Question

How do you calculate a wavelength of light emitted when

Answer 1

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

Explanation:

$Li:3:1s^22s^1$

$Li^{2+}:1:1s^1$

Using Rydberg's Equation for hydrogen and hydrogen like atom:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant

$n_f$ = Higher energy level  

$n_i$= Lower energy level

We have:

$n_f=8, n_i=12$

$R_H=1.097\times 10^7 m^{-1}$

$\frac{1}{\lambda}=1.097\times 10^7 m^{-1}\times \left(\frac{1}{12^2}-\frac{1}{8^2} \right )$

$\frac{1}{\lambda}=1.097\times 10^7 m^{-1}\times \frac{5}{576}$

$\frac{1}{\lambda}=9.523\times 10^4 m$

$\frac{1}{\lambda}=9.523\times 10^4 m$

${\lambda}=0.1050\times 10^{-4}m$

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