## anonymous 5 years ago The half-life of a certain radioactive element is 10 years. How much of a 20g sample will be left after 8 years? (Hint: first find the decay constant.)

1. anonymous

$N(t) = N_0 e^{-\lambda t}. \,$ use this, substitute the values in, and you will get the decay constant

2. anonymous

I already used 10 = ln(2) / k ---> ln(2) / 10 = k. But then what?

3. anonymous

Then I got A = 20e^((ln(2)/10)10). What's next? where do the 8 years come in?

4. anonymous

ok, the amount of radioactive element left after 8 years will be amount remained =20/(2(8/10)) =12.5g

5. anonymous

in my oppinion

6. anonymous

where'd you get the 20 from?

7. anonymous

oh yea i see sorry.

8. anonymous

ok, what does half-life mean in the first pace? it is 10 years, meaning that after 10 years the amount will become smaler by a factor of two, that is we need to divide it by 2, but since we only wait 8 years, which is 8/10 or 0.8 of the half life,we have to multiply the half life by the ratio 0.8 to 1 so, it is 2*0.8=1.6, so after 8 years, the sample reduses by a factor of 1.6

9. dumbcow

Use the equation posted at the top and use t=8, k = ln(2)/10, N_0 = 20 Answer should be around 11.49 rounded