Answer
(a) $Cu_{29}^{66}$
(b) $0.2MeV$
Work Step by Step
(a) We can find the required nucleus as follows:
As it is $\beta ^-$ emission, so $Z=28-(-1)=29$
This atomic number shows that the required element is copper, denoted by Cu.
It can be represented as $Cu^{66}_{29}$.
(b) We can find the required energy as
$E=|\Delta m|c^2$
We plug in the known values to obtain:
$E=(65.9291u-65.9289u)(\frac{931.5MeV/c^2}{1u})c^2=0.2MeV$
