Question

Ratio of the energy of electrons in 1st shell of he and 3rd shell of li+2 is

Answer 1

Answer: 4:1

Explanation:

Answer 2

Given :

Energy of electrons in 1st shell of He and 3rd shell of $Li^2^+$

To Find :

Ratio of the energy of electrons in 1st shell of He and 3rd shell of $Li^{+2}$

Solution :

We know, Energy of an electron in nth orbit = - $\frac{z^{2} }{n^2}$ ×13.6 eV

(where Z = atomic number ; n = Shell number)

For Helium, Z = 2 and n = 1 (as given)

Energy of electron in the First shell of He (E$_1$) = $\frac{2^2}{1}$ × 13.6 eV

                                                                           = 4 × 13.6 eV

For $Li^{+2}$, Z = 3 and n = 3 (as given)

Energy of electron in the third shell of $Li^{+2}$ (E$_{2}$) = $\frac{3^2}{3^2}$ × 13.6

                                                                              = 13.6 eV

Ratio of E$_1$ to E$_2$ = $\frac{4 * 13.6}{13.6}$  = $\frac{4}{1}$

∴ The ratio of the energy of electrons in 1st shell of He and 3rd shell of $Li^{+2}$ is 4 : 1