Question

In Franck-Hertz experiment, the first dip in the current-voltage graph for hydrogen is observed at 10.2 V. The wavelength of light emitted by hydrogen atom when excited to the first excitation level is _____ nm.
(Given $hc = 1245 \, \text{eV nm}, \, e = 1.6 \times 10^{-19} \, \text{C}$).

Answer 1

Correct Answer: 122

In a Franck-Hertz experiment with hydrogen, the initial current dip at 10.2 V indicates the energy required to excite a hydrogen atom. This energy, when emitted as a photon upon de-excitation to the ground state, corresponds to a specific wavelength. We will calculate this wavelength.

Concept Used:

The solution utilizes concepts from the Franck-Hertz experiment and the Bohr model of atomic excitation and photon emission.

1. Franck-Hertz Experiment: Current dips signify electron kinetic energies sufficient for inelastic collisions, exciting gas atoms. The first dip marks the atom's first excitation energy level.
2. Excitation Energy: The energy absorbed by an atom due to an electron accelerated through a potential \( V \) is \( \Delta E = e V \). This is the energy difference between the ground and first excited states.
3. Photon Emission: Upon returning to the ground state from an excited state, the atom emits a photon with energy \( E_{\text{photon}} = \Delta E \).
4. Planck's Relation: The energy of an emitted photon is related to its wavelength \( \lambda \) by \( E_{\text{photon}} = \frac{hc}{\lambda} \), where \( h \) is Planck's constant and \( c \) is the speed of light.

Step-by-Step Solution:

Step 1: Determine the first excitation energy of the hydrogen atom.

The first excitation potential is \( V = 10.2 \, \text{V} \). The energy absorbed by a hydrogen atom is:

\[ \Delta E = e \times V = e \times 10.2 \, \text{V} = 10.2 \, \text{eV} \]

This is the energy difference between the ground state (\(n=1\)) and the first excited state (\(n=2\)) of hydrogen.

Step 2: Determine the energy of the emitted photon.

When the excited hydrogen atom returns to the ground state, it emits a photon with energy equal to the excitation energy.

\[ E_{\text{photon}} = \Delta E = 10.2 \, \text{eV} \]

Step 3: Calculate the wavelength of the emitted photon.

Using Planck's relation, the wavelength \( \lambda \) is:

\[ \lambda = \frac{hc}{E_{\text{photon}}} \]

Given \( hc = 1245 \, \text{eV nm} \).

Step 4: Substitute the values and compute the result.

\[ \lambda = \frac{1245 \, \text{eV nm}}{10.2 \, \text{eV}} \] \[ \lambda \approx 122.0588 \, \text{nm} \]

Rounding to the nearest integer yields:

\[ \lambda \approx 122 \, \text{nm} \]

The wavelength of the light emitted when the hydrogen atom returns to its ground state from the first excitation level is 122 nm.