The number of protons, neutrons and electrons in ${}_{71}{}^{175}\mathrm{Lu}$ , respectively, are: 

Nuclide is represented as ${ }_z{E}^A$.
E= Symbol of element
A= Atomic mass or mass number
Z= Number of protons= number of electrons for neutral atom.
Z= No. of protons = 71= No. of electrons
A–Z = No. of neutrons = 175-71 = 104
Note: For cation ${ }_{\mathrm{Z}}^{\mathrm{A}}{\mathrm{E}}^{+\mathrm{n}}$ , number of electrons
= z-(n)
For anion ${ }_Z^A{X}^{-n}$, number of electrons = z-(-n)
Number of protons for ion= Z