Question

A sample of ${}^{64}\mathrm{Cu}$exhibits 4225 cpm. After 10.0 h, the same sample gave 2898 cpm. Calculate the half-life of localid="1646069107365" ${}^{64}\mathrm{Cu}$

Short answer

The half-life of the sample ${}^{64}\mathrm{Cu}$is $18.4\mathrm{h}$.

Step-by-step solution

Given information 

A sample of ${}^{64}\mathrm{Cu}$exhibits $4225\mathrm{cpm}$. After 10.0 h, the same sample gave 2898 cpm. Calculate the half-life of ${}^{64}\mathrm{Cu}$.

Formula for half life 

The following formula can be used to compute the fraction of sample and time.

$\ln \left(\frac{N}{N_0}\right)=-\lambda t\dots \dots \dots \dots ..\left(1\right)$

And the half-time formula is as follows:

$t_{1/2}=\ln \frac{2}{\lambda }$

or,

$t_{1/2}=\frac{0.693}{\lambda }\dots \dots \dots \dots \dots \left(2\right)$

where,

$$\begin{gathered} t_{1/2}=halftime \\ \lambda =decaycons\tan t \\ N=numberofradioactivenucleiattimet \\ {N}_0=initialnumberofreadioactivenuclie \end{gathered}$$

Calculation for the half life

Calculate the value of constant-decay decay using the supplied values in equation $\left(1\right).$

$\begin{array}{r}\ln \left(\frac{2898}{4225}\right)=-\lambda \times 10h\\ -0.377=-\lambda \times 10h\\ \lambda =0.0377h^{-1}\end{array}$

In equation $\left(2\right)$, enter the value of the sample's decay constant,

localid="1646203986719" $$\begin{gathered} t_{1/2}=\frac{0.693}{0.0377{\mathrm{h}}^{-1}} \\ =18.4\mathrm{h} \end{gathered}$$

Final answer

The half time of the sample ${}^{64}\mathrm{Cu}$ is $18.4h$.