Question

The half-life of $cobalt - 60$ is $5.3$ years. how many years will it take for $\dfrac{1}{4}$ of the original amount of $cobalt - 60$ to remain?

Answer 1

After one-life, $12$ the mass of the first isotope remains. After another half-life, $14$ the mass of the first isotope remains. Furthermore, in the event that a half-life is $5.3$ years long, at that point this is a time of $10.6$ years. $Co - 60$ has a natural half existence of $9.5$ days, and a compelling half existence of $9.5$ days.

Complete step by step answer:
Beta particles are by and large ingested in the skin and don't go through the whole body. Gamma radiation, be that as it may, can enter the body. The time needed for a radioactive substance to lose $50\%$ of its radioactivity by rot is known as the half-life. The half-existence of $cobalt - 60$ is about $5.3{\text{ }}years$.
$cobalt - 60$, radioactive isotope of cobalt utilized in industry and medication. $cobalt - 60$ is the longest-lived isotope of cobalt, with a half-existence of 5.27{\text{ }}$$$$years. It is produced by illuminating the steady isotope $cobalt - 59$ with neutrons in an atomic reactor. $cobalt - 60$ is utilized in the assessment of materials to uncover inner structure, defects, or unfamiliar articles and in the sanitization of food. In medication, it is utilized to treat malignancy and to sanitize clinical hardware.
The most widely recognized radioactive type of cobalt will be $cobalt - 60$. It is produced monetarily and utilized as a tracer and radiotherapeutic specialist. It is produced in a cycle called actuation, when materials in reactors, for example, steel, are presented to neutron radiation. Numerous components are known to be radioactive, implying that their cores crumble with time.
Now take the equation,
$= \dfrac{1}{4}\not{{{A_0}}} = \not{{{A_0}}} \cdot {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{5.3}}}}$
$= \dfrac{1}{4} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{5.3}}}}$
On taking log both sides,
$\log \left( {\dfrac{1}{4}} \right) = \log \left( {{{\left( {\dfrac{1}{2}} \right)}^{\dfrac{t}{{5.3}}}}} \right)$
$= \log \left( {\dfrac{1}{4}} \right) = \dfrac{t}{{5.3}} \cdot \log \left( {\dfrac{1}{2}} \right)$
Divide by $\log \left( {\dfrac{1}{2}} \right)$ on both sides,
$= {\not{{\dfrac{{\log \left( {\dfrac{1}{4}} \right)}}{{\log \left( {\dfrac{1}{2}} \right)}}}}^2} = \dfrac{t}{{5.3}} \cdot \not{{\dfrac{{\log \left( {\dfrac{1}{2}} \right)}}{{\log \left( {\dfrac{1}{2}} \right)}}}}$
$= 2 = \dfrac{t}{{5.3}}$
$\boxed{t = 2 \cdot 5.3 = 10.6yrs.}$

Note:
The job of the cobalt unit has somewhat been supplanted by the straight quickening agent, which can create higher-energy radiation, and doesn't deliver the radioactive waste that radioisotopes do with their chaperon removal issues. Cobalt treatment actually has a valuable task to carry out in specific applications is as yet inescapable around the world, since the hardware is generally dependable and easy to keep up contrasted with the cutting-edge direct quickening agent.