Using Bohr's model , calculate the wavelength of the radiation emitted when an electron in a hydrogen atom make a transition from the fourth energy level to the second energy level

Explanation

For Fourth level `n = 4`
For second energy level `n = 2`
energy of fourth energy level
`E_(2) = (-2.18 xx 10^(-18))/((4)^(2)) J "atom"^(-1)`
Energy of second energy level
`E_(1) = (-2.18 xx 10^(-18))/((2)^(2)) J "atom"^(-1)`
` Delata E = E_(2) - E_(1) = - 2.18 xx 10^(-18) |(1)/((4)^(2)) - (1)/((2)^(2))| J "atom"^(-1)`
`Delta E = 4.08 xx 10^(-18) J "atom"^(-1)`
The energy is given out the from of a single photon or quentum of radiation .Therefore
`:. Delta E = hv = (hc)/(Delta E) `
`:. or lambda = (6.62 xx 10^(-34) J s xx 3.0 xx 10^(8) ms^(-1))/(4.08 xx 10^(-19) J)`
`= 4.86 xx 10^(-7) m = 486 mu m`