Question

Using the Bohr model, find the wavelength in nanometers of the radiation emitted by a hydrogen atom, when it makes a transition from the n = 9 state to the n = 1 state.

Answer 1

`\lambda_(H) = 92.5 nm`

Given:

n = 9

n' = 1

Solution:

In Bohr's Model, the radiation emitted by a hydrogen atom on transition from the energy level, n = 9 to n' = 1 is given by:

`\Delta E_(H) = 13.6[(1)/(n^(2)) - (1)/(n'^(2))]` (1)

Also,

`\Delta E_(H) = (hc)/(\lambda_(H))`

where

h= Planck's constant

c = speed of light

`\lambda_(H)` = wavelength of radiation emitted by hydrogen atom

Therefore, eqn (1) becomes:

`(hc)/(\lambda_(H)) = 13.6[(1)/(n^(2)) - (1)/(n'^(2))]`

`(6.626* 10^(- 34)* 3* 10^(8))/(\lambda_(H)) = 13.6[(1)/(9^(2)) - (1)/(1^(2))]`

`(1.988* 10^(- 25))/(\lambda_(H)) = 13.6* 0.988* 1.6* 10^(- 19)`

(since, 1 eV =
`1.6* 10^(- 19)`)

`(1.988* 10^(- 25))/(\lambda_(H)) = 2.14* 10^(- 18)`

`\lambda_(H) = 9.25* 10^(-8) m = 92.5 nm`