Question

energy of an electron in the first bohr orbit of hydrogen atom is -13.6 electron volt. energy value of electron in the excited state of lithium +2 is

Answer 1

Answer:

3.27

Explanation:

energy formula( 2.18*Z/N) but here n will be equal to 2

Answer 2

Answer: -122.4 eV

Explanation:

Energy of the nth orbit by Bohr was given by:

$E_n=R_H\times \frac{Z^2}{n^2}eV$

where,

$E_n$ = energy of $n^{th}$ orbit

$R_H$ = Rydberg constant

n = number of orbit  

Z = atomic number

Energy of the first shell(n=1) in hydrogen atom:

Z = 1

$-13.6eV=R_H\times \frac{1^2}{1^2}$

$R_H=-13.6eV$

To find  energy value of electron in the excited state of $Li^{2+}$ is:

$Li:1s^22s^1$

$Li^{2+}:1s^1$

Z = 3 , n= 1

$E_n=-13.6\times \frac{3^2}{1^2}eV=-122.4eV$

The energy value of electron in the excited state of $Li^{2+}$ is -122.4 eV