Question

the half life of the radioactive element strontium-90 is 29.1 years. If 16 grams of strontium-90 are initially present, how many grams are present after 58.2 years? After 291 years?

Answer 1

Answer:

4 g after 58.2 years

0.0156 After 291 years

Explanation:

Given data:

Half-life of strontium-90 = 29.1 years

Initially present: 16g

mass present after 58.2 years =?

Mass present after 291 years =?

Solution:

Formula:

how much mass remains =1/ 2n (original mass) ……… (1)

Where “n” is the number of half lives

to find n

For 58.2 years

n = 58.2 years /29.1 years

n= 2

or  291 years

n = 291 years /29.1 years

n= 10

Put values in equation (1)

Mass after 58.2 years

mass remains =1/ 22 (16g)

mass remains =1/ 4 (16g)

 mass remains = 4g

Mass after 58.2 years

mass remains =1/ 210 (16g)

mass remains =1/ 1024 (16g)

mass remains = 0.0156g

Answer 2

Answer: The amount of sample left after 58.2 years is 3.96 grams and after 291 years is 0.015 grams

Explanation:

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

We are given:

$t_{1/2}=29.1yrs$

Putting values in above equation, we get:

$k=\frac{0.693}{29.1}=0.024yrs^{-1}$

Rate law expression for first order kinetics is given by the equation:

$N=N_oe^{-kt}$      ......(1)

k = rate constant

t = time taken for decay process

$N_o$ = initial amount of the reactant

N = amount left after decay process

• When time is 58.2 years:

We are given:

$k=0.024yrs^{-1}\\t=58.2yrs\\N_o=16g$

Putting values in equation 1, we get:

$N=16\times e^{(-0.024yrs^{-1}\times 58.2yrs)}\\\\N=3.96g$

• When time is 291 years:

We are given:

$k=0.024yrs^{-1}\\t=291yrs\\N_o=16g$

Putting values in equation 1, we get:

$N=16\times e^{(-0.024yrs^{-1}\times 291yrs)}\\\\N=0.015g$

Hence, the amount of sample left after 58.2 years is 3.96 grams and after 291 years is 0.015 grams