All Exams Test series for 1 year @ ₹349 only
Question

A nucleus decays by the emission of a gamma ray from an excited state of spin-parity $2^+$ to the ground state with spin-parity $0^+$. What is the type of the corresponding radiation?

The correct answer is
electric quadrupole

Nuclear Decay Radiation Type Analysis

To determine the type of radiation emitted during a gamma decay, we examine the changes in nuclear spin ($J$) and parity ($\pi$) between the initial and final states.

Spin and Parity Change Calculation

  • Initial state spin-parity: $J_i = 2$, $\pi_i = +$
  • Final state spin-parity: $J_f = 0$, $\pi_f = +$
  • Change in angular momentum (spin): $\Delta J = |J_i - J_f| = |2 - 0| = 2$
  • Change in parity: $\Delta \pi = \pi_i \times \pi_f = (+) \times (+) = +$ (Parity is conserved)

Determining Radiation Multipolarity

The multipole order ($L$) corresponds to the change in angular momentum, so $L = \Delta J = 2$. This indicates a Quadrupole transition.

We need to determine if it's Electric (E) or Magnetic (M):

  • Electric transitions (EL): Parity changes according to $(-1)^L$. For $L=2$, the parity change is $(-1)^2 = +1$. This means parity is conserved.
  • Magnetic transitions (ML): Parity changes according to $(-1)^{L+1}$. For $L=2$, the parity change is $(-1)^{2+1} = (-1)^3 = -1$. This means parity is flipped.

Since the parity is conserved in this transition (from $2^+$ to $0^+$), the radiation must be an Electric transition.

Conclusion

With $L=2$ and conserved parity, the transition corresponds to an Electric Quadrupole (E2) radiation.

Was this answer helpful?

Important Questions from Miscellaneous

  1. Which one of the following pairs of international organizations and their headquarters is incorrect?

  2. In a bacterial cell, a protein is synthesized at random location in the cytoplasm. The protein has to reach one pole of the cell for its appropriate function. The protein reaches the pole by
  3. A solid spherical cork of radius $R$ and specific gravity $0.5$ floats on water. The cork is pushed down so that its centre of mass is at a distance $h$ (where $0 < h < R$) below the surface of water, and then released. The volume of the part of the cork above water level is $\pi R^3 \left(\frac{2}{3} - \cos\theta_0 + \frac{1}{3}\cos^3\theta_0\right)$, where $\theta_0$ is the angle as shown in the figure. 

    At the moment of release, the dependence of the upward force on the cork on $h$ is

  4. The diatomic molecule HF has an absorption line in the rotational band at $40\text{ cm}^{-1}$ for the isotope $^{18}\text{F}$. The corresponding line for the isotope $^{19}\text{F}$ will be shifted by approximately
  5. The low lying energy levels due to the vibrational excitations of an even-even nucleus are shown in the figure below.

     

    The spin-parity $J^p$ of the level $E_1$ is

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App