Question

Activity of 1g of radium $(226)$ is $1Ci$ . What is its half-life?
A. $1600year$
B. $1087year$
C. $540year$
D. $5770year$

Answer 1

Radioactive substances are substances that have a tendency to emit radiation without any type of excitation by external means. These substances are either obtained naturally or prepared artificially in the laboratory. They spontaneously undergo disintegration or decay.

Complete step by step answer:
Radioactivity is a process in which nuclei of certain elements emit radiation without any excitation from external sources. In radioactivity the number of atoms disintegrating per second is more important than the total mass of the substance. The unit of radioactivity is Curie. It is defined as the quantity of any radioactive substance that has a decay rate of $3.7 \times {10^{10}}$ disintegrations per second.
$1Curie = 3.7 \times {10^{10}}$ disintegration per second.
Curie is a non SI unit of radioactivity. The SI unit of radioactivity is Becquerel.
Specific activity of a radionuclide is the activity per gram of the substance.
Now the formula to calculate the half-life of a radioactive element is-
${t_{1/2}} = \dfrac{{0.693}}{\lambda }$
where $\lambda$ is the disintegration constant or decay constant. It is defined as the fraction of total number of atoms which disintegrate per second at any time.
Now to find out the half-life of radium we first need to calculate the disintegration constant.
$- \dfrac{{dN}}{{dt}} = \lambda \times N$
Where $- \dfrac{{dN}}{{dt}}$ is the rate of decay of a radioactive substance and $N$ is the number of atoms undergoing decay.
Now in our given problem activity of radium is $1Ci$ . It means that there are $3.7 \times {10^{10}}$ disintegrations per second. So activity or rate of decay $= - \dfrac{{dN}}{{dt}} = 3.7 \times {10^{10}}$ .
Now to calculate $N$ , we need to find out the number of atoms of radium in $1g$ of radium.
Number of atoms $= \dfrac{{given\,mass}}{{atomic\,mass}} \times {N_A}$
$= \dfrac{1}{{226}} \times 6.023 \times {10^{23}} = 2.665 \times {10^{21}}\,atoms$
Now, substituting the values
$3.7 \times {10^{10}} = \lambda \times 2.665 \times {10^{21}}$
$\lambda = \dfrac{{3.7 \times {{10}^{10}}}}{{2.665 \times {{10}^{21}}}} = 1.388 \times {10^{ - 11}}seconds$
Now calculating the half-life,
${t_{1/2}} = \dfrac{{0.693}}{\lambda }$
${t_{1/2}} = \dfrac{{0.693}}{{1.3883 \times {{10}^{ - 11}}}} = 0.499 \times {10^{11}}seconds$
Now the half-life in years will be equal to-
${t_{1/2}} = \dfrac{{0.499 \times {{10}^{11}}}}{{3600 \times 24 \times 365}}years = 1588years \approx 1600years$
So the half-life of the given radioactive sample is $1600years$ .

So, the correct answer is Option A.

Note: Half-life period of any substance is the time required to reduce to half of its initial amount. Remember the half-life period of a radioactive substance does not depend on the initial amount of the radioactive substance but depends only on the disintegration constant of the radioactive element.