The half-life of C14 radioactive is 5760 yr. After how much time will 200 mg C14 sample be reduced to 25 mg?

Ln(200/25)=Ln(2)/5760*t

Ln(8)=Ln(2)/5760*t
Ln(2^3)=Ln(2)/5760*t
3Ln(2)=Ln(2)/5760*t
3=t/5760
t=17280

To determine the time it takes for a 200 mg sample of C14 to be reduced to 25 mg, we can use the concept of half-life.

The half-life of C14 is 5760 years, which means that after every 5760 years, half of the C14 in a sample will decay.

Let's break down the steps:

Step 1: Calculate the number of half-lives that have occurred.

Since each half-life represents a reduction of 50%, we can find the number of half-lives that have occurred by dividing the initial mass (200 mg) by the final mass (25 mg):

200 mg / 25 mg = 8

This means that 8 half-lives have occurred.

Step 2: Calculate the total time.

To determine the total time it took for 8 half-lives to occur, we multiply the half-life (5760 years) by the number of half-lives:

5760 years * 8 = 46080 years

Therefore, it will take approximately 46080 years for a 200 mg sample of C14 to be reduced to 25 mg.

To determine the time it takes for a given sample of C14 to decay to a specific amount, you can use the formula for exponential decay:

N(t) = N₀ * (1/2)^(t / t₁/₂)

Where:
N(t) is the remaining amount of C14 at time t
N₀ is the initial amount of C14
t is the time elapsed
t₁/₂ is the half-life of C14

In this case, we know that the initial amount (N₀) is 200 mg, and we want to find the time (t) when the remaining amount (N(t)) is 25 mg. The half-life (t₁/₂) is given as 5760 years.

Plugging in the values into the formula:

25 mg = 200 mg * (1/2)^(t / 5760)

To solve for t, we can take the logarithm of both sides of the equation to bring the exponent down:

log(25) = log(200) + log(1/2)^(t / 5760)

Using logarithm properties, we can simplify the equation further:

log(25) = log(200) + (t / 5760) * log(1/2)

Now, isolate the variable t:

(t / 5760) = (log(25) - log(200)) / log(1/2)

Simplify and compute the right side:

(t / 5760) = log(25/200) / log(1/2)

(t / 5760) = log(1/8) / log(1/2)

Use the change-of-base formula to convert the logarithms:

(t / 5760) = logₓ(1/8) / logₓ(1/2)

Assuming the logarithm base is 10:

(t / 5760) = log₁₀(1/8) / log₁₀(1/2)

Evaluate the logarithms:

(t / 5760) = -0.9031 / -0.3010

Simplify:

(t / 5760) = 3

Multiply both sides by 5760 to isolate t:

t = 3 * 5760

t = 17,280 years

Therefore, it will take approximately 17,280 years for a 200 mg C14 sample to be reduced to 25 mg.

k = 0.693/t1/2

ln(No/N) = kt
No = 200
N = 25
k from above
Solve for t in years.