A radioactive substance has a half life of 20 days. What fraction of the original radioactive nuclei will remain after 80 days?

Fiyin1550

2 years ago

n=t\T
n=80/20=4
R=2^n
R=2^4=16
Fraction remaining=1/R
=1/16

Eduphy

10 years ago

I really don't get this one

Esseoghene

9 years ago

Pls can some1 giv a clearer explanation

Newton2i

2 years ago

no you guys are wrong heres the real deal
80 60 40 20 thats total number of 4 and we have the formular to be
N/N = (1/2)^n
now considering the nucleir path
1/2)n where n = 4
opening the bracket we have something like 2^4 which is equal to 2 x 2 x 2 x 2 = 16 then sub it in the space of 2 making
1/16

Excel34

9 years ago

am not clear with this answer, pls put more light

Edeke

7 years ago

The solution given has an error in the calculation : it is supposed to be raised to the power of 4 not 2. Thank you.

awusiobi

9 years ago

please d answer shud be C.. after 20 days 1/2 will decay. after 40 days 1/4. after 60 days 1/6 wil decay and after 80 days 1/8 will decay. so d fraction dat remains after 80 days is 1/8