Question

The dimensions of Rydberg’s constant are
A. $\left[ {{M^0}{L^{ - 1}}T} \right]$
B. $\left[ {ML{T^{ - 1}}} \right]$
C. $\left[ {{M^0}{L^{ - 1}}{T^0}} \right]$
D. $\left[ {M{L^0}{T^2}} \right]$

Answer 1

Hint: The Rydberg’s constant is defined in the Neil Bohr equation. By substituting the dimensional values in the Neil Bohr equation, the dimensional values of the Rydberg’s constant are obtained.

Complete step by step answer:
The Rydberg constant equation is obtained by Neil Bohr using some fundamental constants and understanding some relationships through the Bohr model. In this equation, the wavelength of lines in the atomic spectra Rydberg constant is obtained. It is the function of charge of the electron and rest mass, the speed of light, and Planck’s constant.
$R = \dfrac{{{m_e}{e^4}}}{{8{\varepsilon _0}^2{h^3}c}}\,...........................\left( 1 \right)$
Where,
${m_e}$ is the mass of the electron in its rest state
$h$ is the Plank constant
$c$ is the velocity of light in free space or vacuum
${\varepsilon _0}$ is the permittivity of free space
$e$ is the elementary charge of the electron.

The SI unit of Rydberg’s constant is ${m^{ - 1}}$.

Derivation of dimensional unit:
The SI unit of mass is kilogram, $kg$. Thus, the dimensional formula for mass is $\left[ {{M^1}{L^0}{T^0}} \right]$.
The SI unit of charge is $C$,
$e = I \times t$, where, $e$ is the charge, $I$ is the current and $t$ is the time,
The SI unit of current is ampere, $A$, and The SI unit of time is $s$,
 Thus, the dimensional formula for charge is $\left[ {{M^0}{L^0}{A^1}{T^1}} \right]$.
The SI unit of permittivity of free space is $\dfrac{{{C^2}}}{{N{m^2}}} \Rightarrow \dfrac{{{A^2}{s^2}}}{{\left( {\dfrac{{kgm}}{{{s^2}}}} \right) \times {m^2}}} \Rightarrow \dfrac{{{A^2}{s^4}}}{{kg{m^3}}}$
Thus, the dimensional formula for permittivity of free space is $\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]$.
The SI unit of Planck’s constant is, $Nms \Rightarrow \dfrac{{kgm}}{{{s^2}}} \times m \times s \Rightarrow \dfrac{{kg{m^2}}}{s}$
Thus, the dimensional formula for Planck’s constant is $\left[ {{M^1}{L^2}{T^{ - 1}}} \right]$.
The SI unit of speed of light is, $m{s^{ - 1}}$
Thus, the dimensional formula for speed of light is \[\left[ {{L^1}{T^{ - 1}}} \right]\].

By substituting all the dimensional formula in equation (1), we get
$R = \dfrac{{\left[ {{M^1}{L^0}{T^0}} \right]{{\left[ {{M^0}{L^0}{A^1}{T^1}} \right]}^4}}}{{{{\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]}^2}{{\left[ {{M^1}{L^2}{T^{ - 1}}} \right]}^3}\left[ {{L^1}{T^{ - 1}}} \right]}}$
By solving the power term in the above equation,
$
   \Rightarrow R = \dfrac{{\left[ {{M^1}{L^0}{T^0}} \right]\left[ {{M^0}{L^0}{A^4}{T^4}} \right]}}{{\left[ {{M^{ - 2}}{L^{ - 6}}{T^8}{A^4}} \right]\left[ {{M^3}{L^6}{T^{ - 3}}} \right]\left[ {{L^1}{T^{ - 1}}} \right]}} \\
   \Rightarrow R = \dfrac{{\left[ {{M^1} \times {A^4} \times {T^4} \times {M^2} \times {L^6} \times {T^1} \times {T^3}} \right]}}{{\left[ {{T^8} \times {A^4} \times {M^3} \times {L^6} \times {L^1}} \right]}} \\
 $
By multiplying the common terms in the above equation, we get
$ \Rightarrow R = \dfrac{{\left[ {{M^3}{L^6}{T^8}{A^4}} \right]}}{{\left[ {{M^3}{L^7}{T^8}{A^4}} \right]}}$
By resolving the identical terms, we get $R = \left[ {{M^0}{L^{ - 1}}{T^0}} \right]$

Hence, option (C) is correct.


Note:
- The value of the Rydberg’s constant is ${R_\infty } = 1.097 \times {10^7}\,{m^{ - 1}}$. The Rydberg unit of energy $\left( {{R_y}} \right)$ determines the photon energy, which has the wave number equal to Rydberg constant.
- The Rydberg constant value completely depends on the mass of the electron, the charge of the electron, the permittivity of free space, and the speed of light. The dimensional formula for the Rydberg constant has the only term in length and the other terms of mass and time factor tend to be in the power of zero.