Question

the difference in wavelength of second and third lines of Balmer series in the atomic spectrum is​

Answer 1

Answer:

0.6Rh.

Explanation:

Since, the formulae of the wave length λ is given by the Rydberg equation which states that 1/λ = Rh(1/nf^2 - 1/ni^2) where Rh is the Rydberg constant and the nf and the ni are the number of shells, for Balmer the value of nf =2 and from the question we have third spectral line which is ni=5 and second spectral line which has ni = 4.

So, solving the two equations separately we will get that 1/λ = Rh(1/4 - 1/16) or the value of λ = 16/3Rh.

Again on taking ni = 5 we will get the 1/λ = Rh(1/4 - 1/25) or on solving we get the value of λ =100/21Rh.

So, the difference will be (5.33-4.76)Rh = 0.6Rh.