The Bohr formula gives `E=E_0/n^2` for the nth excited state with `E_0=13.6 eV` being the ground state energy.

The relationship between energy and wavelength is

`Delta E=hc/(Delta lambda)`

`Delta lambda=(hc)/(Delta E)=(hc)/(E_0/n_f^2-E_0/n_i^2)=(hc)/E_0*(1/n_f^2-1/n_i^2)^-1`

A nice thing to know is `barh*c~~200 MeV*fm=2*10^-7 eV*m` , where `barh` is the reduced planks constant.

`Delta lambda...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

The Bohr formula gives `E=E_0/n^2` for the nth excited state with `E_0=13.6 eV` being the ground state energy.

The relationship between energy and wavelength is

`Delta E=hc/(Delta lambda)`

`Delta lambda=(hc)/(Delta E)=(hc)/(E_0/n_f^2-E_0/n_i^2)=(hc)/E_0*(1/n_f^2-1/n_i^2)^-1`

A nice thing to know is `barh*c~~200 MeV*fm=2*10^-7 eV*m` , where `barh` is the reduced planks constant.

`Delta lambda ~~(2pi*2*10^-7 eV*m)/(13.6 eV)*(1/4-1/9)^-1`

`Delta lambda =(2pi*2*10^-7 eV*m)/(13.6 eV)*36/5`

`Delta lambda ~~ (6*2*10^-7 eV*m)/(14 eV)*7`

`Delta lambda ~~(12*10^-7 eV*m)/(2 eV)`

`Delta lambda ~~6*10^-7 m`

`Delta lambda =600*nm`

**Further Reading**