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The series limit for the Paschen series of hydrogen spectrum occurs at 8205.8 $\dot{A}$

Calculate:

  1. Ionization energy of hydrogen atom.
  2. Wave length of the photon that would remove the electron in the ground state of the hydrogen atom.

The solution for this is: 1. 13.6 eV 2. 916 $\dot{A}$

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closed as off-topic by airhuff, Nilay Ghosh, MaxW, paracetamol, Loong May 6 '17 at 11:23

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Both questions are related. Use the following expresion:

${\displaystyle E_{2}-E_{1}={\frac {m_{e}e^{4}}{2(4\pi \epsilon _{0})^{2}\hbar ^{2}}}({\frac {1}{n^{2}_1}-\frac {1}{n^{2}_2})}}$.

For the energy of an 1s hydrogen electron then,calculate the difference between states 2 and 1 (where 2 will represent the ion and 1 the ground state).

The symbols are: $m_{e}$, the mass of the electron, $e$ is electron charge, $h$ is Planck constant and $\epsilon$ is the electric constant of vacuum.

Now, if n$_2$=$\infty$ represents a free electron (or the ion H$^+$) as it feels no atraction to the nucleus (although distance is not in the expression), and if n$_1$=$1$ represents the ground state, then replacing in the formula: ${\displaystyle E_{2}-E_{1}=E(ionization)={\frac {m_{e}e^{4}}{2(4\pi \epsilon _{0})^{2}\hbar ^{2}}}{\frac {1}{1^{2}}}}$,

Where i replace n=1 for ground state and n=$\infty$ for the removed electron.

Notice that this energy is positive, as expected.

Then, E(ionization)=h$\nu$=h$\frac{c}{\lambda}$, from which you get the wavelenght.

If you want $\lambda$ in meters, check the units of E(ionization) be in MKS system.

Hope it helps.

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