Question

The half life of a certain isotope is 2 yrs a sample of 3200 kgs of the isotopes is considered if x kgs of sample remains after 12 yrs then x =?

Answer 1

dude ur funny and creepy

Answer 2

Why???

Answer 3

12 years is \(12\div2=6\) half lives

Answer 4

\[A(t)=A_02^{-t/t_{1/2}}\]

Answer 5

solve via \(3200\times \frac{1}{2^6}\)

Answer 6

\[A(t)=A_02^{-t/t_{1/2}}\]
\[\downarrow\]
\[X(12)=3200\cdot2^{-12/2}\]

Answer 7

Can u guys tell me wat u r soing

Answer 8

where that went......

Answer 9

my arithmetic is bad
let me try again
start with \(3200\) and the half life is 2 years so in two years there will be
\[3200\times \frac{1}{2}=1600\]

Answer 10

in another two years there will be half of that, so
\[1600\times \frac{1}{2}=800\]
in another two years half of that, so
\[800\times \frac{1}{2}=400\] and in another two years half that so
\[400\times\frac{1}{2}=200\] and so on
multiply by \(\frac{1}{2}\) for each half life

Answer 11

it is much easier to do this in one step
12 years is 6 half lives, so you can compute in one step via
\[3200\times \left(\frac{1}{2}\right)^6\]

Answer 12

After n half lives, the fraction of radioactive element left is:\[(\frac{1}{2})^n\]

Answer 13

rather than dividing by two repeatedly

Answer 14

2^6 = 64

Answer 15

=50

Answer 16

Yup......thxxx

Answer 17

\[X=3200\cdot2^{-12/2}\]\[=2^5\times10^2\times2^{-6}\]\[=2^{-1}\times10^2\]\[=\frac12\times100\]