# Decay of some particle involved quarks vs mesons as outgoing states

Let's have decay width of some mother particle into the state which involves hadrons. For simplicity, let's assume that creation of hadrons (on diagram) is possible only through electroweak vertices (by the words, I don't know may this situation be real or not):

for quarks they are $$\tag 1 L^{q}_{int} = c\bar{U}_{i}\gamma^{\mu}(1 - \gamma_{5})U^{CKM}_{ij}D_{j}W_{\mu} + B\bar{q}_{i}\gamma^{\mu}(V_{i} - A_{i}\gamma_{5})q_{i}Z_{\mu},$$ while for (vector and scalar) mesons they are $$\tag 2 L^{m}_{int} = m_{m}f_{m}m_{\mu}Z^{\mu} + m'_{m}f'_{m}m_{\mu}W^{\mu} + f''_{m}\partial_{\mu}m Z^{\mu} + f'''_{m}\partial_{\mu}m W^{\mu}.$$ Let's look at two diagrams: the first one contains, for example, two quarks in out-state, while the second contains meson constructed from these two quarks. Let's then sum over all possible quarks in the first diagram and over all possible mesons in the second diagram.

In this analysis I've neglected the quarks and mesons production by gluons.

Questions:

1) Is it true that full decay width of mother particle calculated with $(1)$ will be always bigger than decay width calculated with $(2)$? If not, why (if the statement about quark production only through weak vertices holds true)?

2) Is it true that decay width of mother particle in channel with two quarks $q \bar{q}{'}$ will be always bigger than decay width in channel with meson consisted of $q, \bar{q}'$?