Question

Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from the n = 5 state to the n = 3 state. (Rh= 2.18×10⁻¹⁸ J)

a) 1.55x10-19j

b)2.18x10-18j

c)2.9x10-19j

d3.44x10-18j

Answer 1

E3 - E5    = $\Delta$ E     = -2.18*10^-18(1/n3^2 - 1/n5^2)

                    = -2.18*10^-18(1/(3)^2-1/(5)^2 )

                  = -2.18*10^-18(1/9 -1/25)

                 = -2.18*10^-18(0.111-0.04)

                   = -1.55*10^-19J

$\Delta$E     = hv

$\Delta$E    = hc/$\lambda$

$\lambda$      = hc/$\Delta$E   

          = 6.625*10^-34 *3*10^8/(1.55*10^-19)

          = 1.282*10^-6m

          = 1282*10^-9m

           = 1282nm >>>>answer