Question

The rate of a radioactive disintegration at an instant is \( 10^8 \, {s}^{-1} \). The half-life of the radioactive sample is \( 3.3 \times 10^{12} \, {s} \). The number of radioactive atoms present in the sample at that instant of time is

• A. \( 4.7 \times 10^{10} \)
• B. \( 3.6 \times 10^{20} \)
• C. \( 4.7 \times 10^{20} \)
• D. \( 4.7 \times 10^{25} \)

Hint:

The number of radioactive atoms can be determined using the relationship between the disintegration rate, decay constant, and half-life.

Answer 1

The Correct Option is C

Given: - Disintegration rate, \( \frac{dN}{dt} = 10^8 \, {s}^{-1} \) - Half-life, \( T_{1/2} = 3.3 \times 10^{12} \, {s} \) 
Step 1: Calculate the decay constant \( \lambda \) The decay constant \( \lambda \) is related to the half-life by: \[ \lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{3.3 \times 10^{12}} \approx 2.1 \times 10^{-13} \, {s}^{-1} \] 
Step 2: Calculate the number of radioactive atoms \( N \) The disintegration rate is given by: \[ \frac{dN}{dt} = \lambda N \] Solving for \( N \): \[ N = \frac{\frac{dN}{dt}}{\lambda} = \frac{10^8}{2.1 \times 10^{-13}} \approx 4.76 \times 10^{20} \] 
Final Answer:  \( 4.7 \times 10^{20} \)