Question

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?

Answer 1

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 ppm
• N = amount of substances remained = 1.0 ppm
• t = time passed
• n = 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.

- - - - - - - - - -

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

1. After two half-lives, how many atoms—of any type—remain in the sample? brainly.com/question/7374780
2. Determine the shortest wavelength in electron transition brainly.com/question/4986277
3. Find out the fraction of the space within the atom is occupied by the nucleus brainly.com/question/10818405

Answer 2

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