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Question

The wave number of the limiting line of the series (visible) in hydrogen spectrum is:

The correct answer is

27, 419 cm-1

Calculating Wave Number of Limiting Line in Hydrogen Spectrum's Visible Series

The question asks about the wave number of the limiting line in the visible series of the hydrogen spectrum. Let's break down the key terms:

  • Hydrogen Spectrum: When a hydrogen atom is excited, its electron jumps to higher energy levels. When the electron returns to a lower energy level, it emits light of specific wavelengths, forming a line spectrum.
  • Series: The lines in the hydrogen spectrum are grouped into series based on the final energy level the electron transitions to. The main series are Lyman ($n_1=1$), Balmer ($n_1=2$), Paschen ($n_1=3$), Brackett ($n_1=4$), Pfund ($n_1=5$), etc.
  • Visible Series: The Balmer series corresponds to transitions where the electron falls to the $n_1=2$ energy level from higher levels ($n_2 = 3, 4, 5, \dots$). This series is the only one that lies partly within the visible region of the electromagnetic spectrum.
  • Limiting Line: For any series, the limiting line corresponds to the transition from the highest possible energy level, which is $n_2 = \infty$, to the specific lower level ($n_1$) that defines the series. In this case, it's the transition from $n_2 = \infty$ to $n_1 = 2$.
  • Wave Number ($\bar{\nu}$): Wave number is the reciprocal of wavelength ($\lambda$), typically measured in cm-1. It is directly proportional to the energy of the emitted photon.

Using the Rydberg Formula for Wave Number

The wave number ($\bar{\nu}$) of a spectral line in the hydrogen spectrum is given by the Rydberg formula:

\(\bar{\nu} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\)

Where:

  • $R_H$ is the Rydberg constant for hydrogen, approximately \(109677 \text{ cm}^{-1}\).
  • $n_1$ is the principal quantum number of the lower energy level.
  • $n_2$ is the principal quantum number of the higher energy level from which the electron transitions.

Calculating for the Limiting Line of the Visible Series

For the limiting line of the visible series (Balmer series):

  • The series is the Balmer series, so the lower energy level is $n_1 = 2$.
  • The line is the limiting line, so the higher energy level is $n_2 = \infty$.
  • The Rydberg constant $R_H = 109677 \text{ cm}^{-1}$.

Substitute these values into the Rydberg formula:

\(\bar{\nu} = 109677 \text{ cm}^{-1} \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right)\)

\(\bar{\nu} = 109677 \text{ cm}^{-1} \left( \frac{1}{4} - 0 \right)\)

\(\bar{\nu} = 109677 \text{ cm}^{-1} \times \frac{1}{4}\)

Now, perform the calculation:

\(\bar{\nu} = \frac{109677}{4} \text{ cm}^{-1}\)

\(\bar{\nu} \approx 27419.25 \text{ cm}^{-1}\)

The calculated wave number is approximately \(27419.25 \text{ cm}^{-1}\). Looking at the given options, the closest value is \(27,419 \text{ cm}^{-1}\).

Series Lower Level ($n_1$) Upper Level ($n_2$) Region
Lyman 1 2, 3, 4, ... , &(\infty) Ultraviolet
Balmer (Visible) 2 3, 4, 5, ... , &(\infty) Visible
Paschen 3 4, 5, 6, ... , &(\infty) Infrared
Brackett 4 5, 6, 7, ... , &(\infty) Infrared
Pfund 5 6, 7, 8, ... , &(\infty) Infrared

Thus, the wave number for the limiting line of the visible series (Balmer series, $n_1=2$, $n_2=\infty$) in the hydrogen spectrum is approximately \(27,419 \text{ cm}^{-1}\).

Revision Table: Hydrogen Spectrum Series & Wave Numbers

Understanding the different series and their corresponding transitions is crucial for solving problems related to the hydrogen spectrum.

  • The series are defined by the final energy level ($n_1$) of the electron.
  • Lines within a series correspond to different initial energy levels ($n_2 > n_1$).
  • The limiting line of a series corresponds to $n_2 = \infty$.
  • The Rydberg formula connects the wave number to the energy level transitions.

Additional Information: Energy Levels and Transitions

The energy levels of a hydrogen atom are quantized, meaning they can only exist at specific discrete values. The energy of the \(n^{th}\) level is given by \(E_n = -\frac{13.6}{n^2}\) eV. When an electron moves from a higher energy level \(n_2\) to a lower energy level \(n_1\), it emits a photon with energy \(E = E_{n_2} - E_{n_1}\). This energy is related to the wave number by the equation \(E = hc\bar{\nu}\), where \(h\) is Planck's constant and \(c\) is the speed of light. The Rydberg formula is derived directly from these energy level transitions.

Each series has a shortest wavelength (highest wave number) and a longest wavelength (lowest wave number).

  • Shortest wavelength/highest wave number occurs for the transition from $n_2 = \infty$ (the limiting line).
  • Longest wavelength/lowest wave number occurs for the transition from $n_2 = n_1 + 1$ (the first line of the series).

The visible series (Balmer) is particularly important because it was the first series to be discovered and studied in detail due to its wavelengths falling within the visible part of the spectrum.

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Important Questions from Electromagnetic Spectrum

  1. An electromagnetic wave propagates through a linear, isotropic, and homogeneous material medium.
    If this medium has a relative permittivity of $\varepsilon_r$ and a relative permeability of $\mu_r$, and the speed of light in vacuum is $c$, what is the ratio of the speed of the electromagnetic wave in the medium ($v$) to its speed in vacuum ($c$)?
  2. The correct order of electromagnetic spectrum with decreasing frequency is:

  3. The value of the proportionality constant μo/(4π) is equal to _________ Tm/A.

  4. Match List I with List II

    List – I

    List – II

    US New Military Bands for Microwaves

    Frequency range in GHz

    A.

    H band

    I.

    2.000 ‐ 3.000 GHz

    B.

    J band

    II.

    4.000 ‐ 6.000 GHz

    C.

    G band

    III.

    6.000 ‐ 8.000 GHz

    D.

    E band

    IV.

    10.000 ‐ 20.000 GHz

    Choose the correct answer from the options given below:

  5. The wavelength in the bright-line emission spectrum of an element are

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