Question

Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.

Answer 1

According to Balmer formula

ṽ=1/λ = RH[1/n12-1/n22]

For the Balmer series, ni = 2.

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

Wave number (ṽ) is inversely proportional to 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 × 106 m–1

Answer 2

Answer:

According to Balmer formula

ṽ=1/λ = RH[1/n12-1/n22]

For the Balmer series, ni = 2.

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

Wave number (ṽ) is inversely proportional to 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 × 106 m–1

Explanation: