Question 814289
A 1200 kg of a radioactive element kept decaying for 50 years until 300 kg of the element remained.
 What is the half-life of the element? 
:
The radioactive decay formula: A = Ao*2^(-t/h), WHERE:
A = Amt remaining after t time (300 mg)
Ao = initial amt (t=0) (1200 mg)
t = time of decay (50 yrs)
h = half-life of substance
:
1200*2^(50/h) = 300
2^(-50/h) = {{{300/1200}}}
2^(-50/h) = .25
log[2^(-50/h)] = log(.25)
log equiv of exponents
{{{-50/h}}}log(2) = log(.25)
{{{-50/h}}} = {{{log(.25)/log(2)}}}
using your calc
{{{-50/h}}} = -2
h = {{{(-50)/(-2)}}}
h = +25 yrs is the half life
:
:
Check this on your calc, enter 1200*2^(-50/25) results: 300