
The equation corresponding to the move number of spectral line in the Brackett series:
$
\left( A \right)\;R\left[ {\left( {\dfrac{1}{{{2^2}}}} \right) - {{\left( {\dfrac{1}{4}} \right)}^3}} \right] \\
\left( B \right)\,R\left[ {{{\left( {\dfrac{1}{4}} \right)}^2} - {{\left( {\dfrac{1}{5}} \right)}^2}} \right] \\
\left( C \right)\,R\left[ {{{\left( {\dfrac{1}{3}} \right)}^2} - {{\left( {\dfrac{1}{5}} \right)}^2}} \right] \\
\left( D \right)\,R\left[ {{{\left( {\dfrac{1}{5}} \right)}^2} - \left( {\dfrac{1}{{{6^2}}}} \right)} \right] \\
$
Answer
593.7k+ views
Hint: Brackett series is for electronic transition from n=4 to some higher energy level.
Complete step by step answer:
Spectral series are the set of wavelengths arranged in sequential fashion. Which characterizes light or any electromagnetic radiation emitted by energized atoms.
Hydrogen atom is the simplest atomic system found in nature, thus it produces the simplest of these series.
Rydberg formula relates energy difference between the various levels of Bohr’s model and the wavelength of absorbed or emitted photons.
It is mathematically expressed as:
$\dfrac{1}{7} = R{Z^2}\left( {\dfrac{1}{{n_l^2}} - \dfrac{1}{{n_h^2}}} \right)$
Where,
$\lambda $ is the wavelength
R is the Rydberg constant $ = 1.097 \times {10^7}$
Z is atomic number.
${n_l}$ is the lower energy level.
${n_h}$ is the higher energy.
$h = 1,2,3, \ldots and\;{n_h} > {n_l}$
For Brackett series,
${n_l}$ is 4 and ${n_h} = 5,6,7,8 \ldots $
Brackett series is displayed when electron transition takes place from higher energy states \[\left( {{n_h} = 5,6,7,8 \ldots } \right)\;to\;{n_l} = 4\] energy state. The spectral lines of the Brackett series lie in the infrared band.
So, the answer is (B). $\because \;{n_l} = 4\;and\;{n_h} = 5$
Note:
The lines in the brackett series have wavelengths from 4.05 μm (Brackett-α) towards shorter wavelengths, the spacing between the lines diminishing as they converge on the series limit at 1.46 μm. They are named after the American physicist Frederick Sumner Brackett (1896–1972).
Complete step by step answer:
Spectral series are the set of wavelengths arranged in sequential fashion. Which characterizes light or any electromagnetic radiation emitted by energized atoms.
Hydrogen atom is the simplest atomic system found in nature, thus it produces the simplest of these series.
Rydberg formula relates energy difference between the various levels of Bohr’s model and the wavelength of absorbed or emitted photons.
It is mathematically expressed as:
$\dfrac{1}{7} = R{Z^2}\left( {\dfrac{1}{{n_l^2}} - \dfrac{1}{{n_h^2}}} \right)$
Where,
$\lambda $ is the wavelength
R is the Rydberg constant $ = 1.097 \times {10^7}$
Z is atomic number.
${n_l}$ is the lower energy level.
${n_h}$ is the higher energy.
$h = 1,2,3, \ldots and\;{n_h} > {n_l}$
For Brackett series,
${n_l}$ is 4 and ${n_h} = 5,6,7,8 \ldots $
Brackett series is displayed when electron transition takes place from higher energy states \[\left( {{n_h} = 5,6,7,8 \ldots } \right)\;to\;{n_l} = 4\] energy state. The spectral lines of the Brackett series lie in the infrared band.
So, the answer is (B). $\because \;{n_l} = 4\;and\;{n_h} = 5$
Note:
The lines in the brackett series have wavelengths from 4.05 μm (Brackett-α) towards shorter wavelengths, the spacing between the lines diminishing as they converge on the series limit at 1.46 μm. They are named after the American physicist Frederick Sumner Brackett (1896–1972).
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