Question

The energies of three consecutive energy levels ${L_3}$, ${L_2}$ and ${L_1}$ of hydrogen atom ${E_0}$, $\dfrac{{4{E_0}}}{9}$ and $\dfrac{{{E_1}}}{4}$ respectively. A photon of wavelength $\lambda$ is emitted for a transition ${L_3}$ to ${L_1}$. What will be the wavelength of emission for transition ${L_2}$ to ${L_1}$?

Answer 1

In the solution, we will use the Planck’s equation for the energy of the radiation emitted. Using the Planck’s equation, we will derive the expression for the difference in energies of two consecutive levels to obtain the wavelength.

Complete step by step answer:
Given:
The energy of level ${L_3}$, ${E_{{L_3}}} = {E_0}$
The energy of level ${L_2}$, ${E_{{L_2}}} = \dfrac{{4{E_0}}}{9}$
The energy of level ${L_1}$, ${E_{{L_1}}} = \dfrac{{{E_0}}}{4}$
Now, the relation for the change in the energy of the radiation emitted during transition from one energy level to the other can be written as,
$\Delta E = h\upsilon \\\ \Rightarrow {E_n} - {E_{n - 1}} = h\upsilon$
Here $\Delta E$ is the change in energy, $h$ is the Planck’s constant, $\upsilon$ is the frequency of the photon of radiation emitted, ${E_n}$ is the upper energy level and ${E_{n - 1}}$ is the lower energy level.

Since $c = \upsilon \lambda$, we can write
$\upsilon = \dfrac{c}{\lambda }$.
Here $c$ is the velocity of light and $\lambda$ is the wavelength of the photon emitted.
We now use the relation $\upsilon = \dfrac{c}{\lambda }$ in ${E_n} - {E_{n - 1}} = h\upsilon$ to get,
${E_n} - {E_{n - 1}} = h\dfrac{c}{\lambda }$
Hence, the relation for the change in energy during a transition from energy level ${L_3}$ to ${L_1}$ is,
${E_{{L_3}}} - {E_{{L_1}}} = h\dfrac{c}{\lambda }$
Here $\lambda$ is the wavelength of the photon emitted during a transition from energy level ${L_3}$ to ${L_1}$.

Since ${E_{{L_3}}} = {E_0}$ and ${E_{{L_1}}} = \dfrac{{{E_0}}}{4}$ we can substitute the values of ${E_0}$ and${E_{{L_1}}}$ in the above equation. Hence,
${E_0} - \dfrac{{{E_0}}}{4} = h\dfrac{c}{\lambda }\\\ \Rightarrow \dfrac{{4{E_0} - {E_0}}}{4} = h\dfrac{c}{\lambda }\\\ \Rightarrow \dfrac{{3{E_0}}}{4} = h\dfrac{c}{\lambda }\\\ \Rightarrow {E_0} = \dfrac{{4hc}}{{3\lambda }}$

Now, let’s express the relation for the change in energy during a transition from energy level ${L_2}$ to ${L_1}$.The relation is written as,
${E_{{L_2}}} - {E_{{L_1}}} = h\dfrac{c}{{{\lambda _2}}}$
Here ${\lambda _2}$ is the wavelength of the photon emitted during a transition from the energy level ${L_2}$ to ${L_1}$.
Since ${E_{{L_2}}} = \dfrac{{4{E_0}}}{9}$ and ${E_{{L_1}}} = \dfrac{{{E_0}}}{4}$, we can substitute $\dfrac{{4{E_0}}}{9}$ for and $\dfrac{{{E_0}}}{4}$ for ${E_{{L_1}}}$ in the above equation. So, the equation becomes
$\dfrac{{4{E_0}}}{9} - \dfrac{{{E_0}}}{4} = h\dfrac{c}{{{\lambda _2}}}\\\ \Rightarrow \dfrac{{4{E_0} \times 4 - 9{E_0}}}{{9 \times 4}} = h\dfrac{c}{{{\lambda _2}}}\\\ \Rightarrow \dfrac{{16{E_0} - 9{E_0}}}{{36}} = h\dfrac{c}{{{\lambda _2}}}\\\ \Rightarrow \dfrac{{7{E_0}}}{{36}} = h\dfrac{c}{{{\lambda _2}}}$

Rewriting the equation, we get
${\lambda _2} = \dfrac{{36hc}}{{7{E_0}}}$
Since ${E_0} = \dfrac{{4hc}}{{3\lambda }}$, we can substitute $\dfrac{{4hc}}{{3\lambda }}$ for ${E_0}$ in ${\lambda _2} = \dfrac{{36hc}}{{7{E_0}}}$.
${\lambda _2} = \dfrac{{36hc}}{{7 \times \dfrac{{4hc}}{{3\lambda }}}}\\\ \Rightarrow {\lambda _2} = \dfrac{{9 \times 3\lambda }}{7}\\\ \therefore {\lambda _2} = \dfrac{{27\lambda }}{7}$

Hence, we obtained the wavelength of the photon emitted during a transition from ${L_2}$ to ${L_1}$ as $\dfrac{{27\lambda }}{7}$.

Note: Planck’ equation generally helps in calculating the energy changes involving light radiations. Hence, we can obtain the frequency and the wavelength of the radiation emitted if the energy of the radiation is known.