Solveeit Logo

Question

Question: The expression for Rydberg constant is, A. \(\dfrac{{2{\Pi ^2}m{h^2}c}}{{{{\left( {{e^2}/4\Pi {\va...

The expression for Rydberg constant is,
A. 2Π2mh2c(e2/4Πεo)2\dfrac{{2{\Pi ^2}m{h^2}c}}{{{{\left( {{e^2}/4\Pi {\varepsilon _o}} \right)}^2}}}
B. 2Π2m(e2/4Πεo)2h3c\dfrac{{2{\Pi ^2}m{{\left( {{e^2}/4\Pi {\varepsilon _o}} \right)}^2}}}{{{h^3}c}}
C. (e2/4Πεo)22π2mh3c\dfrac{{{{\left( {{e^2}/4\Pi {\varepsilon _o}} \right)}^2}}}{{2{\pi ^2}m{h^3}c}}
D. 2π2mh3ce2/4Πεo\dfrac{{2{\pi ^2}m{h^3}c}}{{{e^2}/4\Pi {\varepsilon _o}}}

Explanation

Solution

We know that, when an electric current is passed through hydrogen gas, the hydrogen molecule gets dissociated and a high energy electromagnetic radiation will be produced. These electromagnetic radiations correspond to several lines and the formulae for wavenumber of these emitted electromagnetic radiation will be given by a scientist named Rydberg.

Complete step by step solution
The Rydberg formula can be written as;
v=109,677(1n121n22)\mathop v\limits^ - = 109,677\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)
Where n1{n_1}is orbit number where electrons come from the excited state orbit n2{n_2}
Firstly, these hydrogen line spectra were studied by Niels Bohr using Plank's concept of quantisation of energy.
When electrons comes from higher energy level say n2=2,3,4......{n_2} = 2,3,4......to the ground or lower energy state i.e. n1{n_1}is equal to 11 then the electromagnetic transition series emitted are called the Lyman series and the wavelength of this radiations matches with the ultraviolet region. Similarly When electrons comes from higher energy level say n2=3,4,5......{n_2} = 3,4,5......to the ground or lower energy state i.e. n1{n_1}is equal to 22 then the electromagnetic transition series emitted are called the Balmer series and the wavelength of this radiations matches with the visible region.
The value of Rydberg constant is derived by Neil’s Bohr and it is calculated from constants. The Rydberg constant depends on the rest mass, charge of the electron, speed of light, and the Planck’s constant.
The expression for Rydberg constant is given by;
R=me48εo2h3c=2Π2m(e2/4Πεo)2h3c{R_ * } = \dfrac{{m{e^4}}}{{8\varepsilon _o^2{h^3}c}} = \dfrac{{2{\Pi ^2}m{{\left( {{e^2}/4\Pi {\varepsilon _o}} \right)}^2}}}{{{h^3}c}}

**Hence, option (B) is the correct option.

Note: **
Thus, Similarly When electrons comes from higher energy level say n2=4,5,6......{n_2} = 4,5,6......to the ground or lower energy state i.e. n1{n_1}is equal to 33 then the electromagnetic transition series emitted are called the Paschen series and the wavelength of this radiations matches with the Infrared region.