Rankers Physics

Nucleus: Practice Problem & Solution

The activity of a radioactive sample is measured as $N_0$ counts per minute at t = 0 and $N_0/e$ counts per minute at t = 5 minutes. The time (in minutes) at which the activity reduces to half its value is: (2010 Pre)
$5 \log_e 2$
$\log_e \frac{2}{5}$
$\frac{5}{\log_e 2}$
$5 \log_{10} 2$

Solution Explained:

To solve this problem, we apply the core principles of Nucleus. Understanding the underlying formula is key to arriving at the correct answer below:

Using $A = A_0 e^{-\lambda t}$, we get $N_0/e = N_0 e^{-5\lambda}$, yielding $5\lambda = 1$ and $\lambda = 1/5$ per minute. The half-life is $T_{1/2} = \frac{\ln 2}{\lambda} = 5 \ln 2 = 5 \log_e 2$.

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. A star initially has \[10^{40}\] deuterons. It produces energy via the processes ; If the average power radiated by the star is \[10^{16}\] W, the deuteron supply of the star is exhausted in a time of the order of :
2. For nuclear reaction $_{92}U^{235} + _{0}n^{1} \rightarrow _{56}Ba^{144} + ...... + 3 _{0}n^{1}$: (1998)
3. The energy of a hydrogen atom in its ground state is $-13.6 eV$. The energy of the level corresponding to the quantum number $n = 2$ in the hydrogen atom is (1996)
4. A certain mass of Hydrogen is changed to Helium by the process of fusion. The mass defect in fusion reaction is 0.02866 u. The energy liberated per u is (given 1 u = 931 MeV): (2013)
5. Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): \(\text{Fe}^{56}\) nucleus is more stable than \(\text{U}^{235}\) nucleus. Reason (R): Binding energy of \(\text{Fe}^{56}\) nucleus is more than the binding energy of \(\text{U}^{235}\). In the light of the above statements, choose the correct answer from the option given below.