Question

When an electron makes a transition from (${\rm{n}}\,{\rm{ + }}\,{\rm{1}}$) state to n state, the frequency of emitted radiations is related to n according to (${\rm{n}}\, > > \,{\rm{1}}$)
A. ${\rm{ \nu }}\, \propto {{\rm{n}}^{ - 3}}$
B. ${\rm{ \nu }}\, \propto {{\rm{n}}^2}$
C. ${\rm{ \nu }}\, \propto {{\rm{n}}^3}$
D. ${\rm{ \nu }}\, \propto {{\rm{n}}^{2/3}}$

Answer 1

We will use the Rydberg formula of wavelength. Substitute the value of energy level and apply the given condition, we can determine the relationship between frequency and principal quantum number.

Formula used: $\dfrac{{\rm{1}}}{{\rm{\lambda }}} = \,{{\rm{R}}_{\rm{H}}}{{\rm{Z}}^2}\left[ {\dfrac{{\rm{1}}}{{{\rm{n}}_{{\rm{Lower}}}^{\rm{2}}}} - \dfrac{{\rm{1}}}{{{\rm{n}}_{{\rm{higher}}}^2}}} \right]$

Complete answer:
The formula to determine the wavelength of transitions in emission spectrum is as follows:
$\dfrac{{\rm{1}}}{{\rm{\lambda }}} = \,{{\rm{R}}_{\rm{H}}}{{\rm{Z}}^2}\left[ {\dfrac{{\rm{1}}}{{{\rm{n}}_{{\rm{Lower}}}^{\rm{2}}}} - \dfrac{{\rm{1}}}{{{\rm{n}}_{{\rm{higher}}}^2}}} \right]$
Where,
${\rm{\lambda }}$ is the wavelength.
Z is the atomic number.
${{\rm{R}}_{\rm{H}}}$ is the Rydberg constant.
The value of the Rydberg constant is $109678\,{\rm{c}}{{\rm{m}}^{ - 1}}$ .
${{\rm{n}}_{{\rm{lower}}}}$ is the principal quantum of the energy level in which the electron comes.
${{\rm{n}}_{{\rm{higher}}}}$ is the principal quantum of the energy level from which the electron comes.

The relation between frequency and wavelength is as follows:
$\nu {\rm{ = }}\dfrac{{\rm{c}}}{{\rm{\lambda }}}$
Rearrange for wavelength as follows:
$\Rightarrow \dfrac{\nu }{{\rm{c}}}{\rm{ = }}\dfrac{1}{{\rm{\lambda }}}$
$\Rightarrow\dfrac{\nu }{{\rm{c}}} = \,{{\rm{R}}_{\rm{H}}}{Z^2}\left[ {\dfrac{{\rm{1}}}{{{\rm{n}}_{{\rm{Lower}}}^{\rm{2}}}} - \dfrac{{\rm{1}}}{{{\rm{n}}_{{\rm{higher}}}^2}}} \right]$
For the transition ${\rm{n + 1}}\, \to {\rm{n}}$,
Substitute ${\rm{n}}$ for ${{\rm{n}}_{{\rm{lower}}}}$ and ${\rm{n}}\,{\rm{ + }}\,{\rm{1}}$ for ${{\rm{n}}_{{\rm{higher}}}}$.
$\Rightarrow \nu = \,{{\rm{R}}_{\rm{H}}}{\rm{c}}\,{Z^2}\left[ {\dfrac{{\rm{1}}}{{{{\rm{n}}^2}}} - \dfrac{{\rm{1}}}{{{{\left( {{\rm{n}}\,{\rm{ + }}\,{\rm{1}}} \right)}^2}}}} \right]$
$\Rightarrow \nu = \,{{\rm{R}}_{\rm{H}}}{\rm{c}}\,{Z^2}\left[ {\dfrac{{{{\left( {{\rm{n}}\,{\rm{ + }}\,{\rm{1}}} \right)}^2} - {{\rm{n}}^{\rm{2}}}}}{{{{\rm{n}}^2}{{\left( {{\rm{n}}\,{\rm{ + }}\,{\rm{1}}} \right)}^2}}}} \right]$
$\Rightarrow \nu = \,{{\rm{R}}_{\rm{H}}}{\rm{c}}\,{Z^2}\left[ {\dfrac{{\,2{\rm{n}}\,{\rm{ + }}\,{\rm{1}}}}{{{{\rm{n}}^2}{{\left( {{\rm{n}}\,{\rm{ + }}\,{\rm{1}}} \right)}^2}}}} \right]$
${\rm{If \,n}}\,{\rm{ > > 1}}$
$\Rightarrow{\rm{n}}\,{\rm{ + }}\,{\rm{1}}\, \approx \,{\rm{n}}$
$\Rightarrow2{\rm{n}}\,{\rm{ + }}\,{\rm{1}} \approx \,2{\rm{n}}$
$\Rightarrow\nu = \,{{\rm{R}}_{\rm{H}}}{\rm{c}}\,{Z^2}\left[ {\dfrac{{\,2{\rm{n}}}}{{{{\rm{n}}^2}.\,{{\rm{n}}^2}}}} \right]$
$\Rightarrow\nu = \,\dfrac{{\,2{{\rm{R}}_{\rm{H}}}{\rm{c}}\,{Z^2}}}{{{{\rm{n}}^3}}}$
$\Rightarrow \nu \, \propto {{\rm{n}}^{ - 3}}$
So, when an electron makes a transition from (${\rm{n}}\,{\rm{ + }}\,{\rm{1}}$) state to n state, the relation between the frequency of emitted radiations to n is ${\rm{v}}\, \propto {{\rm{ \nu }}^{ - 3}}$.

**Therefore, option (A) ${\rm{v}}\, \propto {{\rm{ \nu }}^{ - 3}}$ is correct.

Note:**
The Rydberg formula for hydrogen atom is $\dfrac{{\rm{1}}}{{\rm{\lambda }}} = \,{{\rm{R}}_{\rm{H}}}\left[ {\dfrac{{\rm{1}}}{{{\rm{n}}_{{\rm{Lower}}}^{\rm{2}}}} - \dfrac{{\rm{1}}}{{{\rm{n}}_{{\rm{higher}}}^2}}} \right]$. When the value of the principal quantum number is greater than one, the frequency is inversely proportional to the principal quantum number. As the principal quantum number increases, the energy of the level decreases so less energy is required for the transition of an electron. Energy and frequency are directly proportional. So, the frequency decreases.