Question

If the mean life of a radioactive substance is \[ \frac{20}{\ln4} \] minutes, then the ratio of the number of atoms remaining undecayed and the number of atoms decayed of the substance at a time of \(30\) minutes is

• A. \(3:8\)
• B. \(7:8\)
• C. \(1:7\)
• D. \(1:8\)

Hint:

For radioactive decay, \[ \boxed{ N=N_0e^{-\lambda t}, \qquad \lambda=\frac1{\tau}, } \] where \[ \boxed{\tau} \] is the mean life.

Answer 1

The Correct Option is C

Step 1: Find the decay constant. Mean life is \[ \tau=\frac1\lambda. \] Given, \[ \tau=\frac{20}{\ln4}. \] Hence, \[ \lambda = \frac{\ln4}{20}. \]

Step 2: Calculate the number of undecayed atoms. The radioactive decay law is \[ N=N_0e^{-\lambda t}. \] For \[ t=30\text{ min}, \] \[ N = N_0e^{-\frac{\ln4}{20}\times30} = N_0e^{-\frac32\ln4}. \] Since \[ 4^{3/2}=8, \] \[ N = \frac{N_0}{8}. \]

Step 3: Find the required ratio. Number of atoms decayed is \[ N_0-N = N_0-\frac{N_0}{8} = \frac{7N_0}{8}. \] Therefore, \[ \text{Undecayed}:\text{Decayed} = \frac{N_0}{8}:\frac{7N_0}{8} = 1:7. \] Hence, \[ \boxed{1:7}. \] Thus, \[ \boxed{(C)} \] is the correct answer.