Question

Question 46:
If the total energy of an electron in third orbit of a hydrogen like atom is -13.6 eV find the atomic number of the
atomn/ion​

Answer 1

Answer:

3($Li^{+2}$)

Explanation:

Total energy of an electron in a hydrogen like atom = $-13.6\frac{Z^{2}}{n^{2}}$  

                                                                                      ;where Z = atomic number

                                                                                            and n = energy level.

Here, Z =?   and   n = 3

=> $-13.6=-13.6\frac{Z^{2}}{3^{2}}$

=> Z = 3.

Therefore, atomic number of the ion = 3