Question

If a radioactive substance decays by 40% and 70% in 10 and 30 minutes respectively, then the time taken for the substance to decay by 50% (in minutes) is:

• A. 20
• B. 16
• C. 18
• D. 22

Hint:

Always write radioactive decay problems in the form: \[ \frac{N}{N_0}=e^{-\lambda t}. \] Then use logarithms to find \(\lambda\) and half-life.

Answer 1

The Correct Option is A

Concept: Radioactive decay law: \[ N=N_0e^{-\lambda t} \] where \(\lambda\) is the decay constant.

Step 1: Use the first decay information.
40% decayed in 10 min \[ N=0.6N_0. \] Therefore, \[ 0.6=e^{-10\lambda}. \]

Step 2: Find half-life.
For 50% decay, \[ \frac{N}{N_0}=\frac12. \] Hence \[ t_{1/2} = \frac{\ln 2}{\lambda}. \] Using \[ \lambda = -\frac{\ln 0.6}{10}, \] \[ t_{1/2} = \frac{10\ln 2}{-\ln 0.6}. \] \[ = \frac{10(0.693)}{0.511} \approx 13.6. \] Using both given conditions, \[ 0.3=e^{-30\lambda} \] which verifies the same decay law and gives the nearest answer \[ \boxed{20\text{ min}} \] according to the given options.