Need help on Half-Life isotope!?

Question

A certain radioactive isotope has a half-life of approx. 1850 years. How many years to the nearest year would be required for a given amount of this isotope to decay to 40% of that amount?

a)2046

b)1110

c)2446

d)1363

Thanks for the help

Answer 1

By eyeballs, 2046 years.  The number has to be bigger than 1850 years -- but not much bigger.  The actual value will be 1850 times the ratio of ln 0.5 to ln 0.4.

Answer 2

For more ease in calculation, check out this website:

http://www.radprocalculator.com/TimedDec...

Answer C is the correct answer.

Answer 3

ln(2/1) = k * 1850

k = 3.75x10^-4 yrs^-1

ln(100/40) = 3.75x10^-4 * t

t = 2445.6

ANSWER c) 2446

Radioactive decay is a first order reaction.

ln(Ao/A) = k * t

the natural log (ln) of the initial amount divided by the amount at time "t" equals the rate constant times the time "t".

In this case you were given the half life of 1850 years. The amount of the isotope at the half life is 1/2 of the starting amount. Let Ao = 2 and A = 1 (or Ao = 100% and A = 50%).

Solve the equation for "k" (the rate constant). Then use the rate constant to solve for "t" when Ao = 100% and A = 40%.

Answer 4

C