The gamma decay brings a excited nucleus to a lower energy state by emitting a photon. Photon carry angular momentum with intrinsic spin of 1. Its parity is positive. Thus, we have three constrain from conversation laws immediately.

$E_f = E_i + \hbar \omega$

$J_f^{\pi_f} = J_i^{\pi_i} + L$

where $E$ is the energy of the nucleus, $\hbar \omega$ and $L$ are the photon energy and angular momentum respectively. We also have to consider the parities of electric and magnetic transition are different.

To calculate the transition rate, we can start from a classical equation of power emission from an antenna, since the photon energy is quantized, the transition rate [number of photon emitted per time] is the power divided by a photon energy.

$T(qL) = \frac{2(2L+1)}{\epsilon_0 L [(2L+1)!!]^2 \hbar} (\frac{\omega}{c})^{2L+1} B(qL)$

where $qL$ is the electromagnetic multipole with angular momentum $L$, and $B(qL)$ the the reduced transition probability, it is equal to the square of the magnitude of the transition matrix element $M_{fi}(qL)$.

In the electric transition, the multipole is

$qL = e r^L$

we assume the transition is conduced by a single nucleon and the rest of the nucleus is unaffected. The transition matrix element than can be written as

$M_{fi}(qL) = \left$

The single particle wave function can be written as

$\left|j m\right>= R_{nl}(r) [Y_l \times \chi_{1/2}]_{jm}$

The matrix elements becomes,

$M_{fi}(qL) = e \int_{0}^{\infty} R_{n_f l_f}^* r^L R_{n_i l_i} r^2 dr \times \left$

To evaluate the radial integral, we make another assumption that the nucleus is a sphere of uniform density with radius $R=r_0 A^{1/3}$,

$R_{nl}(r) = \frac{\sqrt{3}}{R^{3/2}}$, for $r, so that $\int_{0}^{R} |R_{nl}(r)|^2 r^2 dr = 1$

$\left=\frac{3}{R}\int_{0}^{R} r^{L+2} dr = \frac{3}{L+3} r_0^L A^{L/3}$

The reduced transitional probability

$B_{sp}(qL)=\sum \limits_{M m_f} |\left|^2$

$= e^{2} \left< r^{L} \right> ^{2} \sum \limits_{M m_f} \left$

the angular part could be assumed as $1/4\pi$ as the total solid angle is $4\pi$. Thus, with these three assumptions, we have the Weisskopf single particle estimation for the L-pole reduced electric transition probability

$B_W(EL) = \frac{1}{4\pi}(\frac{3}{L+3})^2 r_0^{2L} A^{2L/3} [e^2 fm^{2L}]$

For the magnetic transition, we have to take into account of the spin and orbit angular momentum. The single particle reduced transition probability

$B_{sp}(ML) = \sum \limits_{M m_f} |\left|^2$

the result,

$B_{sp}(ML)=L(2L+1) \left< r^{(L-1)} \right> ^2$

$\sum \limits_{M m_f} ((g_s - \frac{2g_l}{L+1}) \left< [ Y_{L-1} \times \vec{s} ]_{LM} \right> \left< [ Y_{L-1} \times \vec{j} ]_{LM} \right> )^2$

The term

$L(2L+1)(g_s \frac{2g_l}{L+1})^2 \sim 10$

and the rest of the angular part assumed to be $1/4\pi$ again, then

$B_W(ML) = \frac{10}{\pi}(\frac{3}{L+3})^2 r_0^{(2L-2)} A^{(2L-2)/3} \mu_N^2 fm^{2L-2}$

and notice that $\mu_N = e\hbar / (2m_p)$.

Some results can be deduced form the calculation

$B_{sp}(ML)/B_{sp}(EL) \sim 0.3 A^{-2/3}$

$B_{sp}(EL)/B_{sp}(E(L-1)) \sim \frac{1}{7} \times 10^7 A^{-2/3} E_\gamma^{-2}$

$T(E1) = 1.0 \times 10^{14} A^{2/3} E_\gamma^3$

$T(E2) = 7.3 \times 10^{7} A^{4/3} E_\gamma^5$

$T(E3) = 34 A^{2} E_\gamma^7$

$T(M1) = 3.1 \times 10^{13} A^{0} E_\gamma^3$

$T(M2) = 2.2 \times 10^{7} A^{2/3} E_\gamma^5$

$T(M3) = 10 A^{4/3} E_\gamma^7$

The deviation from the single particle limit indicates a strong collective state.

$0 \rightarrow 0$, forbidden

$1^+ \rightarrow 0^+$, M1

$2^+ \rightarrow 0^+$,  E2