Chemistry, 21.07.2019 07:30 marknjenbennetp3j1v1

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. how many years would pass before the strontium-90 concentration would drop to 1.0 ppm?

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Answer from: dieulynx6111

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 :

$ln \frac{N}{N_0} = - k*t$ ----------- 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 :

$T _\frac{1}{2} = \frac{ln2}{k}$ ---------equation (2)

Given : Half life of Strontium -90 = 28.8 years

Plugging value of $T_\frac{1}{2}$ in above formula (equation 2) :

$28.8 yrs = \frac{ln 2}{ k }$

Multiply both side by k

$28.8 yrs * k = \frac{ln 2 }{k} * k$

Dividing both side by 28.8 yrs

$\frac{28.8 yrs * k}{28.8 yrs} = \frac{ln 2}{28.8 yrs}$

(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 :

$ln (\frac{1.0 ppm}{10.3 ppm} ) = - 0.0241 yrs^-^1 * t$

ln (0.0971 ) = -0.0241 yrs ^-^1 * t

(ln 0.0971 = - 2.33 )

Dividing both side by - 0.0241 yrs⁻¹

$\frac{-2.33}{-0.0241 yrs^-^1} = \frac{-0.0241 yrs^-^1 * t}{-0.0241 yrs^-^1}$

t = 96.68 yrs

Hence the concentration of Strontium-90 will drop from 10.3 ppm to 1.0 ppm is 96.68 yrs

Answer from: culbertchristi6111

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:

$\boxed{ \ N = \frac{N_o}{2^n} \ }$

where $\boxed{ \ n = \frac{t}{t_{1/2}} \ }$ are used. In these expressions;

N₀ = initial number = 10.3 ppmN = amount of substances remained = 1.0 ppmt = time passedn = the number of half-lives $t_{1/2} =$ half-live = 28.8 years

Step-1: find out the number of half-lives (n)

$\boxed{ \ N = \frac{N_o}{2^n} \ } \rightarrow \boxed{ \ 2^n = \frac{N_o}{N} \ }$

$\boxed{ \ 2^n = \frac{10.3}{1.0} \ }$

$\boxed{ \ 2^n = 10.3 \ }$

$\boxed{ \ n \cdot \ln{2} = \ln{10.3} \ } \rightarrow \boxed{ \ n = \frac{\ln{10.3}}{\ln{2}} \ }$

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

$\boxed{ \ n = \frac{t}{t_{1/2}} \ } \rightarrow \boxed{ \ t = n \times t_{1/2} \ }$

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 moreAfter two half-lives, how many atoms—of any type—remain in the sample? Determine the shortest wavelength in electron transition Find out the fraction of the space within the atom is occupied by the nucleus
The half-life for beta decay of strontium-90 is 28.8 years. a milk sample is found to contain 10.3 p

The half-life for beta decay of strontium-90 is 28.8 years. a milk sample is found to contain 10.3 p

Answer from: Quest

Ineed with this also i hope you understand it

Answer from: Quest

Answer two atoms of oxygen;

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The half-life for beta decay of strontium-90 is 28.8 years. a milk sample is found to contain 10.3 p...