Question

Cobalt-60 has a half-life of about 5 years.

How many grams of a 300 g sample will remain after 20 years?

A. 0.0625 g

B. 9.375 g

C. 18.75 g

D. 37.5 g

Answer 1

C?

Answer 2

I'd be happy to give you feedback if you'd share the work you've done that indicates that C is the answer.

Answer 3

Generally, you need to figure out / obtain the pertinent "decay constant" when discussing "half life."

Exponential decay:\[A=A _{0}e ^{kt}\]
were "k" is the decay constant and is negative.

Answer 4

Let \[A=\frac{ 1 }{ 2 }A _{0}\] Subst. 5 years for t. Find k.

Answer 5

or this formula
k = ln (.5) / half-life

Answer 6

Yes, that'd be a bit faster. Thank you.

Answer 7

k = -0.1386294361

Answer 8

the observed relationship , ie "Cobalt-60 has a half-life of about 5 years"is:

\(C(t) \approx C_o \left( \dfrac{1}{2} \right)^{t/5}\)

where t is measured in years

that is the observed and fundamental relationship. it is easy to transfer that into variations of e and so on.

but you may also just say that

\(C(20) = C_o \left( \dfrac{1}{2} \right)^{20/5}\)

Answer 9

So for 300g and 20y we test it

\[C(20) = 300 \left( \dfrac{1}{2} \right)^{20/5}\]

which is C

Answer 10

ending amt = bgn * (2.71828^(k * time))
ending amt = 300 * (2.71828^( -0.1386294361 * 20))
ending amt = 300 * (2.71828^(-2.7725887222))
ending amt = 300 * 0.0625
ending amt = 18.75

Answer 11

Thank you both.

Answer 12

u r welcome

Answer 13

I wish I could medal both of you, though.

Answer 14

meddle the wolf!!!!!

Answer 15

I guess we're both good at this exponential growth / decay stuff,huh?

Answer 16

nah

@Yuii

fan up :)

Answer 17

LOL IrishBoy