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Q12.3:
A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?
Solution :
Given Variables & Initial Setup
To determine the frequency of the emitted radiation during an electronic transition, we identify the following parameters provided by the problem:
- Energy difference between the two levels: $\Delta E = 2.3 \text{ eV}$
- Planck's constant: $h \approx 6.626 \times 10^{-34} \text{ J}\cdot\text{s}$
- Elementary charge (conversion factor): $1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}$
Step 1: Conversion of Energy to SI Units
The energy difference must be expressed in Joules ($\text{J}$) to maintain consistency with the SI units of Planck’s constant. [Per the definition of the electronvolt as a unit of energy].
$\Delta E (\text{in Joules}) = 2.3 \text{ eV} \times (1.602 \times 10^{-19} \text{ J/eV})$
$\Delta E = 3.6846 \times 10^{-19} \text{ J}$
Step 2: Theoretical Foundation of Emission Frequency
According to the Bohr postulate for atomic transitions, the frequency ($\nu$) of the emitted photon is directly proportional to the energy difference between the two states. [Per the Bohr Frequency Condition: $\Delta E = h\nu$].
Rearranging for frequency ($\nu$):
$\nu = \frac{\Delta E}{h}$
Step 3: Numerical Calculation
Substituting the values calculated in Step 1 and the known constant for Planck's constant:
$\nu = \frac{3.6846 \times 10^{-19} \text{ J}}{6.626 \times 10^{-34} \text{ J}\cdot\text{s}}$
$\nu \approx 0.55608 \times 10^{15} \text{ Hz}$
$\nu \approx 5.56 \times 10^{14} \text{ Hz}$
Analytical Summary
The transition results in the release of a photon with frequency $\nu$. By applying the relationship $\nu = \frac{\Delta E}{h}$, we establish that the energy gap of $2.3 \text{ eV}$ corresponds to the visible spectrum region (specifically green-yellow light).
Final Solution: The frequency of the emitted radiation is $5.56 \times 10^{14} \text{ Hz}$.
More Questions from Class 12 Physics Atoms EXERCISES
- Q12.1(a): Choose the correct alternative from the clues given at the end of the each statement: (a) The size of the atom in Thomson’s model is .......... the atomic size in Rutherford’s model.
- Q12.1(a): Choose the correct alternative from the clues given at the end of the each statement: The size of the atom in Thomson’s model is .......... the atomic size in Rutherford’s model.
- Q12.1(b): Choose the correct alternative from the clues given at the end of the each statement: (b) In the ground state of .......... electrons are in stable equilibrium, while in .......... electrons always experience a net force.
- Q12.1(b): Choose the correct alternative from the clues given at the end of the each statement: In the ground state of .......... electrons are in stable equilibrium, while in .......... electrons always experience a net force.
- Q12.1(c): Choose the correct alternative from the clues given at the end of the each statement: (c) A classical atom based on .......... is doomed to collapse.
- Q12.1(c): Choose the correct alternative from the clues given at the end of the each statement: A classical atom based on .......... is doomed to collapse.
- Q12.1(d): Choose the correct alternative from the clues given at the end of the each statement: (d) An atom has a nearly continuous mass distribution in a .......... but has a highly non-uniform mass distribution in ..........
- Q12.1(d): Choose the correct alternative from the clues given at the end of the each statement: An atom has a nearly continuous mass distribution in a .......... but has a highly non-uniform mass distribution in ..........
- Q12.1(e): Choose the correct alternative from the clues given at the end of the each statement: (e) The positively charged part of the atom possesses most of the mass in ..........
- Q12.1(e): Choose the correct alternative from the clues given at the end of the each statement: The positively charged part of the atom possesses most of the mass in ..........
- Q12.2: Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?
- Q12.2: Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?
- Q12.3: A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?
- Q12.4: The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?
- Q12.4: The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?
- Q12.5: A hydrogen atom initially in the ground level absorbs a photon, which excites it to the $n = 4$ level. Determine the wavelength and frequency of photon.
- Q12.5: A hydrogen atom initially in the ground level absorbs a photon, which excites it to the ((n)) = 4 level. Determine the wavelength and frequency of photon.
- Q12.6(a): (a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the $n = 1, 2$, and $3$ levels.
- Q12.6(a): (a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the ((n)) = 1, 2, and 3 levels.
- Q12.6(b): (b) Calculate the orbital period in each of these levels.
- Q12.6(b): (b) Calculate the orbital period in each of these levels.
- Q12.7: The radius of the innermost electron orbit of a hydrogen atom is $5.3 \times 10^{-11}$ m. What are the radii of the $n = 2$ and $n = 3$ orbits?
- Q12.7: The radius of the innermost electron orbit of a hydrogen atom is ((5.3 \times 10^{-11})) m. What are the radii of the ((n)) = 2 and ((n)) = 3 orbits?
- Q12.8: A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
- Q12.8: A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
- Q12.9: In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius $1.5 \times 10^{11}$ m with orbital speed $3 \times 10^{4}$ m/s. (Mass of earth = $6.0 \times 10^{24}$ kg.)
- Q12.9: In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius ((1.5 \times 10^{11})) m with orbital speed ((3 \times 10^{4})) m/s. (Mass of earth = ((6.0 \times 10^{24})) kg.)
CBSE Solutions for Class 12 Physics Atoms
Chapters in CBSE - Class 12 Physics
- Electric Charges and Fields
- Electrostatic Potential and Capacitance
- Current Electricity
- Moving Charges and Magnetism
- Magnetism and Matter
- Electromagnetic Induction
- Alternating Current
- Electromagnetic Waves
- Ray Optics and Optical Instruments
- Wave Optics
- Dual Nature of Radiation and Matter
- Nuclei
- Semiconductor Electronics
- Communication Systems
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