Question

What is the wavelength (in nm) of the photon absorbed for a transition of an electron from ninitial = 2 that results in the least energetic spectral line in the visible series of the H atom?

Answer 1

Answer: 656.6 nm.

Step-by-step explanation:

Using Rydberg's Equation for hydrogen atom:

`(1)/(\lambda)=R_H\left((1)/(n_i^2)-(1)/(n_f^2) \right )`

Where,

`\lambda` = Wavelength of radiation

`R_H` = Rydberg's Constant

`n_f` = Higher energy level = 3 (least energetic for visible series)

`n_i`= Lower energy level = 2

We have:

`n_f=3, n_i=2`

`R_H=1.097* 10^7 m^(-1)`

`(1)/(\lambda)=1.097* 10^7 m^(-1)* \left((1)/(2^2)-(1)/(3^2) \right )`

`(1)/(\lambda)=1.097* 10^7 m^(-1)* (5)/(36)`

`(1)/(\lambda)=0.1523* 10^(7) m`

`\lambda=6.566* 10^(-7)m=656.6nm`

(
`1 m= 10^9nm`)

The wavelength of the photon emitted when the hydrogen atom undergoes a transition from n = 2 to n = 3 is 656.6 nm.