For neutral atom, number of protons = number of electrons `=29`
Thus, atomic number of the element `=29`
Electronic configuration of element with `Z=29` will be :
`1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(10) 4s^(1) or [Ar]^(18) 3d^(10) 4s^(1)`, i.e., `._(29)Cu`.
Mass number = No. of protons + No. of neutrons `=29+35=64`