Two different radioactive isotopes decay to 10% of their respective original amounts. isotope a does this in 33 days, while isotope b does this in 43 days. what is the approximate difference in the half-lives of the isotopes? a. 3 b. 10 c. 13 d. 33

Decay equation is given by

---------------------(1)

where A is amount at time t, P is original amount, k is decay constant, t is time

So using this on isotope A first. It decays to 10% of original. So lets say its original amount was , so now its amount will be 10% of which will be

So plug in A place, in P place, then 33 as time in t place and let its decay constant be . So plug all these in equation (1)

Divide both sides by

Solve for

for that take ln on both sides

-------------(2)

Similarly for isotope B, if its original amount was then 10 % of will be . And if its decay constant be and its time is 43

so plug these in equation (1)

Divide both sides by

------------------------(3)

again solve for

-------------(3)

Now half life means amount is 1/2 of original amount. So that means if P is original amount then A will become or 0.5P

Using this in equation (1) we get

Divide both sides by P

------------------------(3)

-----------(4)

thats the equation for half life

So plug result of as -0.069775 in k place in equation (4) we get its half life as

----------------------(5)

So life for isotope A is 9.933989

Similarly find half life of isotope B.

For that plug as -0.053548 in k place in equation (4) we get its half life as

-----------(6)

so half life of isotope B is 12.9443

Now find difference betwen half lives of two isotopes by subtracting results (5) and (6)

so we get final answer as 12.9443- 9.933989 = 3.01

So choice A as 3 is the right answer.

12 cups   is the answer

step-by-step explanation:

write out the polynomial, with leading coefficient 1:

y = 1 (x+2)^1 (x-7)^1 (x-5)^2

next time, share the answer choices.   .