Answer
(a) For $_{\space \space 5}^{11}B$, $E_B = 76.21 MeV $
For $_{\space \space 6}^{11}C$, $E_B =73.44 MeV$
(b) The binding energy in $_{\space \space 5}^{11}B$ is larger than $_{\space \space 6}^{11}C$.
Work Step by Step
$(a)$ The binding energy of $_{\space \space 5}^{11}B$
$E_B = (Zm_H + (A-Z)m_n -m_{B-11})c^2$
$E_B = ((5)(1.007825 u) + (11-5)(1.008665 u) - 11.009305 u)931.5 MeV/u$
$E_B = 76.21 MeV $
The binding energy of $_{\space \space 6}^{11}C$
$E_B = (Zm_H + (A-Z)m_n -m_{C-11})c^2$
$E_B = ((6)(1.007825 u) + (11-6)(1.008665 u) - 11.011434 u)931.5 MeV/u$
$E_B =73.44 MeV$
$(b)$ The binding energy in $_{\space \space 5}^{11}B$ is larger than $_{\space \space 6}^{11}C$ because $_{\space \space 5}^{11}B$ has one less proton, which means it also has one fewer electron. These make $_{\space \space 5}^{11}B$ more dense and packed tightly close to the nucleus, we can also see that the atomic mass of $_{\space \space 5}^{11}B$ is also lighter, hence results in stronger binding energy.
