calculate the wavelength of the light emitted when an electron in a hydrogen atom n=6 to n=2

To calculate the wavelength of the emitted light when an electron  move from n=6 to n= 3,

 we start by           

calculating the energy difference between these two levels

∆E = E₆ – E₃

E₆= E₀ /n²

but E₀  = -13.6 eV

 = -13.6/36

=-0.377

E₃ = E₀ /n²

 = -13.6/9

 =-1.511

 the energy difference is

∆E = E₅ – E₁

=-0.377-(-1.511)

= 1.1332 eV

from E = hf = hc/λ

λ =hc/E

all symbols have their usual meaning

λ = (6.626× 10^-34 × 3×10⁸)/ 1.1332 eV

but eV  =1.6×10-¹⁹c

therefore λ = (6.626× 10^-34 × 3×10⁸)/ 1.1332 ×1.6×10-¹⁹c

 use your calculator