Question

In the emission spectrum of hydrogen the transitions observed in this experiment are in the visible region corresponding to the Balmer series in other series emmission lines are present in different regions of the eletromagnetic spectrum
Calculate the wavelenth of the n=4 to n=1 and the n=4 to n=3 transitions. Indicate in which regions of the electromagnetic spectrum these transitions would occur.

Answer 1

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  

$n_i$ = Lower energy level

1)  The wavelength of the n=4 to n=1.

$n_i=1,n_f=4$

Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.097\times 10^7m^{-1}\left(\frac{1}{1^2}-\frac{1}{4^2} \right )\\\\\lambda =9.273\times 10^{-8}m=92.73 nm$

$1m=10^9nm$

The region of this electromagnetic transition will be in a ultraviolet region.

2) The wavelength of the n=4 to n=1.

$n_i=3,n_f=4$

Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.097\times 10^7m^{-1}\left(\frac{1}{3^2}-\frac{1}{4^2} \right )\\\\\lambda =1.875\times 10^{-6}m=1875 nm$

$1m=10^9nm$

The region of this electromagnetic transition will be in an infrared region.