Question

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

Answer 1

Explanation:

The given data is as follows.

           $n_{1}$ = 9,          $n_{2}$ = ?

              Z for hydrogen = 1

As we know that,

                       Energy (E) = $\frac{hc}{\lambda}$

where,   h = planck's constant = $6.626 \times 10^{-34}$ Js

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

         $\lambda$ = wavelength

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

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

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

Putting the given values into the above equation as follows.

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

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

                   n = 2

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