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 $hc = 1245 \, \text{eV nm}, \, e = 1.6 \times 10^{-19} \, \text{C}$).
(Given $hc = 1245 \, \text{eV nm}, \, e = 1.6 \times 10^{-19} \, \text{C}$).
Correct Answer: 122
Approach Solution - 1
The problem states that in a Franck-Hertz experiment with hydrogen, the first dip in the current-voltage graph occurs at 10.2 V. We need to calculate the wavelength of the light emitted when a hydrogen atom, excited by this energy, returns to its ground state.
Concept Used:
The solution is based on the principles demonstrated by the Franck-Hertz experiment and the Bohr model of the atom.
- Franck-Hertz Experiment: The dips in the current-voltage graph correspond to the voltages at which the electrons have just enough kinetic energy to cause an inelastic collision with the gas atoms. This energy is transferred to the atom, exciting it from its ground state to a higher energy level. The first dip corresponds to the first excitation energy of the atom.
- Excitation Energy: The energy (\( \Delta E \)) absorbed by an atom from an electron accelerated through a potential \( V \) is given by \( \Delta E = e V \), where \( e \) is the charge of an electron. This energy corresponds to the difference between the ground state energy (\(E_1\)) and the first excited state energy (\(E_2\)).
- Photon Emission: When the excited atom de-excites and returns to its ground state, it emits a photon. The energy of this photon (\( E_{\text{photon}} \)) is equal to the energy it absorbed during excitation, \( E_{\text{photon}} = \Delta E \).
- Planck's Relation: The energy of a photon is related to its wavelength (\( \lambda \)) by the equation: \[ E_{\text{photon}} = \frac{hc}{\lambda} \] 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 dip in the current is observed at an accelerating voltage of \( V = 10.2 \, \text{V} \). This voltage is the first excitation potential. The energy absorbed by a hydrogen atom during the inelastic collision is:
\[ \Delta E = e \times V = e \times 10.2 \, \text{V} = 10.2 \, \text{eV} \]This is the energy required to excite a hydrogen atom from its ground state (\(n=1\)) to its first excited state (\(n=2\)).
Step 2: Determine the energy of the emitted photon.
When the excited hydrogen atom de-excites from the first excited state back to the ground state, it releases this energy in the form of a single photon. Therefore, the energy of the emitted photon is 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, we can find the wavelength \( \lambda \) corresponding to the photon's energy:
\[ \lambda = \frac{hc}{E_{\text{photon}}} \]We are given the value of the product \( hc = 1245 \, \text{eV nm} \).
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 the result to the nearest integer, as is common for such problems, we get:
\[ \lambda \approx 122 \, \text{nm} \]Thus, the wavelength of light emitted by the hydrogen atom when excited to the first excitation level is 122 nm.
Approach Solution -2
The energy corresponding to the first dip is given as:
\[10.2 \, \text{eV} = \frac{hc}{\lambda}\]
Rearrange to solve for \( \lambda \):
\[\lambda = \frac{1245 \, \text{eV} \cdot \text{nm}}{10.2 \, \text{eV}}\]
\[\lambda = 122.06 \, \text{nm}\]
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