The ratio of the energies of the hydrogen atom in its first to second excited state is

Options

A. ` 1//4 `
B. `4//9 `
C. `9//4 `
D. ` 4 `

Answer

C

Explanation

As we know that the energy of an electron in `n^ ( th ) ` orbit of the hydrogen atom is given by
` therefore E _n = ( -13.6 ) /( n^2 ) eV `
For the first excited state , ` n = 2 `
` therefore E _ e = ( -13.6 ) /( 2^ 2 ) eV `
For the second excited state ` n = 3 `
` therefore E _ 3 = ( - 13.6 ) /( 3 ^ 2 ) eV `
` therfore ( E _ 2 ) /( E _ 3 ) = ( 3^2 ) /( 2^ 2 ) = (9 ) /( 4 ) `