If the fraction of the atoms of a radioactive material left after 120years is \(\frac{1}{64}\), what is the half-life of the material? \(\frac{1}{64}\)
20 years
10 years
2 years
24 years
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Discussions (12)
Chai
U guys want to confuse people the more, u are already making me sick,
the easiest way is to use Zhepwo rule,
since is dealing with remainder.
1/N=1/R-----zhepwo rule1
R=2^n--------zhepwo rule2
n=t/T-----Zhepwo rule3
1/N=1/R .. rule1 to get R.
1/64=1/R
R=64
but R=2^n..rule2 to get n.
64=2^n
2^6=2^n
n=6
n = t/T...rule3 to get half life T
time t = 120yrs
6 = 120/T
T= 20yrs// which is A. pls 
the method u guys brought is correctoo but is difficult to understand. I use zhepwo method to solve any half life question and it has been helping matters. I rest my case.
Answer is A
in 20 years = 1/2
40 =1/4
60 = 1/8
80 = 1/16
100 = 1/32
120 = 1/64
So its 20 years
I20 YEARS = 1/64
It takes 6 lives to arrive at 1/2 from 1/64
120 years/6 = 20 years.
thus the 1/2 is 20 years.
The best solution
Fraction remaining = 1/64 = 1/2^6
T= 120yrs and t(half life) =?
but Fr = 1/2^n therefore , n = 5
and.
n = T/t
5 = 120/t
t = 24yrs
CORRECT ANSWER - D
I think the answer is D. Because counting down you have 1/64, 1/32, 1/16, 1/8, 1/4, then 1/2. were where 120years gives you 1/64. 1/64 = 5 half life. Therefore t>1/2= 120/5 = 24years
