Question

g A hydrogen atom initially in the n = 4 states emits a photon and makes a transition to the n = 2 level. What is the wavelength of the photon in nm? (report 4 sig figs) What part of the electromagnetic spectrum does this correspond to?

Answer 1

Answer: The wavelength of the photon is 486.2 nm and it lies in the visible region

Step-by-step explanation:

To calculate the wavelength of light, we use Rydberg's Equation:

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

Where,

`\lambda` = Wavelength of radiation

`R_H` = Rydberg's Constant =
`1.097* 10^7m^(-1)`

`n_f` = Higher energy level = 4

`n_i` = Lower energy level = 2

Putting the values in above equation, we get:

`(1)/(\lambda )=1.097* 10^7m^(-1)\left((1)/(2^2)-(1)/(4^2) \right )\\\\\lambda =4.862* 10^(-7)m`

Converting this into nanometers, we use the conversion factor:

`1m=10^9nm`

So,
`4.862* 10^(-7)m* ((10^9nm)/(1m))=486.2nm`

As, the range of wavelength of visible light is 400 nm - 700 nm. So, the wavelength of the given photon lies in the visible region

Hence, the wavelength of the photon is 486.2 nm and it lies in the visible region