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# 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: ${E_3} = - 1.51{\text{ }}eV$, ${E_1} = - 13.6eV$)

Last updated date: 17th Apr 2024
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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.
$\Rightarrow {E_1} - {E_3} = \dfrac{{hc}}{\lambda }$.

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.