Question

A monochromatic light of wavelength \[\lambda \] is incident on a hydrogen atom that lifts it to $3^{rd}$ orbit from ground level. Find the wavelength and frequency of the incident photon.
(Given: \[{E_3} = - 1.51{\text{ }}eV\], \[{E_1} = - 13.6eV\])

Answer 1

Hint: Use the Bohr’s theory due to the incidence of monochromatic light on the hydrogen atom that lifts it from ground level to $3^{rd}$ orbit in which there is a relationship between energy change and frequency. We also get the energy equation in terms of wavelength.

Formula used:
The change in energy of the photon,
\[\Delta E = h\nu \]
\[ \Rightarrow {E_1} - {E_3} = h\nu \].
where $h$ = planck constant and $\nu $ = frequency of the incident photon.
\[\Delta E = \dfrac{{hc}}{\lambda }\] where $\lambda $= wavelength of the incident photon and \[c\]is the speed of light.
\[ \Rightarrow {E_1} - {E_3} = \dfrac{{hc}}{\lambda }\].

Complete step by step answer:
Monochromatic light of wavelength $\lambda $ incident on a hydrogen atom. Then it is lifted to the 3rd orbit from the ground level.hence from the Bohr’s theory a photon is absorbed by the hydrogen atom.
The energy of 3rd orbit is \[{E_3}\] and the energy of ground level is \[{E_1}\].
Given, the energy of the incident photon at ground level,\[{E_1} = - 13.6eV\]
And, the energy of the incident photon at the 3rd orbit, \[{E_3} = - 1.51{\text{ }}eV\]
According to Bohr’s theory,
\[\Delta E = h\nu \]
\[ \Rightarrow {E_3} - {E_1} = h\nu \]…………………(1)
where $h$ = plank constant = $6.625 \times {10^{ - 34}}$
$\nu $ = frequency of the incident photon.
\[{E_3} - {E_1} = ( - 1.51) - ( - 13.6) = 12.09eV\]
\[ \Rightarrow {E_3} - {E_1} = 12.09 \times 1.6 \times {10^{ - 19}}\]\[J\]
\[ \Rightarrow {E_3} - {E_1} = 19.344 \times {10^{ - 19}}\]\[J\]
\[\therefore \nu = \dfrac{{{E_3} - {E_1}}}{h}\]
\[ \Rightarrow \nu = \dfrac{{19.344 \times {{10}^{ - 19}}}}{{6.625 \times {{10}^{ - 34}}}}\]
\[ \Rightarrow \nu = 2.919 \times {10^{15}}\]
Eq (1) can be written as, \[{E_3} - {E_1} = \dfrac{{hc}}{\lambda }\],
Since the frequency \[\nu = \dfrac{c}{\lambda }\] where $\lambda $ = wavelength of the incident photon
and \[c\]= speed of light = \[3 \times {10^8}m/s\]
\[\therefore \lambda = \dfrac{{hc}}{{({E_3} - {E_1})}}\]
\[ \Rightarrow \lambda = \dfrac{{6.625 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{19.344 \times {{10}^{ - 19}}}}\]
\[ \Rightarrow \lambda = 1.027 \times {10^{ - 7}}m\]

Hence the wavelength of the incident photon is \[\lambda = 1.027 \times {10^{ - 7}}m\] and the frequency of the photon is \[\nu = 2.919 \times {10^{15}}\].

Note: When the electron of a hydrogen atom comes down from a higher energy level(\[{E_i}\]) to a lower energy level (\[{E_f}\]) a photon of wavelength $\lambda $ and frequency $\nu $ is emitted from the atom, hence from the Bohr’ theory we get, \[{E_i} - {E_f} = h\nu = \dfrac{{hc}}{\lambda }\] .
And in the opposite case, When the electron of a hydrogen atom lifts from a lower energy level (\[{E_f}\]) to a higher energy level (\[{E_i}\]), a photon of wavelength $\lambda $ and frequency $\nu $ is absorbed by the atom, hence from the Bohr’ theory we get, \[{E_i} - {E_f} = h\nu = \dfrac{{hc}}{\lambda }\] .
Due to the absorption of the photon, some hydrogen spectrums become dark – these are called an absorption spectrum.