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The clectron in a hydrogen atom, originally in level n=9, undergoes a transition to a lower level by emitting a photon

of wavelength 384 nm. What is the final level of the electron?

c = 3.00 x 10 m/s,

h = 6.626 X 10^-34J-s,

Rh= 2.179 X 10^-18

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Respuesta :

Explanation:

The given data is as follows.

           [tex]n_{1}[/tex] = 9,          [tex]n_{2}[/tex] = ?

              Z for hydrogen = 1

As we know that,

                       Energy (E) = [tex]\frac{hc}{\lambda}[/tex]

where,   h = planck's constant = [tex]6.626 \times 10^{-34}[/tex] Js

              c = speed of light = [tex]3 \times 10^{8}[/tex] m/s

         [tex]\lambda[/tex] = wavelength

According to Reydberg's equation, we will calculate the energy emitted by the photon as follows.

         [tex]\Delta E = -2.179 \times 10^{-18} J \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}][/tex]

or,     [tex]\frac{hc}{\lambda}[/tex] = [tex]-2.179 \times 10^{-18} J \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}][/tex]

Putting the given values into the above equation as follows.

         [tex]\frac{hc}{\lambda}[/tex] = [tex]-2.179 \times 10^{-18} J \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}][/tex]

        [tex]\frac{6.626 \times 10^{-34} Js \times 3 \times 10^{8}m/s}{384 \times 10^{-9} m}[/tex] = [tex]-2.179 \times 10^{-18} J \times (1)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{(9)^{2}}][/tex]  

                   n = 2

Thus, we can conclude that the final level of the electron is 2.

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