Question

Americium-241 is a radioactive substance used in smoke detectors. The half life of americium is 432 years. If a smoke detector initially contains 1 gram of Americium 241, how much will remain in 432 years?

Question 4 options:

0.5 grams


1.0 grams


1.5 grams


2.0 grams

Answer 1

Answer : The correct option is, (2) 0.5 grams

Solution : Given,

As we know that the radioactive decays follow first order kinetics.

First we have to calculate the half life of Americium-241.

Formula used :
`t_(1/2)=(0.693)/(k)`

Putting value of half-life in this formula, we get the rate constant.

`432years=(0.693)/(k)`

`k=1.6* 10^(-3)year^(-1)`

The expression for rate law for first order kinetics is given by :

`k=(2.303)/(t)\log(a)/(a-x)`

where,

k = rate constant =
`1.6* 10^(-3)year^(-1)`

t = time taken for decay process = 432 years

a = initial amount of the Americium-241 = 1 g

a - x = amount left after decay process = ?

Putting values in above equation, we get

`1.6* 10^(-3)=(2.303)/(432)\log(1)/(a-x)`

`a-x=0.502g=0.5g`

Therefore, the amount remain in 432 years will be, 0.5 grams