radioactive radium has a half-life of 1620 yrs, if the initial quantity is 5g how much remains after 600yrs?

Question

Using the equation y=Ce^kt

Answer 1

k=(-ln 2)/(half life)=-(ln 2)/1620.
C=5g
t=600

so y=5*e^(-ln2/1620*600)=3.8679g

Answer 2

a slightly different way of doing it. It is like having a balance(shares) in bank with -ive intersest rate(compound) that reduces principle to half in 1620 years so if
p = 100
a =50
t = 1620
r=?

a = p(1-r/100)^t

or 50/100 = (1-r/100)^1620 eqn -1

==>>You may jump to short cut given near end or

taking log
log (0.5) = 1620 log( 1-0.01r)
0.01r = 1-antilog ( (log 0.5)/1620)

>>so r = 100X { 1- antilog [ (log 0.5)/1620] }

use tables to get the value of r
now p =5gm
a= ?
r = obtained above
t= 600

a = 5X( 1- r/100)^600 u have the answer (solve using log)


===--->>>short cut
0.2 a = (1-r/100) ^ 600 ---eqn 1
0.5 = (1-r/100)^1620 ---eqn 2
do the following
( eqn --1)^27
(eqn --2)^10
you have
(0.2 a )^27 = (1-r/100)^16200
and (0.5)^10 = (1-r/100) ^16200

Divide

(0.2 a)^27 = 0.5^10

27 log (2 /10) +log a = 10 log (5/10)

log a = 10 log5 -10 -27log2 +27
antilog gives the correct answer.