The nucleus $\stackrel{235}{92}\mathrm{Y}$, initially at rest, decays into ${}_{90}{}^{231}\mathrm{X}$ by emitting an $\mathrm{\alpha }$-particle

The binding energies per nucleon of the parent nucleus, the daughter nucleus and $\mathrm{\alpha }$-particle are 7·8 MeV, 7·835 MeV and 7·07 MeV, respectively. Assuming the daughter nucleus to be formed in the unexcited state and neglecting its share in the energy of the reaction, find the speed of the emitted -particle. (Mass of $\mathrm{\alpha }$-particle = 6·68 = 10^–27 kg)
KE of α particle 𝐸_𝑘𝛼
$$\begin{gathered} {\mathrm{E}}_{\mathrm{ka}}=({\mathrm{m}}_{\mathrm{y}}-{\mathrm{m}}_{\mathrm{x}}-{\mathrm{m}}_{\mathrm{\alpha }}){\mathrm{c}}^2 \\ ={\mathrm{m}}_{\mathrm{y}}{\mathrm{c}}^2-{\mathrm{m}}_{\mathrm{x}}{\mathrm{c}}^2-{\mathrm{m}}_{\mathrm{\alpha }}{\mathrm{c}}^2 \end{gathered}$$
(235 × 7.8 – 231 × 7.835 – 4 × 7.07) MeV = 1833 – 1809.885 – 28.28 = 1833 – 1838.165 = -5.165 MeV
Ek < 0 wrong information