Respuesta :
Answer:
[tex]\lambda_{H} = 92.5 nm[/tex]
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:
[tex]\Delta E_{H} = 13.6[\frac{1}{n^{2}} - \frac{1}{n'^{2}}][/tex] (1)
Also,
[tex]\Delta E_{H} = \frac{hc}{\lambda_{H}}[/tex]
where
h= Planck's constant
c = speed of light
[tex]\lambda_{H}[/tex] = wavelength of radiation emitted by hydrogen atom
Therefore, eqn (1) becomes:
[tex]\frac{hc}{\lambda_{H}} = 13.6[\frac{1}{n^{2}} - \frac{1}{n'^{2}}][/tex]
[tex] \frac{6.626\times 10^{- 34}\times 3\times 10^{8}}{\lambda_{H}} = 13.6[\frac{1}{9^{2}} - \frac{1}{1^{2}}][/tex]
[tex]\frac{1.988\times 10^{- 25}}{\lambda_{H}} = 13.6\times 0.988\times 1.6\times 10^{- 19}[/tex]
(since, 1 eV = [tex]1.6\times 10^{- 19}[/tex])
[tex]\frac{1.988\times 10^{- 25}}{\lambda_{H}} = 2.14\times 10^{- 18}[/tex]
[tex]\lambda_{H} = 9.25\times 10^{-8} m = 92.5 nm[/tex]
