Question

Using expression for energy of electron, obtain the Bohr’s formula for hydrogen spectral line.

Answer 1

Hint: We have already been given the expression for energy of electron as $hv = {E_n} - {E_p}$ . Also, we know that momentum equals mass times velocity and speed of light equals frequency times wavelength. Using these substitutions we can find the Bohr’s formula for the hydrogen spectral line. Bohr’s model explains the spectral lines of the hydrogen atomic emission spectrum. While the electron of the atom remains in the ground state, its energy is unchanged. When the atom absorbs one or more quanta of energy, the electron moves from the ground state orbit to an excited state orbit that is further away.

Complete step by step answer:
Suppose an electron jumps from ${n^{th}}$ higher orbit to the ${p^{th}}$ lower orbit.
Let ${E_p}$ and ${E_n}$ be the energies of the ${p^{th}}$ and ${n^{th}}$ orbit respectively.
Energy of electron in the ${n^{th}}$ orbit,
${E_n} = - \dfrac{{m{e^4}}}{{8\varepsilon _0^2{h^2}}}\dfrac{1}{{{n^2}}}$
Energy of electron in the ${p^{th}}$ orbit,
${E_p} = - \dfrac{{m{e^4}}}{{8\varepsilon _0^2{h^2}}}\dfrac{1}{{{p^2}}}$

According to Bohr’s third postulate, energy emitted is given by
$hv = {E_n} - {E_p}$
\[
\Rightarrow hv = - \dfrac{{m{e^4}}}{{8\varepsilon _0^2{h^2}}}\dfrac{1}{{{n^2}}} - \left( {
\dfrac{{m{e^4}}}{{8\varepsilon _0^2{h^2}}}\dfrac{1}{{{p^2}}}} \right) \\
\Rightarrow v = \dfrac{{m{e^4}}}{{8\varepsilon _0^2{h^3}}}\left( {\dfrac{1}{{{p^2}}} - \dfrac{1}{{{n^2}}}} \right) \\
\]
Now, $v = \dfrac{c}{\lambda }$ where c is the velocity and $\lambda $ is the wavelength of the electron.
\[
\dfrac{c}{\lambda } = \dfrac{{m{e^4}}}{{8\varepsilon _0^2{h^3}}}\left( {\dfrac{1}{{{p^2}}} - \dfrac{1}{{{n^2}}}} \right) \\
\Rightarrow \dfrac{1}{\lambda } = \dfrac{{m{e^4}}}{{8\varepsilon _0^2{h^3}c}}\left( {\dfrac{1}{{{p^2}}} - \dfrac{1}{{{n^2}}}} \right) \\
\therefore \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{p^2}}} - \dfrac{1}{{{n^2}}}} \right) \\
\]
where $R = \dfrac{{m{e^4}}}{{8\varepsilon _0^2{h^3}c}}$ is called the Rydberg’s constant.

Note: In the original formula we also have a term of Z i.e. the atomic number of atom i.e.
\[\dfrac{1}{\lambda } = {Z^2}R\left( {\dfrac{1}{{{p^2}}} - \dfrac{1}{{{n^2}}}} \right)\]
But this equation too is valid for all hydrogen-like species, i.e. atoms having only a single electron, and the particular case of hydrogen spectral lines is given by Z=1.