Answer
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Hint: Balmer series occurs only when electrons shift from nth energy level to the second energy level. We substitute values of the energy level in Rydberg’s wavelength formula and find the relation between $k$ and Rydberg’s constant.
Formula used
Rydberg’s wavelength equation $\dfrac{1}{\lambda } = R(\dfrac{1}{{{n_f}}} - \dfrac{1}{{{n_i}}})$
Here,
Wavelength is represented by $\lambda $
Rydberg’s constant is represented by $R$
Initial and final energy levels are represented by ${n_i},{n_f}$ respectively
Complete step by step solution:
Rydberg’s equation for visible radiation of the hydrogen spectrum is given as
$\dfrac{1}{\lambda } = R(\dfrac{1}{{{n_f}^2}} - \dfrac{1}{{{n_i}^2}})$
For the Balmer series, the energy level difference is from n to the second energy level
So,
Initial energy level is ${n_i} = n$
Final energy level is ${n_f} = 2$
$\dfrac{1}{\lambda } = R(\dfrac{1}{4} - \dfrac{1}{{{n^2}}})$. . . . . (1)
From question $\lambda = \dfrac{{k{n^2}}}{{{n^2} - 4}}$. . . . . (2)
Equating (1) and (2)
$ \lambda = \dfrac{{k{n^2}}}{{{n^2} - 4}} = \dfrac{{4{n^2}}}{{R({n^2} - 4)}}$
$ \Rightarrow k = \dfrac{4}{R} $
Hence option (D) $k = \dfrac{4}{R}$ is the correct option
Additional information The Balmer series, or Balmer lines in atomic physics, is one of a set of six named series describing the spectral line emissions of the hydrogen atom. Spectral line emissions take place because an electron is moved from a higher energy level to a lower energy level the difference in energy is released as spectral line emissions.
Note: Often a mistake is made while taking the initial and final energy levels. For questions like these, the energy levels of the different spectrums are required to be known. Students might also make a mistake while taking the initial and final energy levels. An energy spectrum is released when an electron is moved from a higher energy level to a lower energy level. Hence the final energy level is always lower than the initial.
Formula used
Rydberg’s wavelength equation $\dfrac{1}{\lambda } = R(\dfrac{1}{{{n_f}}} - \dfrac{1}{{{n_i}}})$
Here,
Wavelength is represented by $\lambda $
Rydberg’s constant is represented by $R$
Initial and final energy levels are represented by ${n_i},{n_f}$ respectively
Complete step by step solution:
Rydberg’s equation for visible radiation of the hydrogen spectrum is given as
$\dfrac{1}{\lambda } = R(\dfrac{1}{{{n_f}^2}} - \dfrac{1}{{{n_i}^2}})$
For the Balmer series, the energy level difference is from n to the second energy level
So,
Initial energy level is ${n_i} = n$
Final energy level is ${n_f} = 2$
$\dfrac{1}{\lambda } = R(\dfrac{1}{4} - \dfrac{1}{{{n^2}}})$. . . . . (1)
From question $\lambda = \dfrac{{k{n^2}}}{{{n^2} - 4}}$. . . . . (2)
Equating (1) and (2)
$ \lambda = \dfrac{{k{n^2}}}{{{n^2} - 4}} = \dfrac{{4{n^2}}}{{R({n^2} - 4)}}$
$ \Rightarrow k = \dfrac{4}{R} $
Hence option (D) $k = \dfrac{4}{R}$ is the correct option
Additional information The Balmer series, or Balmer lines in atomic physics, is one of a set of six named series describing the spectral line emissions of the hydrogen atom. Spectral line emissions take place because an electron is moved from a higher energy level to a lower energy level the difference in energy is released as spectral line emissions.
Note: Often a mistake is made while taking the initial and final energy levels. For questions like these, the energy levels of the different spectrums are required to be known. Students might also make a mistake while taking the initial and final energy levels. An energy spectrum is released when an electron is moved from a higher energy level to a lower energy level. Hence the final energy level is always lower than the initial.
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