Respuesta :
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: 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:
[tex]t_{1/2}=\frac{0.693}{k}[/tex]
We are given:
[tex]t_{1/2}=29.1yrs[/tex]
Putting values in above equation, we get:
[tex]k=\frac{0.693}{29.1}=0.024yrs^{-1}[/tex]
Rate law expression for first order kinetics is given by the equation:
[tex]N=N_oe^{-kt}[/tex] ......(1)
k = rate constant
t = time taken for decay process
[tex]N_o[/tex] = initial amount of the reactant
N = amount left after decay process
- When time is 58.2 years:
We are given:
[tex]k=0.024yrs^{-1}\\t=58.2yrs\\N_o=16g[/tex]
Putting values in equation 1, we get:
[tex]N=16\times e^{(-0.024yrs^{-1}\times 58.2yrs)}\\\\N=3.96g[/tex]
- When time is 291 years:
We are given:
[tex]k=0.024yrs^{-1}\\t=291yrs\\N_o=16g[/tex]
Putting values in equation 1, we get:
[tex]N=16\times e^{(-0.024yrs^{-1}\times 291yrs)}\\\\N=0.015g[/tex]
Hence, the amount of sample left after 58.2 years is 3.96 grams and after 291 years is 0.015 grams
