Question

In the videos we successfully arrived at Rydberg's formula for Hydrogen atom from the Bohr's Postulates.
What do you think will the modified version of Rydberg's formula for Hydrogen like atoms.

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Form Hydrogen like atom with atomic number Z the energy of the atom if electron is in any orbit n is given by
$E=-13.6\frac{z^2}{n^2}$
Now if electron jumps from $n_2$ to $n_1$ then $\Delta E=E_2-E_1=\frac{-13.6z^2}{n_2{}^2}-\left(\frac{-13.6z^2}{n_1{}^2}\right)$
$\Delta E=13.6z^2(\frac{1}{n_1{}^2}-\frac{1}{n_2{}^2})$
This loss in energy is given out as a photon whose wavelength is related to its energy by $\Delta E=\frac{hc}{\lambda }$
$\Rightarrow \frac{hc}{\lambda }=13.6z^2(\frac{1}{n_1{}^2}-\frac{1}{n_2{}^2})$
$\Rightarrow \frac{1}{\lambda }=\frac{13.6}{hc}{z}^2(\frac{1}{n_1{}^2}-\frac{1}{n_2{}^2})$
We know from the videos that
$R=\frac{13.6}{hc}$
$\Rightarrow \frac{1}{\lambda }=R{z}^2(\frac{1}{n_1{}^2}-\frac{1}{n_2{}^2})$