Question

Find the value of Rydberg’s constant if the energy of the electron in the second orbit in the Hydrogen atom is -3.4eV.

Answer 1

Hint: Rydberg's formula for finding the wavelength of photons released or absorbed during a transition from a certain level in an atom to another level in an atom is necessary to solve this problem. In this problem, the Balmer series transitions in the Hydrogen atom is talked about.

Complete step-by-step answer:
Rydberg's formula is based on Bohr's atomic model. This model is accurate for single electron models. Hence, it was able to accurately describe the electronic transitions shown in Hydrogen atoms along with the electronic transition levels.
Thus, when an electron makes a transition from a level$({{n}_{1}})$ to another higher level $({{n}_{2}})$, where the energy of the electron in ${{n}_{1}}$level is ${{E}_{1}}$and the energy of the${{n}_{2}}$level is${{E}_{2}}$. Then, due to the transition a photon of wavelength$(\lambda )$is absorbed or emitted.
Therefore, ${{E}_{1}}-{{E}_{2}}=h\vartheta =\dfrac{hc}{\lambda }.$
The corresponding Rydberg formula is given by,$\dfrac{1}{\lambda }=R{{Z}^{2}}\\{\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}}\\}$. Here, Z refers to the atomic number of the atom in which these transitions are taking place.
As per the question, an electron is in the second orbit, hence (n=2) of Hydrogen atoms. Hence, Z=1. Further, the energy level in (n=2) is -3.4eV. Hence, the amount of energy required to ionize the atom is 3.4eV.
Therefore,$E=\dfrac{hc}{\lambda }\Rightarrow \dfrac{1}{\lambda }=\dfrac{E}{hc}.$
$E=3.4eV=3.4\times 1.6\times {{10}^{-19}}J.$
$h=6.6\times {{10}^{-34}}Js$and$c=3\times {{10}^{8}}m{{s}^{-1}}.$
Putting in these values to the above equation we get, $\dfrac{1}{\lambda }=\dfrac{3.4\times 1.6\times {{10}^{-19}}J}{(6.6\times {{10}^{-34}}Js)(3\times {{10}^{8}}m{{s}^{-1}})}=\dfrac{5.44\times {{10}^{7}}}{19.8}{{m}^{-1}}.$
Upon ionization, the final level of the electron will be infinity. Hence, in this case${{n}_{1}}=2$and${{n}_{2}}=\infty .$
Putting these values into the Rydberg’s equation, $\dfrac{1}{\lambda }=R{{(1)}^{2}}\\{\dfrac{1}{2_{{}}^{2}}-\dfrac{1}{\infty _{{}}^{2}}\\}$
We will put the value of the inverse of wavelength from the value we found out above. Therefore, $\dfrac{1}{\lambda }=R(\dfrac{1}{4})\Rightarrow R=\dfrac{4}{\lambda }\Rightarrow R=4\times \dfrac{5.44\times {{10}^{7}}}{19.8}{{m}^{-1}}\Rightarrow R=1.094\times {{10}^{7}}{{m}^{-1}}.$
Hence, the value of Rydberg’s constant is, $R=1.094\times {{10}^{7}}{{m}^{-1}}.$

Note: Ionizing an atom, means to completely remove electrons from an atom. In a Hydrogen atom, consisting of only a single electron, it has a single ionization energy itself. The ionization energy depends upon the level or orbit of the atom in which the electron is situated at.