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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?
Solution :
Step 1: Theoretical Foundation
According to the Bohr model of the hydrogen atom, the radius of the $n^{th}$ orbit is determined by the quantization of angular momentum. The expression for the radius $r_n$ of the $n^{th}$ orbit is given by:
$r_n = n^2 r_1$
[Per Bohr’s Quantization Postulate: $L = mvr = \frac{nh}{2\pi}$]
Where:
- $n$ is the principal quantum number (integer).
- $r_1$ is the radius of the innermost (ground state) orbit, defined as the Bohr radius $a_0$.
- $r_1 = 5.3 \times 10^{-11} \, \text{m}$.
Step 2: Calculation for the $n=2$ Orbit
Substituting $n = 2$ into the derived expression:
$r_2 = (2)^2 \times r_1$
$r_2 = 4 \times (5.3 \times 10^{-11} \, \text{m})$
$r_2 = 21.2 \times 10^{-11} \, \text{m}$
$r_2 = 2.12 \times 10^{-10} \, \text{m}$
Step 3: Calculation for the $n=3$ Orbit
Substituting $n = 3$ into the derived expression:
$r_3 = (3)^2 \times r_1$
$r_3 = 9 \times (5.3 \times 10^{-11} \, \text{m})$
$r_3 = 47.7 \times 10^{-11} \, \text{m}$
$r_3 = 4.77 \times 10^{-10} \, \text{m}$
Visual Representation of Orbital Radii
Comparative Data Summary
| Orbit Number ($n$) | Radius Calculation ($n^2 \times r_1$) | Calculated Radius ($m$) |
|---|---|---|
| 1 | $1^2 \times 5.3 \times 10^{-11}$ | $5.3 \times 10^{-11}$ |
| 2 | $4 \times 5.3 \times 10^{-11}$ | $2.12 \times 10^{-10}$ |
| 3 | $9 \times 5.3 \times 10^{-11}$ | $4.77 \times 10^{-10}$ |
Final Solution: The radius of the $n=2$ orbit is $2.12 \times 10^{-10} \, \text{m}$ and the radius of the $n=3$ orbit is $4.77 \times 10^{-10} \, \text{m}$.
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?
- Q12.3: 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?
- Q12.8: 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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