calculate the wavelength of the light emitted when an electron in a hydrogen atom n=4 to n=3

To calculate the wavelength of the emit light when an electron  move from n=4 to n= 3,

 we start by

calculating the energy difference between these two levels

∆E = E₄ – E₃

E₆= E₀ /n²

but E₀  = -13.6 eV

 = -13.6/16

=-0.85 eV

E₃ = E₀ /n²

 = -13.6/9

 =-1.511

 the energy difference is

∆E = E₄ – E₃

=-0.377-(-1.511)

= 1.134 eV

from E = hf = hc/λ

λ =hc/E

all symbols have their usual meaning

λ = (6.626× 10^-34 × 3×10⁸)/ 1.134eV

but eV  =1.6×10-¹⁹c

therefore λ = (6.626× 10^-34 × 3×10⁸)/ (1.134×1.6×10-¹⁹)

 use your calculator