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

Rydberg constant describes the wavelengths or frequencies of light in various series of related spectral lines, mainly those emitted by hydrogen atoms in the Balmer series.

Complete step-by-step answer:
We know,
The wavelength of an electron is represented by $\text{ }\!\!\lambda\!\!\text{ }$
${{\text{ }\!\!\eta\!\!\text{ }}_{\text{1}}}$ is the orbit number which is 2 in this case
${{\text{ }\!\!\eta\!\!\text{ }}_{\text{2}}}$ is \[\infty \] because the destination orbit number is undefined
\[\dfrac{\text{1}}{\text{ }\!\!\lambda\!\!\text{ }}\text{=R}\left( \dfrac{\text{1}}{{{\text{ }\!\!\eta\!\!\text{ }}_{\text{1}}}^{\text{2}}}-\dfrac{\text{1}}{{{\text{ }\!\!\eta\!\!\text{ }}_{\text{2}}}^{\text{2}}} \right)\]
Where R is Rydberg constant.
\[\Rightarrow \dfrac{\text{hc}}{\text{ }\!\!\lambda\!\!\text{ }}\text{=hcR}\left( \dfrac{\text{1}}{{{\text{ }\!\!\eta\!\!\text{ }}_{\text{1}}}^{\text{2}}}-\dfrac{\text{1}}{{{\text{ }\!\!\eta\!\!\text{ }}_{\text{2}}}^{\text{2}}} \right)\]
\[\Rightarrow \Delta \text{E=hcR}\left( \dfrac{\text{1}}{{{\text{ }\!\!\eta\!\!\text{ }}_{\text{1}}}^{\text{2}}}-\dfrac{\text{1}}{{{\text{ }\!\!\eta\!\!\text{ }}_{\text{2}}}^{\text{2}}} \right)\]
Given,
$\text{E = }-\text{3}\text{.4eV}$
So,
$\Delta \text{E = 0}-(-\text{3}\text{.4eV})$
$=\text{ 3}\text{.4eV = 3}\text{.4}\times \text{1}\text{.6}\times \text{1}{{\text{0}}^{-19}}$
$=\text{ 5}\text{.44}\times \text{1}{{\text{0}}^{-19}}\text{ J}$
Where,
$\text{h = 6}\text{.63}\times \text{1}{{\text{0}}^{-34}}\text{ }{{\text{m}}^{2}}\text{kg/s}$
$\text{c = 3}\times \text{1}{{\text{0}}^{8}}\text{ m/s}$
So,
$5.44\times {{10}^{-19}}\text{ = }\dfrac{\text{R}\times \text{6}\text{.63}\times \text{1}{{\text{0}}^{-34}}\times 3\times {{10}^{8}}}{4}$
$\Rightarrow \text{R = 1}\text{.09}\times \text{1}{{\text{0}}^{6}}\text{ }{{\text{m}}^{-1}}$
Therefore, the value of Rydberg’s constant is $\text{1}\text{.09}\times \text{1}{{\text{0}}^{6}}\text{ }{{\text{m}}^{-1}}$ if the energy of electron in the second orbit of hydrogen atom is -3.4eV.

Note: We should have knowledge about the Rydberg’s constant.
1) In spectroscopy, the Rydberg constant, symbol for heavy atoms or for hydrogen, named after the Swedish physicist Johannes Rydberg, is a physical constant relating to the electromagnetic spectrum of an atom.
2) When used in this form in the mathematical description of a series of spectral lines, the result is the number of waves per unit length, of the week numbers. Multiplication by the speed of light yields the frequencies of the spectral lines.
3) Kinetic and potential energy of atoms results from the motion of electrons. When the electrons are excited they move to a higher energy orbital farther away from the atom. The further the orbital from the nucleus, the higher the potential energy of the electron at the energy level.