When alpha decay happens the daughter nucleus is usually in ground state.
When beta decay happens the daughter nucleus is usually in excited state which is then de-excited by emitting gamma photon.
Why is this?
[Physics] Why alpha decay usually leads to ground state while beta decay usually leads to excited state
nuclear-physicsradiation
Related Solutions
At first, consider two particles decay:
$A\rightarrow B + e^-$ Where A is initially rest. So $\vec{p_B}+\vec{p_{e^-}}=0$
now \begin{align} \frac{p_B^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}&=E_{released} \\ \frac{p_{e^-}^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}&=E_{released} \tag{1} \end{align}
see here you have uncoupled equation (equ.1) for $p_{e^-}$ .. So, solving above (equ.1) you will get a fixed $p_{e^-}$. Hence the energy($\sqrt{p_{e^-}^2+m_{e^-}^2}$) of the $\beta$ particle is always fixed in the two body $\beta$ decay. (you can find $p_{B}$ too using the momentum conservation formula)
At first, consider three particles decay:
$A\rightarrow B + e^-+\nu_e$ Where A is initially rest. So $\vec{p_B}+\vec{p_{e^-}}+\vec{\nu_e}=0$
so \begin{align} \frac{p_B^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}+p_{\nu_e}&=E_{released} \\ \frac{(\vec{p_{e^-}}+\vec {p_{\nu_e}})^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}+p_{\nu_e}&=E_{released} \tag{2} \end{align}
Now see in contrast to the two particles decay equation, here we have five unknowns $\big{(}p_{e^-},p_{\nu},p_{B},\theta\ (\rm the\ angle\ between\ \vec{p_{e^-}},\vec{p_{\nu}}),\phi\ (\rm the\ angle\ between\ \vec{p_{e^-}},\vec{p_{B}})\big)$ but four coupled equations *. so we can't solve them uniquely. That's also what happens physically. You will get different values of $p_{e^-},p_{\nu},p_{B},\theta,\phi$ satisfying the four coupled equations. Hence different $\beta$ particles will have different energy($\sqrt{p_{e^-}^2+m_{e^-}^2}$) maintaining the statistics of decay process. Hence the continuous spectra.
*The four coupled equations are equ.2 and three equations which we can get by taking Dot products of $\vec{p_{e^-}},\vec{p_{\nu}}\rm\ and\ \vec{p_{B}}$ with the momentum conservation($\vec{p_B}+\vec{p_{e^-}}+\vec{\nu_e}=0$ ) and remember they lie in a plane so two angles ($\theta\rm\ and\ \phi$) are sufficient.
In general, decays which release a lot of energy are faster than than decays which release only a little energy.
As Carl Witthoft points out, the first excited state is unlikely to decay to the ground state partially because the energy difference is small, and partially because the spin difference is large. (The photon for the $5^+\to1^+$ transition within $\rm ^{116}In$ would have to carry angular momentum $4\hbar$: a so-called electric hexadecapole (E4) transition.)
We can make the same combination angular momentum / energy argument for the beta decays. Beta decays are parity-violating: the electron tends to come out left-handed, and the antineutrino comes out right-handed. The two leptons are most likely to leave the nucleus without orbital angular momentum (that is, in an $s$-wave state), and if you detect them in roughly opposite directions then together they carry one unit $\hbar$ of spin. We therefore expect the beta transition to strongly prefer $\Delta J=1$ transitions. If the beta decay were to have $\Delta J > 1$, the leptons would have to be emitted with some $p$-wave or higher orbital angular momentum; those wavefunctions have much less overlap with the nucleus than the $s$-wave.
So the energetic decay $\rm^{116}In(5^+) \not\to {}^{116}Sn(0^+)$ is strongly suppressed by angular momentum considerations, and the preferred decays for the isomer $\rm^{116}In(5^+) \to {}^{116}Sn(4^+)$ only have an energy of about $\rm1.5\,MeV$. That low-energy decay proceeds more slowly than the $\rm3.9\,MeV$ decay $\rm^{116}In(1^+) \to {}^{116}Sn(0^+)$.
For internal consistency in this seat-of-my-pants, hand-waving argument, notice that the decay to the lowest-energy $4^+$ state is about five times more likely than the decay to the highest-energy $4^+$ state.
My golden-rule skills are too weak to make this argument quantitative; I'd love to see a definitive answer.
Best Answer
The question uses the term "Usually" which is not a correct description , however the decay schemes can be understood by analzing the process in detail.
An alpha particle is identical to a helium nucleus, being made up of two protons and two neutrons bound together.
There are models in which a nucleus can be seen as cluster of alpha-particles; say Carbon -12 as composed of three alpha particles.
In the decay process it comes out from the nucleus of its parent atom, (invariably one of the heaviest elements) by quantum mechanical process of tunneling and is repelled further from it by electric force , as both the alpha particle and the nucleus are positively charged.
The process changes the original atom (its mass number decreasing by 4 and atomic number by 2) from which the alpha particle is emitted into a different element called daughter nucleus.
Sometimes one of these daughter nuclides will also be radioactive, usually decaying further by one of the other processes.
This tunneling through the barrier depends on the barrier potential defined by strong nuclear interaction and as the decay process is intended for stabilizing the nucleus to lowermost energy levels therefore many a time the daughter nucleus is found in the ground state but this energy transfer also leads to daughter being in an excited state and later reaching the ground state by emitting a beta particle or a gamma radiation.
Beta electron emission occurs by the transformation of one of the nucleus’s neutrons into a proton, an electron and an antineutrino.
Beta positron decays is a similar process, but involves a proton changing into a neutron, a positron and a neutrino.
The above process gets into motion for unbalanced nucleus where excess proton or neutron is found. The decay process is guided by weak interaction and the parity and angular momentum conservation/non-conservation are guiding principles which determine the transition to be allowed or forbidden.
The Q value of such reactions plays an important role and the presence of a free proton after its conversion and its spin relations with the associated electron plays a significant role as to the decay leading to a Fermi-transition or GT-allowed or a mixture of the two.
The beta decay process usually lands a daughter nucleus in an excited state from where it goes to ground or lower energy state by gamma transition.
The detail analysis of the transition is essential to find the final energy of the daughter nucleus. One can attribute it to the complexity of the beta decay process.
After a nucleus undergoes alpha or beta decay, it is often left in an excited state with excess energy and goes to stabilizing itself by gamma emission.
For a detailed analysis one can see: Chapter 8 Beta Decay (pdf) from a Nuclear Chemistry course by Loveland.