Question:
A $^{60}$Co nucleus emits a $\beta$-particle and is converted to $^{60}$Ni$^{*}$ with $J^P = 4^+$, which in turn decays to the $^{60}$Ni ground state with $J^P = 0^+$ by emitting two photons in succession, as shown in the figure. Which one of the following statements is CORRECT?
A $^{60}$Co nucleus emits a $\beta$-particle and is converted to $^{60}$Ni$^{*}$ with $J^P = 4^+$, which in turn decays to the $^{60}$Ni ground state with $J^P = 0^+$ by emitting two photons in succession, as shown in the figure. Which one of the following statements is CORRECT?
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For transitions between nuclear energy levels, electric and magnetic quadrupole transitions are common for $\Delta J = 2$ and $\Delta J = 1$, respectively.
Updated On: Aug 30, 2025
- $4^+ ⇒ 2^+$ is an electric octupole transition
- $4^+ ⇒ 2^+$ is a magnetic quadrupole transition
- $2^+ ⇒ 0^+$ is an electric quadrupole transition
- $2^+ ⇒ 0^+$ is a magnetic quadrupole transition
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The Correct Option is C
Solution and Explanation
- The $2^+ ⇒ 0^+$ transition is an electric quadrupole transition. This is because electric quadrupole transitions occur when a change in the angular momentum quantum number of a nucleus is $\Delta L = 2$ and there is an even parity change between the initial and final states. The energy difference between the $2^+$ and $0^+$ states also supports this transition.
- The $4^+ ⇒ 2^+$ transition involves a change of $\Delta J = 2$, and therefore, it is an electric quadrupole transition, not octupole.
Thus, the correct answer is (C) $2^+ ⇒ 0^+$ is an electric quadrupole transition.
- The $4^+ ⇒ 2^+$ transition involves a change of $\Delta J = 2$, and therefore, it is an electric quadrupole transition, not octupole.
Thus, the correct answer is (C) $2^+ ⇒ 0^+$ is an electric quadrupole transition.
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