
The difference in the wavelength of the 1 st line of Lyman series and the 2 nd line of Balmer series in a hydrogen atom:
A. \[\dfrac{9}{2R}\]
B. \[\dfrac{4}{R}\]
C. \[\dfrac{88}{15R}\]
D. none
Answer
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Hint: The range of hydrogen is significant in stargazing on the grounds that a large portion of the universe is made of hydrogen. Emanation or retention measures in hydrogen offer ascent to arrangement, which are successions of lines relating to nuclear changes, each finishing or starting with a similar nuclear state in hydrogen.
Step by step answer: The Lyman series involve jumps to or from the ground state (n=1); the Balmer series (in which all the lines are in the visible region) corresponds to n=2, the Paschen series to n=3, the Brackett series to n=4, and the Pfund series to n=5.
As we know, wavelength of photon emitted when electron jumps from \[{{n}_{1}}\] to \[{{n}_{2}}\] is
\[\dfrac{1}{\lambda }={{R}_{H}}*{{Z}^{^{2}}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)\]
In the hydrogen atom, for the first line of the Lyman series \[{{n}_{1}}=1\] and . \[{{n}_{2}}=2\]
\[\dfrac{1}{\lambda }={{R}_{H}}*{{1}^{2}}\left( \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{2}^{2}}} \right)={{R}_{H}}*\left( \dfrac{3}{4} \right)\]
Similarly, for the 2nd line of the balmer series \[{{n}_{2}}=2\] and \[{{n}_{2}}=4\] .
\[\dfrac{1}{\lambda }={{R}_{H}}*{{1}^{2}}\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{4}^{2}}}={{R}_{H}}*\dfrac{3}{16} \right)\]
So, difference between wavelength is \[\dfrac{12}{3}*\dfrac{1}{R}=\dfrac{4}{R}\].
Additional Information: At the point when light from energized particles is seen through a spectroscope, pictures of the cut show up along the size of the instrument as a progression of hued lines. The different tones relate to light of unmistakable frequencies, and the arrangement of lines is known as a line range.
Note: The Lyman arrangement is in the bright while the Balmer arrangement is in the obvious and the Paschen, Brackett, Pfund, and Humphreys arrangement are in the infrared. In 1914, Niels Bohr proposed a hypothesis of the hydrogen molecule which clarified the birthplace of its range and which likewise prompted a totally new idea of nuclear structure.
Step by step answer: The Lyman series involve jumps to or from the ground state (n=1); the Balmer series (in which all the lines are in the visible region) corresponds to n=2, the Paschen series to n=3, the Brackett series to n=4, and the Pfund series to n=5.
As we know, wavelength of photon emitted when electron jumps from \[{{n}_{1}}\] to \[{{n}_{2}}\] is
\[\dfrac{1}{\lambda }={{R}_{H}}*{{Z}^{^{2}}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)\]
In the hydrogen atom, for the first line of the Lyman series \[{{n}_{1}}=1\] and . \[{{n}_{2}}=2\]
\[\dfrac{1}{\lambda }={{R}_{H}}*{{1}^{2}}\left( \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{2}^{2}}} \right)={{R}_{H}}*\left( \dfrac{3}{4} \right)\]
Similarly, for the 2nd line of the balmer series \[{{n}_{2}}=2\] and \[{{n}_{2}}=4\] .
\[\dfrac{1}{\lambda }={{R}_{H}}*{{1}^{2}}\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{4}^{2}}}={{R}_{H}}*\dfrac{3}{16} \right)\]
So, difference between wavelength is \[\dfrac{12}{3}*\dfrac{1}{R}=\dfrac{4}{R}\].
Additional Information: At the point when light from energized particles is seen through a spectroscope, pictures of the cut show up along the size of the instrument as a progression of hued lines. The different tones relate to light of unmistakable frequencies, and the arrangement of lines is known as a line range.
Note: The Lyman arrangement is in the bright while the Balmer arrangement is in the obvious and the Paschen, Brackett, Pfund, and Humphreys arrangement are in the infrared. In 1914, Niels Bohr proposed a hypothesis of the hydrogen molecule which clarified the birthplace of its range and which likewise prompted a totally new idea of nuclear structure.
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