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Question

In the Rydberg formula for the spectrum of the hydrogen atom, the wavenumber is _____.

The correct answer is

directly proportional to the fourth power of an electron charge

Rydberg Formula and Wavenumber Relationship

The Rydberg formula is a fundamental equation in atomic physics that describes the wavelengths, or more commonly, the wavenumbers, of spectral lines emitted by a hydrogen atom. These lines are observed when an electron transitions between different energy levels within the atom.

The general Rydberg formula for the wavenumber ($\tilde{\nu}$) of a spectral line is given by:

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

Where:

  • $\tilde{\nu}$ is the wavenumber (inversely proportional to wavelength, $\lambda$)
  • $R_H$ is the Rydberg constant for hydrogen
  • $n_1$ and $n_2$ are integers representing the principal quantum numbers of the final and initial electron energy levels, respectively, with $n_2 > n_1$.

Rydberg Constant and Electron Charge

The Rydberg constant ($R_H$) is a crucial physical constant derived from fundamental constants of nature. Its precise value determines the exact positions of the spectral lines. The formula for the Rydberg constant is:

$$R_H = \frac{m_e e^4}{8 \epsilon_0^2 h^3 c}$$

Let's break down the components of this formula in a table:

Symbol Meaning
$m_e$ Mass of the electron
$e$ Elementary charge (magnitude of the electron charge)
$\epsilon_0$ Permittivity of free space
$h$ Planck's constant
$c$ Speed of light in vacuum

From the formula for the Rydberg constant ($R_H = \frac{m_e e^4}{8 \epsilon_0^2 h^3 c}$), we can clearly see its dependence on the elementary charge ($e$). The Rydberg constant is directly proportional to the fourth power of the electron charge, i.e., $R_H \propto e^4$.

Wavenumber Proportionality to Electron Charge

Now, let's combine this understanding with the Rydberg formula for the wavenumber:

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

Since the term $\left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ is a constant for a specific electron transition, the wavenumber $\tilde{\nu}$ is directly proportional to the Rydberg constant $R_H$.

Given that $R_H \propto e^4$, it logically follows that the wavenumber $\tilde{\nu}$ is also directly proportional to the fourth power of the electron charge.

Therefore, we can state:

$$\tilde{\nu} \propto e^4$$

This means that if the value of the elementary charge were to change, the wavenumber of the spectral lines in the hydrogen atom would change by its fourth power. This relationship is a direct consequence of the fundamental constants that govern atomic structure and transitions.

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Important Questions from Dual Nature: Photon and Matter Waves

  1. In a photoelectric experiment, both sodium (work function = 2.3 eV) and tungsten (work function = 4.5 eV) metals are illuminated by an ultraviolet light of same wavelength. If the stopping potential for tungsten is measured to be 1.8 V, then the value of the stopping potential for sodium will be

  2. Rapid electron acceleration and deceleration in a conducting wire can generate _______ with frequencies ranging from ______.

  3. The photoelectric current is directly proportional to the number of photo electrons emitted per second. This implies that
  4. The de-Broglie wavelength associated with a ball of mass 150 g traveling at 30.0 m/s would be
  5. A proton accelerated through a potential difference of V volts has a de-Broglie wavelength $\lambda$ associated with it. In order to get the same wavelength associated with an $\alpha$-particle, the required accelerating potential is
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