How long will it take for 20% of the u−238 atoms in a sample of u−238 to decay?

2 answers:

8 0

Answer is: it takes 1,448 billion years.
The half-life for the radioactive decay of U-238 is 4.5 billion years and is independent of initial concentration.
c₀ - initial concentration of U-238.
c - concentration of U-238 remaining at time.
t = 4,5·10⁹ y.
First calculate the radioactive decay rate constant λ:
λ = 0,693 ÷ t = 0,693 ÷ 4,5·10⁹ y = 1,54·10⁻¹⁰ 1/y.
ln(c/c₀) = -λ·t₁.
ln(0,8/1) = -1,54·10⁻¹⁰ 1/y · t₁.
t₁ = 1,448·10⁹ y.

Elodia [21]3 years ago

8 0

It will require for 20 % of U-238 atoms in a sample of U-238 to decay.

Further Explanation:

Radioactive decay involves stabilization of unstable atomic nucleus and is accompanied by the release of energy. This emission of energy can be in form of different particles like alpha, beta and gamma particles.

Half-life is time period in which half of the radioactive species is consumed. It is denoted by .

The expression for half-life is given as follows:

$\lambda = \dfrac{{0.693}}{{{t_{{\text{1/2}}}}}}$ …… (1)

Where,

${t_{{\text{1/2}}}}$ is half-life period

$\lambda$ is the decay constant.

The half-life period for decay of U-238 is $4.5 \times {10^9}{\text{ yrs}}$ .

Substitute $4.5 \times {10^9}{\text{ yrs}}$ for ${t_{{\text{1/2}}}}$ in equation (1).

$\begin{aligned}\lambda&= \dfrac{{0.693}}{{4.5 \times {{10}^9}{\text{ yrs}}}} \\&= 1.54 \times {10^{ - 10}}{\text{ yr}}{{\text{s}}^{ - 1}} \\\end{galigned}$

Since it is radioactive decay, it is first-order reaction. Therefore the expression for rate of decay of U-238is given as follows:

$\lambda = \dfrac{{2.303}}{t}\log \left( {\dfrac{a}{{a - x}}} \right)$

…… (2)

Where,

$\lambda$ is the decay or rate constant.

t is the time taken for decay process.

a is the initial amount of sample.

x is the amount of sample that has been decayed.

Rearrange equation (2) to calculate t.

$t = \dfrac{{2.303}}{\lambda }\log \left( {\dfrac{a}{{a - x}}} \right)$ …… (3)

Consider 100 g to be initial amount of U-238. Since 20 % of it is decayed in radioactive process, 20 g of U-238 is decayed and therefore 80 g of the sample is left behind.

Substitute 100 g for a, 80 g for (a–x) and $1.54 \times {10^{ - 10}}{\text{ yr}}{{\text{s}}^{ - 1}}$ for $\lambda$ in equation (3).

$\begin{aligned}t &= \dfrac{{2.303}}{{1.54 \times {{10}^{ - 10}}{\text{ yr}}{{\text{s}}^{ - 1}}}}\log \left( {\dfrac{{100{\text{ g}}}}{{80{\text{ g}}}}} \right)\\&= 1.449 \times {10^9}{\text{ yrs}}\\\end{aligned}$