Question

Given the value of Rydberg constant is, ${{10}^{7}}{{m}^{-1}}$the waves number of the last line of the Balmer series in hydrogen spectrum will be:
(A). $0.025$×${{10}^{4}}{{m}^{-1}}$
(B). 0.5×${{10}^{7}}{{m}^{-1}}$
(C). 0.25×${{10}^{7}}{{m}^{-1}}$
(D). 2.5×${{10}^{7}}{{m}^{-1}}$

Answer 1

Hint: Reciprocal of wavelength is wave number. Wave number gives number of lines per distance. The Hydrogen spectrum has four series. Namely: Lyman series, Balmer series, Paschen series, Brackett series, Pfund series. Use the formula of wave number.

Complete step by step answer:
We know that the Balmer series is the second line of the hydrogen spectrum.
So we have formula of wave number:
\[\overline{\nu }=\dfrac{1}{\lambda }=R{{Z}^{2}}(\dfrac{1}{n_{2}^{1}}-\dfrac{1}{n_{2}^{2}})\] --------(1)
Where,
$\overline{\nu }$ = wave number
$\lambda $ = wavelength
R = Rydberg constant
Z = Atomic number
${{n}_{1}}$=First transition state
\[{{n}_{2}}\]= next transition state
Given that it is a Balmer series. Therefore in Balmer series:
${{n}_{1}}=2$ and \[{{n}_{2}}=\text{ }infinity\text{ }(\infty )\]
R = Rydberg constant = ${{10}^{-7}}$
It is a hydrogen atom, therefore the atomic number must be equal to one.
Z = 1
Put value of \[R,Z,{{n}_{1}},{{n}_{2}}\] in equation (1)
We get,
\[\overline{\nu }=\dfrac{1}{\lambda }={{10}^{7}}\times {{1}^{2}}(\dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{\infty }^{2}}})\]
\[\begin{align}
  & \overline{\nu }={{10}^{7}}\times {{1}^{2}}(\dfrac{1}{4}-0) \\
 & \overline{\nu }={{10}^{7}}\times {{1}^{2}}(0.25) \\
 & \overline{\nu }=0.25\times {{10}^{7}}{{m}^{-1}} \\
\end{align}\]
Therefore the wave number is \[\overline{\nu }=0.25\times {{10}^{7}}{{m}^{-1}}\]

Hence option (C) is correct.

Note: Students need to understand the terms given in the formula of wave number. Once you will understand terms and derivation in formula, students will be able to learn this formula. Do not get confused between the first line of the Balmer series and the last line of the Balmer series. Value of reciprocal of infinity is zero always. Bohr’s proposed atomic model, which satisfactorily explained the stability of atoms of atomic structure and hydrogen line spectra.