Question

A radioactive element with $6\times10^{5}$ atoms initially is left with $0.75\times10^{5}$ undecayed atoms in $48$ years. The half-life is:

• A. 16 years
• B. 24 years
• C. 12 years
• D. 6 years
• E. 18 years

Hint:

$0.75$ is $1/8$ of $6$. Since $1/8 = (1/2)^3$, exactly 3 half-lives have passed.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem deals with radioactive decay and the concept of half-life. The half-life is the time required for half of the radioactive nuclei in a sample to decay.
Step 2: Key Formula or Approach:
The law of radioactive decay gives the number of undecayed nuclei (N) at time t: \[ N(t) = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}} \] where \( N_0 \) is the initial number of nuclei and \( T_{1/2} \) is the half-life. Alternatively, we can express the number of half-lives that have passed as \( n = t/T_{1/2} \), so the formula becomes \( N = N_0(1/2)^n \). Step 3: Detailed Explanation:
We are given: - Initial number of atoms, \( N_0 = 6 \times 10^5 \) - Number of undecayed atoms at time t, \( N = 0.75 \times 10^5 \) - Time elapsed, \( t = 48 \) years Let's use the formula \( N = N_0(1/2)^n \) to find the number of half-lives, n, that have occurred. \[ 0.75 \times 10^5 = (6 \times 10^5) \left(\frac{1}{2}\right)^n \] First, solve for the fraction \( N/N_0 \): \[ \frac{N}{N_0} = \frac{0.75 \times 10^5}{6 \times 10^5} = \frac{0.75}{6} \] To simplify this fraction, we can write \( 0.75 \) as \( 3/4 \). \[ \frac{N}{N_0} = \frac{3/4}{6} = \frac{3}{4 \times 6} = \frac{3}{24} = \frac{1}{8} \] Now we have: \[ \frac{1}{8} = \left(\frac{1}{2}\right)^n \] We can write 8 as a power of 2: \( 8 = 2^3 \). \[ \frac{1}{2^3} = \left(\frac{1}{2}\right)^n \] \[ \left(\frac{1}{2}\right)^3 = \left(\frac{1}{2}\right)^n \] By comparing the exponents, we find that the number of half-lives is \( n=3 \). We know that the total time elapsed is the number of half-lives multiplied by the half-life period: \[ t = n \times T_{1/2} \] We are given \( t = 48 \) years and we found \( n = 3 \). \[ 48 = 3 \times T_{1/2} \] Solve for the half-life, \( T_{1/2} \): \[ T_{1/2} = \frac{48}{3} = 16 \text{ years} \] Step 4: Final Answer:
The half-life time of the radioactive element is 16 years.