Question

in a hydrogen atom,if energy of an electron in groundstate is 13.6eV, then the energy in the 2nd excited state is

Answer 1

E = 13.6/n^2
for 2nd excited state, value of n = 3
therefore, energy becomes E = 13.6/9 = 1.51eV

Answer 2

The energy in the 2nd excited state in a hydrogen atom is -1.51 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 nth orbit

$R_H$= Rydberg constant  

n = number of orbit  

Z = atomic number

Energy of the ground state (n=1) in hydrogen atom (Z = 1)

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

$R_H=-13.6eV$

Energy of the second excited state (n=3) in hydrogen atom (Z = 1)

$E_3=R_H\times \frac{1^2}{3^2}$

$E_3=-13.6eV\times \frac{1}{9}$

$E_3=-1.51eV$

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