Question

Cal. wave no. for longest wavelength tranistion in Balmer series of atomic hydrogen​

Answer 1

complete question :-

Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen

Answer:

According to Balmer formula

v = λ1=R H

[ n 121 − n 221 ]

For the Balmer series, ni = 2.

Thus, the expression of wavenumber(ṽ) is given by,

Wave number (ṽ) is inversely proportional to the wavelength of transition. Hence, for the longest wavelength transition, ṽ has to be the smallest.

For ṽ to be minimum, nf should be minimum. For the Balmer series, a transition from ni = 2 to nf = 3 is allowed. Hence, taking nf = 3,we get:

ṽ= 1.5236 × 10(6) m(–1)