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Question: A monochromatic light of wavelength \[\lambda \] is incident on a hydrogen atom that lifts it to \(3...

A monochromatic light of wavelength λ\lambda is incident on a hydrogen atom that lifts it to 3rd3^{rd} orbit from ground level. Find the wavelength and frequency of the incident photon.
(Given: E3=−1.51 eV{E_3} = - 1.51{\text{ }}eV, E1=−13.6eV{E_1} = - 13.6eV)

Explanation

Solution

Use the Bohr’s theory due to the incidence of monochromatic light on the hydrogen atom that lifts it from ground level to 3rd3^{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,
ΔE=hν\Delta E = h\nu
⇒E1−E3=hν\Rightarrow {E_1} - {E_3} = h\nu.
where hh = planck constant and ν\nu = frequency of the incident photon.
ΔE=hcλ\Delta E = \dfrac{{hc}}{\lambda } where λ\lambda = wavelength of the incident photon and ccis the speed of light.
⇒E1−E3=hcλ\Rightarrow {E_1} - {E_3} = \dfrac{{hc}}{\lambda }.

Complete step by step answer:
Monochromatic light of wavelength λ\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 E3{E_3} and the energy of ground level is E1{E_1}.
Given, the energy of the incident photon at ground level,E1=−13.6eV{E_1} = - 13.6eV
And, the energy of the incident photon at the 3rd orbit, E3=−1.51 eV{E_3} = - 1.51{\text{ }}eV
According to Bohr’s theory,
ΔE=hν\Delta E = h\nu
⇒E3−E1=hν\Rightarrow {E_3} - {E_1} = h\nu…………………(1)
where hh = plank constant = 6.625×10−346.625 \times {10^{ - 34}}
ν\nu = frequency of the incident photon.
E3−E1=(−1.51)−(−13.6)=12.09eV{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
∴ν=E3−E1h\therefore \nu = \dfrac{{{E_3} - {E_1}}}{h}
⇒ν=19.344×10−196.625×10−34\Rightarrow \nu = \dfrac{{19.344 \times {{10}^{ - 19}}}}{{6.625 \times {{10}^{ - 34}}}}
⇒ν=2.919×1015\Rightarrow \nu = 2.919 \times {10^{15}}
Eq (1) can be written as, E3−E1=hcλ{E_3} - {E_1} = \dfrac{{hc}}{\lambda },
Since the frequency ν=cλ\nu = \dfrac{c}{\lambda } where λ\lambda = wavelength of the incident photon
and cc= speed of light = 3×108m/s3 \times {10^8}m/s
∴λ=hc(E3−E1)\therefore \lambda = \dfrac{{hc}}{{({E_3} - {E_1})}}
⇒λ=6.625×10−34×3×10819.344×10−19\Rightarrow \lambda = \dfrac{{6.625 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{19.344 \times {{10}^{ - 19}}}}
⇒λ=1.027×10−7m\Rightarrow \lambda = 1.027 \times {10^{ - 7}}m

Hence the wavelength of the incident photon is λ=1.027×10−7m\lambda = 1.027 \times {10^{ - 7}}m and the frequency of the photon is ν=2.919×1015\nu = 2.919 \times {10^{15}}.

Note: When the electron of a hydrogen atom comes down from a higher energy level(Ei{E_i}) to a lower energy level (Ef{E_f}) a photon of wavelength λ\lambda and frequency ν\nu is emitted from the atom, hence from the Bohr’ theory we get, Ei−Ef=hν=hcλ{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 (Ef{E_f}) to a higher energy level (Ei{E_i}), a photon of wavelength λ\lambda and frequency ν\nu is absorbed by the atom, hence from the Bohr’ theory we get, Ei−Ef=hν=hcλ{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.