quad \times \quad 0.2455$
$\implies \lambda \quad =\quad 0.1131/min$
Therefore, the decay constant is 0.1131/minute.

(b) Now, we know that the mean life of the sample is denoted by $\tau$ and is given as the reciprocal of the decay constant.
$ \tau \quad =\quad \cfrac { 1 }{ \lambda }$
And we found the value of $\lambda$ as 0.1131/min. Substituting the value of decay constant in the above equation, we get
$ \tau \quad =\quad \cfrac { 1 }{ 0.1131 }$
$\implies \tau \quad =\quad 8.85\quad min$
Now, half-life of the sample is given as,
$ T\quad =\quad 0.693\quad \times \quad \tau$
Substituting the value of $\tau$, we get
$ T\quad =\quad 0.693\quad \times \quad 8.85$
$\implies T\quad =\quad 6.13\quad min$
Therefore, the half-life of the sample is 6.13 minutes.

Note: While calculating the decay constant, do make sure that the base of the log is 10 else you might end up getting the wrong answer. The half-life of a sample is 69.3% of the mean life. It is applicable for any sample.