Hamiltonian for $B\rightarrow  K^{*} \nu \bar{\nu} $ decay is given [@BB] as follows:

$$\begin{aligned}
   H^{\nu   \bar{\nu}}_{eff}   &   =  &
\frac{4G_{F}}{\sqrt{2}}\left(         \frac{\alpha}{2         \pi
\sin^{2}\theta_{W}}    \right)    V^{*}_{ts}    V_{tb}   X(x_{t})
(\bar{s}\gamma_{\mu  }  L   b)(\bar{\nu}\gamma_{\mu  }  L  \nu  )
\label{Heff} \end{aligned}$$

where $x_{t}=m^{2}_{t}/M^{2}_{W}$ and $$\begin{aligned}
 X(x) & = & X_{0}(x)+\frac{\alpha_{s}}{4 \pi } \,
X_{1}(x) \end{aligned}$$ Here, $X_{0}$ represents pure electroweak one-loop contributions and $X_{1}$ results from $O(g^{4}_{2}\alpha_{s})$ two-loop diagrams. We do not display here the explicit forms of $X_{0}(x_{t})$ and $X_{1}(x_{t})$, which can be found in ref.[@BB]. At $\mu =  M_{W}$ and $m_{t}=175$ GeV, we find that $X(x_{t})=1.47$.

The differential decay rate for $B\rightarrow K^{*} \nu \bar{\nu}
$ at zero momentum transfer is $$\begin{aligned}
\frac{d\Gamma(B\rightarrow K^{*} \nu \bar{\nu})}{d q^{2}} \mid_{q^{2}=0} 
& = &
\frac{G^{2}_{F}}{192   \,  \pi^{3}   \,   m^{3}_{B}}   \,  \left(
\frac{\alpha }{2 \pi \sin^{2}\theta_{W}}  \right)^{2} |V^{*}_{ts}
V_{tb}|^{2} \, (m^{2}_{B}-m^{2}_{K^{*}})^{3}  \nonumber \\[0.4cm]
& \cdot &  |A^{B\rightarrow K^{*}}_{0}(0)|^{2} |X(x_{t})|^{2} 
 \label{eq:OT4}\end{aligned}$$

Taking $\tau_{B}=1.5       \times       10^{-12}$sec$^{-1}$, $\sin^{2}\theta_{W}=0.23$, $|V_{tb}|=1$ and $A^{B\rightarrow
K^{*}}_{0}(0)=0.4$ [@OXT] and varying $V_{ts}$ in the range $0.030 \leq |V_{ts}|  \leq 0.048$ we find

$$\begin{aligned}
4.81 \times   10^{-8}  \,   GeV^{-2}   \;\;\;  \leq   \;\;\; 
\frac{1}{\Gamma_{Tot.}}
\frac{d \Gamma(B\rightarrow K^{*} \nu \bar{\nu}) }{dq^{2}} \;\;\; 
\leq \;\;\; 1.23 \times 10^{-7} \, GeV^{-2} \end{aligned}$$

The differential decay rate for $B\rightarrow   \rho  \ell
