Question

How do you use the Rydberg equation to determine the color of a flame test for an element. For example how do we know that potassium gives off a violet color during a flame test using the Rydberg equation?

Answer 1

When the electrons of hydrogen or hydrogenic atom absorb energy they get excited towards higher energy level. However, when these electrons fall down to lower energy levels they emit radiations and result in a line spectrum with four lines of different colours i.e. red, blue-green, blue and voilet at different wavelenths specific for each line.

Red= 656nm

Blue-green= 486nm

Blue= 434nm

Voilet= 410nm

These wavelegths are determined using Rydberg equation which is stated as below-

delta E = Rh (1/i2 - 1/f2) = hc/

where, Rh = Rydberg constant = 2.179* 10-18J

i= Lower energy level

j= Higher Energy level

h= Plank Constant = 6.626*10-34 Js

c= Speed of light = 3*108 m/s

= wavelenght

Now, if we talk about the role of this equation to determine the colour of flame test for an element then, according to the value of wavelength determined using the equation we will be able to know the colour of flame.

In case of potassium ion on heating excitation of electron occurs and on de-excitation it results in emission of light. If de-excitation occurs from j=6 to i=2 we will have-

Rh (1/i2 - 1/f2) = hc/

1/ = [Rh (1/i2 - 1/f2) ] / hc

So,

1/​​​​​​​ = 1.097*107 ( 1/22 - 1/62)

1/​​​​​​​ = 0.243*107 m

= 4.115*10-7 m =411nm which relates to voilet line.