An  $\alpha -particle$ with K.E. T= 5.3 Mev initiates a nuclear reaction  $B{e}^9{+}_2{\alpha }^4\to C^{12}{+}_0{n}^1$  with energy yield Q=+5.7 Mev. Then the K.E. of neutron outgoing at the right angle to the motion direction of the  $\alpha -particle$ will be [initially Be⁹ is at rest]  $\frac{68}{n}$ (in MeV), then n is____