Question

How do you calculate a wavelength of light emitted when

theelcetron in doubly ionized lithium makes a transiton
fromE12 to E8?

Answer 1

Answer: Wavelength of light emitted when the electron in doubly ionized lithium makes a transition from E12 to E8 is
`0.1050* 10^{-4m`

Step-by-step explanation:

`Li:3:1s^22s^1`

`Li^(2+):1:1s^1`

Using Rydberg's Equation for hydrogen and hydrogen like 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

`n_i`= Lower energy level

We have:

`n_f=8, n_i=12`

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

`(1)/(\lambda)=1.097* 10^7 m^(-1)* \left((1)/(12^2)-(1)/(8^2) \right )`

`(1)/(\lambda)=1.097* 10^7 m^(-1)* (5)/(576)`

`(1)/(\lambda)=9.523* 10^4 m`

`(1)/(\lambda)=9.523* 10^4 m`

`{\lambda}=0.1050* 10^(-4)m`

The wavelength of the photon emitted when the hydrogen atom undergoes a transition is
`0.1050* 10^{-4m`