Question

What is the lowest energy of the spectral line emitted by the hydrogen atom in the Lyman series? (h=Plank constant; C=Velocity of light; R=Rydberg constant)
(A) $ \dfrac{{5h{R_h}c}}{{36}} $
(B) $ \dfrac{{4h{R_h}c}}{3} $
(C) $ \dfrac{{3h{R_h}c}}{4} $
(D) $ \dfrac{{7h{R_h}}}{{144}} $

Answer 1

Hint: The Lyman series is a hydrogen spectral series of transitions and consequent ultraviolet emission lines when an electron moves from n 2 to n = 1 (where n is the main quantum number), the electron's lowest energy state, in physics and chemistry. Lyman-alpha is the transition from n = 2 to n = 1, Lyman-beta is the transition from 3 to 1, Lyman-gamma is the transition from 4 to 1, and so on. Theodore Lyman, the series' creator, is honoured with its name. The higher the energy of the electromagnetic emission, the larger the difference in the main quantum numbers.

Complete answer:
The atomic hydrogen emission spectrum has been split into a variety of spectral series, with wavelengths calculated using the Rydberg formula. These spectral lines are caused by electrons in an atom transitioning between two energy levels. The Rydberg formula was crucial in the development of quantum mechanics since it classified the series. In astronomical spectroscopy, spectral series are used to identify the presence of hydrogen and calculate red shifts. The Lyman series was produced using version $ \dfrac{1}{\lambda } = {R_{\text{H}}}\left( {1 - \dfrac{1}{{{n^2}}}} \right)\qquad \left( {{R_{\text{H}}} \approx 1.0968 \times {{10}^7}{\mkern 1mu} {{\text{m}}^{ - 1}} \approx \dfrac{{13.6{\mkern 1mu} {\text{eV}}}}{{hc}}} \right) $ of the Rydberg formula. Where n is a natural number that is larger than or equal to 2 (e.g., 2, 3, 4,...). As a result, the lines in the picture above correspond to the wavelengths of n = 2 on the right and n = $ \infty $ on the left. There are an unlimited number of spectral lines, but as they approach n = $ \infty $ (the Lyman limit), they get exceedingly thick, so only part of the initial and final lines are visible.
So we use $ \Delta E = Ehc[\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}] $
The lowest energy transition in the Lyman series is from n=1 to n=2.
Substituting $ {n_2} = 2 $
 $ v = \dfrac{1}{\lambda } = {R_h}c(1 - \dfrac{1}{4}) $
 $ v = \dfrac{{3{R_h}c}}{4} $
We know that
 $ E = hv $
 $ E = \dfrac{{3h{R_h}c}}{4} $
Hence option c is correct.

Note:
Harvard scientist Theodore Lyman found the first line in the Lyman series spectrum in 1906 while investigating the ultraviolet spectrum of electrically excited hydrogen gas. Lyman found the remaining lines of the spectrum (all in the ultraviolet) between 1906 and 1914. The spectrum of radiation produced by hydrogen is discrete or non-continuous.