The half-life for the radioactive decay of C-14 is 5730 years and is independent of the initial concentration. How long does it take for 25% of the C-14 atoms in a sample of the C-14 to decay

It will take 2378 years for 25% of the C-14 atoms in a sample of the C-14 to decay.

Explanation:

Radioactive decays/reactions always follow a first order reaction dynamic.

Let the initial amount of C-14 atoms be A₀ and the amount of atoms at any time be A

The general expression for rate of reaction for a first order reaction is

(dA/dt) = -kA (Minus sign because it's a rate of reduction)

k = rate constant

(dA/dt) = -kA

(dA/A) = -kdt

 ∫ (dA/A) = -k ∫ dt 

Solving the two sides as definite integrals by integrating the left hand side from A₀ to A and the right hand side from 0 to t.

We get

In (A/A₀) = -kt

(A/A₀) = e⁻ᵏᵗ

A(t) = A₀ e⁻ᵏᵗ

Although, we can obtain k from the information on half life.

For a first order reaction, the rate constant (k) and the half life (T(1/2)) are related thus

T(1/2) = (In2)/k

T(1/2) = 5730 years

k = (In 2)/5730 = 0.000120968 = 0.000121 /year.

So, the amount of C-14 atoms left at any time is given as

A(t) = A₀ e⁻⁰•⁰⁰⁰¹²¹ᵗ

How long does it take for 25% of the C-14 atoms in a sample of the C-14 to decay?

When 25% of C-14 atoms in a sample decay, 75% of C-14 atoms in the sample remain.

Hence,

A(t) = 75%

A₀ = 100%

100 = 75 e⁻⁰•⁰⁰⁰¹²¹ᵗ

e⁻⁰•⁰⁰⁰¹²¹ᵗ = (75/100) = 0.75

In e⁻⁰•⁰⁰⁰¹²¹ᵗ = In 0.75 = - 0.28768

-0.000121t = -0.28768

t = (0.28768/0.000121) = 2,377.54 = 2378 years

Hope this Helps!!!

particle   p is an electron and particle q is a neutron