Respuesta :
Answer : The correct answer is 96.68 yrs
Radioactivity Decay :
it is a process in which a nucleus of unstable atom emit energy in form of radiations like alpha particle , beta particle etc .
Radioactive decay follows first order kinetics , so its rate , rate constant , amount o isotopes can be calculated using first order equations .
The first order equation for radioactive decay can be expressed as :
[tex] ln \frac{N}{N_0} = - k*t [/tex] ----------- equation (1)
Where : N = amount of radioisotope after time "t"
N₀ = Initial amount of radioisotope
k = decay constant and t = time
Following steps can be used to find time :
1) To find deacy constant :
Decay constant can be calculated using half life . Decay constant and half life can be related as :
[tex] T _\frac{1}{2} = \frac{ln2}{k} [/tex] ---------equation (2)
Given : Half life of Strontium -90 = 28.8 years
Plugging value of [tex] T_\frac{1}{2} [/tex] in above formula (equation 2) :
[tex] 28.8 yrs = \frac{ln 2}{ k } [/tex]
Multiply both side by k
[tex] 28.8 yrs * k = \frac{ln 2 }{k} * k [/tex]
Dividing both side by 28.8 yrs
[tex] \frac{28.8 yrs * k}{28.8 yrs} = \frac{ln 2}{28.8 yrs} [/tex]
(ln 2 = 0.693 )
k = 0.0241 yrs⁻¹
Step 2 : To find time :
Given : N₀ = 10.3 ppm N = 1.0 ppm k = 0.0241 yrs⁻¹
Plugging these value in equation (1) as :
[tex] ln (\frac{1.0 ppm}{10.3 ppm} ) = - 0.0241 yrs^-^1 * t [/tex]
[tex] ln (0.0971 ) = -0.0241 yrs ^-^1 * t [/tex]
(ln 0.0971 = - 2.33 )
Dividing both side by - 0.0241 yrs⁻¹
[tex] \frac{-2.33}{-0.0241 yrs^-^1} = \frac{-0.0241 yrs^-^1 * t}{-0.0241 yrs^-^1} [/tex]
t = 96.68 yrs
Hence the concentration of Strontium-90 will drop from 10.3 ppm to 1.0 ppm is 96.68 yrs
96.9 years
Further explanation
Given:
- The half-life for beta decay of strontium-90 is 28.8 years.
- A milk sample is found to contain 10.3 ppm strontium-90.
Question:
How many years would pass before the strontium-90 concentration would drop to 1.0 ppm?
The Process:
In the calculations of half-lives, the expressions are the following:
[tex]\boxed{ \ N = \frac{N_o}{2^n} \ }[/tex]
where [tex]\boxed{ \ n = \frac{t}{t_{1/2}} \ }[/tex] are used. In these expressions;
- N₀ = initial number = 10.3 ppm
- N = amount of substances remained = 1.0 ppm
- t = time passed
- n = the number of half-lives
- [tex]t_{1/2} =[/tex] half-live = 28.8 years
Step-1: find out the number of half-lives (n)
[tex]\boxed{ \ N = \frac{N_o}{2^n} \ } \rightarrow \boxed{ \ 2^n = \frac{N_o}{N} \ }[/tex]
[tex]\boxed{ \ 2^n = \frac{10.3}{1.0} \ }[/tex]
[tex]\boxed{ \ 2^n = 10.3 \ }[/tex]
[tex]\boxed{ \ n \cdot \ln{2} = \ln{10.3} \ } \rightarrow \boxed{ \ n = \frac{\ln{10.3}}{\ln{2}} \ }[/tex]
We get n = 3.364572
Step-2: find out in how many years would pass before the strontium-90 concentration would drop to 1.0 ppm
[tex]\boxed{ \ n = \frac{t}{t_{1/2}} \ } \rightarrow \boxed{ \ t = n \times t_{1/2} \ }[/tex]
t = 3.364572 x 28.8
t = 96.9
Thus in 96.9 years will pass before the concentration of strontium-90 will drop to 1.0 ppm.
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Notes:
The half-life of radioactive decay is the period of time required for half of the initial amount of the substance to disintegrate. The shorter the half-life of radioactive decay, the higher the rate of radioactive decay and the more radioactivity. The half-life is the characteristic property of each element.
Learn more
- After two half-lives, how many atoms—of any type—remain in the sample? https://brainly.com/question/7374780
- Determine the shortest wavelength in electron transition https://brainly.com/question/4986277
- Find out the fraction of the space within the atom is occupied by the nucleus https://brainly.com/question/10818405
