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 : 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 :

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

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

where,

k = rate constant  =

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

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

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considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.