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?

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Answer 1 · 2017-10-22T09:26:08

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

After one-life, $1/2$ the mass of the original isotope remains. After another half-life, $1/4$ 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 $1/16$ of the original mass remains?

Answer 2 · 2017-10-22T09:36:28

$10.6yrs$

The half-life equation is :

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

What we know:

$t_("1/2")->"half-life"=5.3yrs.$

$A_0->"Initial quantity"$

$A(t)->"Amount left after t years"=1/4 A_0$

$t->"time undergone"=color(blue)(?$

Substituting in the equation:

$=1/4cancel(A_0)=cancel(A_0)*(1/2)^(color(blue)t/5.3)$

$=1/4=(1/2)^(color(blue)t/5.3)$

Take the $log$ of both sides:

$log(1/4)=log((1/2)^(color(blue)t/5.3))$

$=log(1/4)=color(blue)t/5.3*log(1/2)$

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

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

$=2=color(blue)t/5.3$

$=>color(blue)t=2*5.3=10.6yrs.$