91Ó°ÊÓ

1. Half-life of${}^{49}\mathrm{Sc},{\mathrm{T}}_{1/2}=57.0\min$,
2. At, the counting rate of sample is.${\mathrm{R}}_0=6000\frac{\mathrm{counts}}{\min }$
3. The counting rate of background activity is.$\mathrm{R}=30\frac{\mathrm{counts}}{\min }$

The strength of a radioactive source is called its activity, which is defined as the rate at which the isotope decays. Specifically, it is the number of atoms that decay and emit radiation in one second to the number of counts per min.

The activity of undecayed sample remaining after a given time is as follows

$A=A_0{e}^{-\left(\frac{\ln 2}{T_{1/2}}\right)t}$ …… (i)

From equation (i), the fraction of the initial activity to the present activity of the sample after the decay can be used to get the time of the decay.

Now, according to the given problem, A = 30counts/min and${\mathrm{A}}_0=6000\frac{\mathrm{counts}}{\min }$

Thus, the fractional value of the activity is given using equation (i) as:

$$\begin{gathered} \frac{A_0}{A}=\frac{6000}{30} \\ {e}^{\left(\frac{\ln 2}{T_{{\displaystyle \frac{1}{2}}}}\right)}=200 \end{gathered}$$

From the above relation, it can be seen that the exponential value is 200. Again, for the given half-life of the sample to be $57\min ≈1\mathrm{hour}\mathrm{and}\ln 2=0.693\mathrm{or}≈1$

Obtain the exponential value of the time of decay that is the time at which the activity of the sample is equal to the background activity $\left(\mathrm{A}=30\frac{\mathrm{counts}}{\min }\right)$is 200. That implies

$$\begin{gathered} e^t≈200 \\ t≈5\mathrm{hr} \end{gathered}$$

Hence, the counting rate of the sample will be equal to the background rate in 3h, 7h, 10h.