Vacuum ultraviolet photolysis of ethyl bromide was studied at 104.8-106.7 nm (11.4-11.6 eV) in the pressure range of 0.2-18.6 torr at $25^{circ}$ using an argon resonance lamp with and without additives, i.e., NO and He. Since the ionization potential of $CH_3CH_2Br$ is lower than the photon energy, the competitive processes between the photoionization and the photodecomposition were also investigated. The observations indicated that 50% of absorbed light leads to the former process and the rest to the latter one. In the absence of NO the principal reaction products for the latter process were found to be $CH_4, C_2H_2, C_2H_4, C_2H_6, and C_3H_8$. The product quantum yields of these reaction products showed two strikingly different phenomena with an increase in reactant pressure. The major products, $C_2H_4$ and $C_2H_6$, showed positive effects with pressure whereas the effects on minor products were negative in both cases, i.e., He and reactant pressures. Addition of NO completely suppresses the formation of all products except $C_2H_4$ and reduces the $C_2H_4$ quantum yield. These observations are interpreted in view of existence of two different electronically excited states. The initial formation of short-lived Rydberg transition state undergoes HBr molecular elimination and this state can across over by collisional induction to a second excited state which decomposes exclusively by carbon-bromine bond fission. The estimated lifetime of the initial excited state was ${sim}4{ imes}10^{-10}$ sec. The extinction coefficient for $CH_3CH_2Br$ at 104.8-106.7 nm and $25{circ}$ was determined to be ${varepsilon} = (1/PL)ln(I_0/I_t) = 2061{pm}160atm^{-1}cm6{-1}$ with 95% confidence level.