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

Explanation:

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

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

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = $1.097\times 10^7m^{-1}$

$n_f$ = Higher energy level = 4

$n_i$ = Lower energy level = 2

Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.097\times 10^7m^{-1}\left(\frac{1}{2^2}-\frac{1}{4^2} \right )\\\\\lambda =4.862\times 10^{-7}m$

Converting this into nanometers, we use the conversion factor:

$1m=10^9nm$

So, $4.862\times 10^{-7}m\times (\frac{10^9nm}{1m})=486.2nm$

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