Question

Ionisation energy of hydrogen atom is 13.6 eV. Calculate the energy of electron for Li2+ and Be3+ in the first excited state.

Answer 1

Answer:

The energy of electron in 1st orbit of Li2+ and Be3+ is (13.6*9) and (13.6*16) respectively

Explanation:

The formula for total energy of electron is 13.6z^2/n^2

Answer 2

Answer:

Explanation:

Energy in each state = 13.6 x $\frac{Z^2}{n^2}$

First excited state = n = 2

Z = Li = 3 (Atomic Number)

Z = Be = 4 (Atomic Number)

Substitute values for each case:

Lithium case

13.6 x $\frac{3^2}{2^2}$  = 30.6 eV

Beryllium case

13.6 x $\frac{4^2}{2^2}$  = 54.4 eV