Question

Half-life for radioactive $^{14}C$ is 5760 yr. In how many years, 200 mg of $^{14}C$ will be reduced to 25 mg?

• A. 5760 yr
• B. 11520 yr
• C. 17280 yr
• D. 23040 yr

Answer 1

The Correct Option is C

 To determine how long it takes for 200 mg of radioactive \(^{14}C\) to decay to 25 mg, we can use the concept of half-life. The half-life of a substance is the time taken for the quantity of that substance to reduce to half its initial amount due to radioactive decay.

The half-life \( T_{1/2} \) of \(^{14}C\) is given as 5760 years. The formula to find the time \( t \) required for a substance to reduce to a certain amount is derived from the exponential decay formula:

\[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]

Where:

• N(t) is the remaining quantity after time t.
• N₀ is the initial quantity.
• T_1/2 is the half-life.

 

Substituting the given values into the formula:

• Initial quantity, \( N_0 = 200 \text{ mg} \)
• Final quantity, \( N(t) = 25 \text{ mg} \)
• Half-life, \( T_{1/2} = 5760 \text{ years} \)

Therefore, the equation becomes:

\[ 25 = 200 \left( \frac{1}{2} \right)^{\frac{t}{5760}} \]

To solve for \( t \), first divide both sides by 200:

\[ \frac{25}{200} = \left( \frac{1}{2} \right)^{\frac{t}{5760}} \]

Simplifying gives:

\[ \frac{1}{8} = \left( \frac{1}{2} \right)^{\frac{t}{5760}} \]

Express \(\frac{1}{8}\) as a power of \(\frac{1}{2}\):

\[ \frac{1}{8} = \left( \frac{1}{2} \right)^3 \]

So, equating the exponents:

\[ \frac{t}{5760} = 3 \]

Thus, \( t = 3 \times 5760 = 17280 \) years.

Therefore, it takes 17280 years for 200 mg of \(^{14}C\) to decay to 25 mg.