Question

a radioactive isotope has a half-life of 2 weeks. if the original of the radiation is 1000 counts per second, how many weeks must elapse until the strength ot the radiation is less than 1 count per second

Answer 1

1000*(1/2)^x = 1, where x is how many 'cycles' of 2 weeks you need to wait. I get x = 9.965, so you'd have to wait about 19 weeks

Answer 2

how do you get the 19 weeks

Answer 3

you do the value of x, so 9.965 * 2 = 19.93. should've said about 20 week, my mistake, but that is how you get the number of weeks

Answer 4

thanks. so the equation is the one at the top

Answer 5

please can you come back here

Answer 6

what did you want to know?

Answer 7

how did you get the 9.965

Answer 8

i solved for x in the equation 1000*(1/2)^x = 1. i had to use a calculator, but you can also use logarithms.

Answer 9

to do that you would do:
(1/2)^x = 1/1000
ln ((1/2)^x) = ln(1/1000)
x*ln ((1/2)) = ln(1/1000)
x = \(ln (1/2)/ln(1/1000)\)

Answer 10

oh sorry, i should have said:
x = ln(1/1000)/ln (1/2)

Answer 11

where 'ln' is the natural logarithm

Answer 12

than I multiply 9.9657by 2 is that right

Answer 13

yes, that will give you how long you would have to wait

Answer 14

to get a count rate of 1

Answer 15

ok understood.