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What is ionization energy? Assuming expression for energy of electron, derive an expression for wavelength of spectral lines in hydrogen atom.

Vidyadip · Accepted Answer

The ionization energy of an atom is the minimum amount of energy required to be given to an electron in the ground state of that atom to set the electron free.
Suppose an electron jumps from $n ^{\text {th }}$ higher orbit to $m ^{\text {th }}$ lower orbit.
Let $E_n$ and $E_m$ be the energies of electron in $n ^{\text {th }}$ and $m ^{\text {th }}$ orbit respectively.
$\therefore $ Energy of electron in $n^{\text {th }}$ orbit is, $-\frac{m_e Z^2 e^4}{8 \varepsilon_o h^2 n^2}$
$\therefore $ Energy of electron in $m ^{ th }$ orbit is, $-\frac{m_e Z^2 e^4}{8 \varepsilon_o h^2 m^2}$
According to Bohr's $3^{\text {rd }}$ postulate, energy emitted is given by.
$h v=E_n-E_m$
$\begin{array}{ll}\therefore & h v=\frac{m_e Z^2 e^4}{8 z_o h^2}\left(\frac{1}{m^2}-\frac{1}{n^2}\right) \\
\therefore & v=\frac{m_e Z^2 e^4}{8 \varepsilon_e h^3}\left(\frac{1}{m^2}-\frac{1}{n^2}\right) \\
\therefore & \frac{c}{\lambda}=\frac{m_e z^2 e^4}{8 \varepsilon_e h^3}\left(\frac{1}{m^2}-\frac{1}{n^2}\right) \ \ldots \ldots \ldots \ldots(v=\end{array}$
= $\frac{c}{\lambda}$, where $c$ is velocity of light in vacuum)
For hydrogen, $Z=1$,
$\therefore \ \frac{1}{\lambda}=\frac{m_e e^4}{8 c c_0 h^3}\left(\frac{1}{m^2}-\frac{1}{n^2}\right)$
$\therefore \ \frac{1}{\lambda}=R\left(\frac{1}{m^2}-\frac{1}{n^2}\right)$
Where, $R=\frac{m_e e^4}{8 c \varepsilon_o h^3}$ is called Rydberg's constant.
Above expression is for wavelength of spectral lines in hydrogen atom.

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