Question

The time gap between 44% decay and 93% decay of a radioactive substance is 81 minutes. The half-life of the radioactive substance in minutes is:

• A. 18
• B. 54
• C. 27
• D. 9

Hint:

To find the half-life from two different percentages of decay over a specific time, use the natural logarithm of the ratios and solve for \( k \) first.

Answer 1

The Correct Option is C

Using the decay formula \( N = N_0 e^{-kt} \) where \( N \) is the remaining amount, \( N_0 \) is the initial amount, \( k \) is the decay constant, and \( t \) is time. 
We set up equations for the 44% and 93% decay points: For 44% decay (56% remaining): \[ N_1 = 0.56 N_0 = N_0 e^{-k t_1} \] \[ \ln(0.56) = -k t_1 \] For 93% decay (7% remaining): \[ N_2 = 0.07 N_0 = N_0 e^{-k t_2} \] \[ \ln(0.07) = -k t_2 \] The time gap between these two decays is given as 81 minutes: \[ t_2 - t_1 = 81 \] \[ -k t_2 + k t_1 = 81k \] \[ \ln(0.07) - \ln(0.56) = 81k \] \[ k = \frac{\ln(0.07/0.56)}{81} \] The half-life \( T \) of the substance is given by \( T = \frac{\ln(2)}{k} \): \[ T = \frac{\ln(2)}{\frac{\ln(0.07/0.56)}{81}} \] Calculating \( T \) gives us the half-life: \[ T = \frac{81 \ln(2)}{\ln(0.07/0.56)} \]