# The half-life of cobalt-60 is 5.3 years. How many years will it take for 1/4 of the original amount of coblat-60 to remain?

Oct 22, 2017

It this not TWO half-lives.....?

#### Explanation:

After one-life, $\frac{1}{2}$ the mass of the original isotope remains. After another half-life, $\frac{1}{4}$ the mass of the original isotope remains.

And if a half-life is $5.3$ years long, then this is a period of 10.6 years.....

How long until $\frac{1}{16}$ of the original mass remains?

Oct 22, 2017

$10.6 y r s$

#### Explanation:

The half-life equation is :

A(t)=A_0*(1/2)^(t/(t_("1/2"))

What we know:

• t_("1/2")->"half-life"=5.3yrs.
• ${A}_{0} \to \text{Initial quantity}$
• $A \left(t\right) \to \text{Amount left after t years} = \frac{1}{4} {A}_{0}$
• t->"time undergone"=color(blue)(?

Substituting in the equation:

$= \frac{1}{4} \cancel{{A}_{0}} = \cancel{{A}_{0}} \cdot {\left(\frac{1}{2}\right)}^{\frac{\textcolor{b l u e}{t}}{5.3}}$

$= \frac{1}{4} = {\left(\frac{1}{2}\right)}^{\frac{\textcolor{b l u e}{t}}{5.3}}$

Take the $\log$ of both sides:

$\log \left(\frac{1}{4}\right) = \log \left({\left(\frac{1}{2}\right)}^{\frac{\textcolor{b l u e}{t}}{5.3}}\right)$

$= \log \left(\frac{1}{4}\right) = \frac{\textcolor{b l u e}{t}}{5.3} \cdot \log \left(\frac{1}{2}\right)$

Dividend both sides by color(red)(log(1/2)

=cancel(log(1/4)/color(red)(log(1/2)))^2=color(blue)t/5.3*cancel(log(1/2)/color(red)(log(1/2))

$= 2 = \frac{\textcolor{b l u e}{t}}{5.3}$

$\implies \textcolor{b l u e}{t} = 2 \cdot 5.3 = 10.6 y r s .$