Question

The limiting line in Balmer series will have a frequency of :
(A) $ 32.29 \times {10^{15}}{s^{ - 1}} $
(B) $ 3.65 \times {10^{14}}{s^{ - 1}} $
(C) $ - 8.22 \times {10^{14}}{s^{ - 1}} $
(D) $ 8.22 \times {10^{14}}{s^{ - 1}} $

Answer 1

Hint: In atomic physics, the Balmer series (or Balmer lines) is one of six named series that describe the spectral line emissions of the hydrogen atom. The Balmer series is determined using the Balmer formula, which was developed by Johann Balmer in 1885 and is an empirical equation.

Complete answer:
The electron transitions from n $ \geqslant $ 3 to n = 2 in the Balmer series, where n refers to the electron's radial quantum number or primary quantum number. $ {{H - \alpha }} $ denotes the transition from n = 3 to n = 2, $ {{\;H - \beta }} $ denotes the transition from 4 to 2, $ {{H - \gamma }} $ denotes the transition from 5 to 2, and $ {{H - \delta }} $ denotes the transition from 6 to 2. The initial spectral lines connected with this series are historically referred to as "H-alpha," "H-beta," "H-gamma," and so on, because they are situated in the visible region of the electromagnetic spectrum.
Johannes Rydberg, a scientist, extended the Balmer equation for all hydrogen transitions in 1888. The equation typically used to compute the Balmer series is a specific example of the Rydberg formula, and it is written as a simple reciprocal mathematical rearrangement of the formula above (using m for n as the only integral constant required):
  $ \dfrac{1}{\lambda } = \dfrac{4}{B}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{n^2}}}} \right) = {R_{{H}}}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{n^2}}}} \right)\quad {{for}}\;n = 3,4,5,... $
Now we know that
  $ \bar v = \dfrac{1}{\lambda } $
In the Balmer series, $ {n_1} $ =2 and $ {n_2} $ = 3,4,5..... The series limit is the last line in the spectrum. When $ {n_2} = \infty $ , the limiting line is the one with the shortest wavelength and highest energy.
  $ \therefore \bar v = \dfrac{1}{\lambda } = \dfrac{{{R_H}}}{{n_1^2}} = \dfrac{{3.29 \times {{10}^{15}}}}{{{2^2}}} = \dfrac{{3.29 \times {{10}^{15}}}}{4} $
  $ \Rightarrow \bar v = 8.22 \times {10^{14}}{s^{ - 1}} $
Hence option d is correct.

Note:
The Balmer series is particularly helpful in astronomy because, due to the abundance of hydrogen in the cosmos, the Balmer lines exist in many celestial objects and are therefore often detected and relatively strong compared to lines from other elements. The relative intensity of spectral lines is highly significant in the spectral classification of stars, which is largely a determination of surface temperature. The Balmer series in particular is very important. Surface gravity (linked to physical size) and composition are two more properties of a star that may be identified by examining its spectrum closely.