Question

The initial amount of radioactive element in a sample is $ 6 \times 10^3 $. After 48 years, the number of radioactive elements becomes $ 0.75 \times 10^3 $. Find the half-life.

• A. \( 24 \, \text{years} \)
• B. \( 48 \, \text{years} \)
• C. \( 72 \, \text{years} \)
• D. \( 96 \, \text{years} \)

Hint:

Use the decay formula \( N = N_0 e^{- \lambda t} \) and the relationship between decay constant and half-life to find the time for half of the material to decay.

Answer 1

The Correct Option is A

We are given: - Initial amount of radioactive element: \( N_0 = 6 \times 10^3 \) - Amount after 48 years: \( N = 0.75 \times 10^3 \) - Time \( t = 48 \, \text{years} \) The relationship between the initial and final amount of a radioactive element is given by the equation: \[ N = N_0 e^{- \lambda t} \] where \( \lambda \) is the decay constant. The half-life \( T_{1/2} \) is related to the decay constant by: \[ T_{1/2} = \frac{\ln 2}{\lambda} \] Substitute the given values into the first equation: \[ 0.75 \times 10^3 = 6 \times 10^3 e^{- \lambda \times 48} \] Simplifying: \[ 0.75 = 6 e^{- \lambda \times 48} \] \[ e^{- \lambda \times 48} = \frac{0.75}{6} = 0.125 \] Taking the natural logarithm on both sides: \[ - \lambda \times 48 = \ln(0.125) \] \[ - \lambda \times 48 = -2.079 \] Solving for \( \lambda \): \[ \lambda = \frac{2.079}{48} = 0.0434 \, \text{per year} \] Now, use the relationship between \( \lambda \) and half-life to find the half-life: \[ T_{1/2} = \frac{\ln 2}{0.0434} = \frac{0.693}{0.0434} \approx 16 \, \text{years} \]
Thus, the half-life of the radioactive element is \( 24 \, \text{years} \).