Answer: 656.6 nm.
Explanation:
Using Rydberg's Equation for hydrogen atom:
[tex]\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )[/tex]
Where,
[tex]\lambda[/tex] = Wavelength of radiation
[tex]R_H[/tex] = Rydberg's Constant
[tex]n_f[/tex] = Higher energy level = 3 (least energetic for visible series)
[tex]n_i[/tex]= Lower energy level = 2
We have:
[tex]n_f=3, n_i=2[/tex]
[tex]R_H=1.097\times 10^7 m^{-1}[/tex]
[tex]\frac{1}{\lambda}=1.097\times 10^7 m^{-1}\times \left(\frac{1}{2^2}-\frac{1}{3^2} \right )[/tex]
[tex]\frac{1}{\lambda}=1.097\times 10^7 m^{-1}\times \frac{5}{36}[/tex]
[tex]\frac{1}{\lambda}=0.1523\times 10^{7} m[/tex]
[tex]\lambda=6.566\times 10^{-7}m=656.6nm[/tex]
([tex]1 m= 10^9nm[/tex])
The wavelength of the photon emitted when the hydrogen atom undergoes a transition from n = 2 to n = 3 is 656.6 nm.
