Atomic Spectra Numerical Problems - Chapter 20
Problem 20.1: A hydrogen atom is in its ground state (n = 1). Using Bohr's theory, calculate (a) the radius of the orbit (b) the linear momentum of the electron (c) the angular momentum of the electron (d) the kinetic energy (e) the potential energy and (f) the total energy.
Given Data:
- Quantum number, \( n = 1 \)
- Mass of electron, \( m = 9.1 \times 10^{-31} \, \text{kg} \)
To Determine:
- (a) Radius \( r = ? \)
- (b) Linear Momentum \( p = ? \)
- (c) Angular Momentum \( L = ? \)
- (d) Kinetic Energy \( K = ? \)
- (e) Potential Energy \( U = ? \)
- (f) Total Energy \( T = ? \)
Calculations:
(a) Radius of orbit:
\[ r_n = \frac{n^2 h^2}{4 \pi^2 K e^2 m} \]
\[ r_1 = \frac{(1)^2 (6.63 \times 10^{-34})^2}{4 \times (3.14)^2 \times 9 \times 10^9 \times (1.6 \times 10^{-19})^2 \times 9.1 \times 10^{-31}} = 5.39 \times 10^{-11} \, \text{m} \]
(b) Linear momentum:
\[ p_n = m v_n = m \left( \frac{2 \pi K e^2}{n h} \right) \]
\[ p_1 = 9.1 \times 10^{-31} \times \left( \frac{2 \times 3.14 \times 9 \times 10^9 \times (1.6 \times 10^{-19})^2}{1 \times 6.63 \times 10^{-34}} \right) = 1.99 \times 10^{-24} \, \text{Ns} \]
(c) Angular momentum:
\[ L_1 = m v_1 r_1 = p_1 r_1 = 1.99 \times 10^{-24} \times 5.39 \times 10^{-11} = 1.05 \times 10^{-34} \, \text{Js} \]
(d) Kinetic energy:
\[ K_n = \frac{1}{2} m v_n^2 = \frac{1}{2} m \left( \frac{2 \pi K e^2}{n h} \right)^2 \]
\[ K_1 = \frac{1}{2} \times 9.1 \times 10^{-31} \times \left( \frac{2 \times 3.14 \times 9 \times 10^9 \times (1.6 \times 10^{-19})^2}{1 \times 6.63 \times 10^{-34}} \right)^2 = 2.17 \times 10^{-18} \, \text{J} \]
\[ \Rightarrow K_1 = \frac{2.17 \times 10^{-18}}{1.6 \times 10^{-19}} = 13.6 \, \text{eV} \]
(e) Potential energy:
\[ U = -\frac{K e^2}{r_n} = -\frac{K e^2}{r_1} = -\frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{5.39 \times 10^{-11}} = -4.35 \times 10^{-18} \, \text{J} \]
\[ \Rightarrow U = \frac{-4.35 \times 10^{-18}}{1.6 \times 10^{-19}} = -27.2 \, \text{eV} \]
(f) Total energy:
\[ T = K + U = 13.6 \, \text{eV} + (-27.2 \, \text{eV}) = -13.6 \, \text{eV} \]
Problem 20.2: What are the energies in eV of quanta of wavelength? λ = 400, 500 and 700 nm.
Given Data:
- Wavelengths: (a) \( \lambda_1 = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \)
- (b) \( \lambda_2 = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} \)
- (c) \( \lambda_3 = 700 \, \text{nm} = 700 \times 10^{-9} \, \text{m} \)
To Determine:
- Energies: (a) \( E_1 = ? \)
- (b) \( E_2 = ? \)
- (c) \( E_3 = ? \)
Calculations:
(a) Energy for 400 nm:
\[ E_1 = \frac{h c}{\lambda_1} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} = 4.97 \times 10^{-19} \, \text{J} \]
\[ \Rightarrow E_1 = \frac{4.97 \times 10^{-19}}{1.6 \times 10^{-19}} = 3.10 \, \text{eV} \]
(b) Energy for 500 nm:
\[ E_2 = \frac{h c}{\lambda_2} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{500 \times 10^{-9}} = 3.98 \times 10^{-19} \, \text{J} \]
\[ \Rightarrow E_2 = \frac{3.98 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.49 \, \text{eV} \]
(c) Energy for 700 nm:
\[ E_3 = \frac{h c}{\lambda_3} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{700 \times 10^{-9}} = 2.84 \times 10^{-19} \, \text{J} \]
\[ \Rightarrow E_3 = \frac{2.84 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.77 \, \text{eV} \]
Problem 20.3: An electron jumps from a level \( E_i = -3.5 \times 10^{-19} \, \text{J} \) to \( E_f = -1.20 \times 10^{-18} \, \text{J} \). What is the wavelength of the emitted light?
Given Data:
- Initial energy, \( E_i = -3.5 \times 10^{-19} \, \text{J} \)
- Final energy, \( E_f = -1.20 \times 10^{-18} \, \text{J} \)
To Determine:
- Wavelength, \( \lambda = ? \)
Calculations:
\[ \Delta E = E_f - E_i = -1.20 \times 10^{-18} - (-3.5 \times 10^{-19}) = -8.5 \times 10^{-19} \, \text{J} \]
\[ \lambda = \frac{h c}{\Delta E} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{8.5 \times 10^{-19}} = 234 \times 10^{-9} \, \text{m} = 234 \, \text{nm} \]
Problem 20.4: Find the wavelength of the spectral line corresponding to the transition in hydrogen from n = 6 state to n = 3 state.
Given Data:
- Initial state, \( n = 6 \)
- Final state, \( p = 3 \)
- Rydberg constant, \( R_H = 1.0974 \times 10^7 \, \text{m}^{-1} \)
To Determine:
- Wavelength, \( \lambda = ? \)
Calculations:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{p^2} - \frac{1}{n^2} \right) = 1.0974 \times 10^7 \left( \frac{1}{3^2} - \frac{1}{6^2} \right) \]
\[ = 1.0974 \times 10^7 \left( \frac{1}{9} - \frac{1}{36} \right) = 1.0974 \times 10^7 \left( \frac{4 - 1}{36} \right) = 1.0974 \times 10^7 \times \frac{3}{36} \]
\[ \Rightarrow \lambda = \frac{36}{1.0974 \times 10^7 \times 3} = 1.093 \times 10^{-6} \, \text{m} = 1093 \, \text{nm} \]
Problem 20.5: Compute the shortest wavelength radiation in the Balmer series. What value of n must be used?
Given Data:
- For Balmer series, \( p = 2 \)
- For shortest wavelength, \( n = \infty \)
- Rydberg constant, \( R_H = 1.0974 \times 10^7 \, \text{m}^{-1} \)
To Determine:
- Shortest wavelength, \( \lambda_s = ? \)
Calculations:
\[ \frac{1}{\lambda_s} = R_H \left( \frac{1}{p^2} - \frac{1}{n^2} \right) = 1.0974 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = 1.0974 \times 10^7 \times \frac{1}{4} \]
\[ \Rightarrow \lambda_s = \frac{4}{1.0974 \times 10^7} = 3.645 \times 10^{-7} \, \text{m} = 364.5 \, \text{nm} \]
Problem 20.6: Calculate the longest wavelength of radiation for the Paschen series.
Given Data:
- For Paschen series, \( p = 3 \)
- For longest wavelength, \( n = 4 \)
- Rydberg constant, \( R_H = 1.0974 \times 10^7 \, \text{m}^{-1} \)
To Determine:
- Longest wavelength, \( \lambda_L = ? \)
Calculations:
\[ \frac{1}{\lambda_L} = R_H \left( \frac{1}{p^2} - \frac{1}{n^2} \right) = 1.0974 \times 10^7 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \]
\[ = 1.0974 \times 10^7 \left( \frac{1}{9} - \frac{1}{16} \right) = 1.0974 \times 10^7 \left( \frac{16 - 9}{144} \right) = 1.0974 \times 10^7 \times \frac{7}{144} \]
\[ \Rightarrow \lambda_L = \frac{144}{1.0974 \times 10^7 \times 7} = 1.875 \times 10^{-6} \, \text{m} = 1875 \, \text{nm} \]
Problem 20.7: Electrons in an X-ray tube are accelerated through a potential difference of 3000 V. If these electrons were slowed down in a target, what will be the minimum wavelength of X-rays produced?
Given Data:
- Potential difference, \( V_0 = 3000 \, \text{V} \)
To Determine:
- Minimum wavelength, \( \lambda_{\text{min}} = ? \)
Calculations:
\[ \lambda_{\text{min}} = \frac{h c}{V_0 e} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{3000 \times 1.6 \times 10^{-19}} = 4.14 \times 10^{-10} \, \text{m} \]
Problem 20.8: The wavelength of K X-ray from copper is \( 1.377 \times 10^{-10} \, \text{m} \). What is the energy difference between the two levels from which this transition results?
Given Data:
- Wavelength, \( \lambda = 1.377 \times 10^{-10} \, \text{m} \)
To Determine:
- Energy difference, \( \Delta E = ? \)
Calculations:
\[ \Delta E = \frac{h c}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.377 \times 10^{-10}} = 1.444 \times 10^{-15} \, \text{J} \]
\[ \Rightarrow \Delta E = \frac{1.444 \times 10^{-15}}{1.6 \times 10^{-19}} = 9.025 \times 10^3 \, \text{eV} \]
Problem 20.9: A tungsten target is struck by electrons that have been accelerated from rest through 40 kV potential difference. Find the shortest wavelength of the bremsstrahlung radiation emitted.
Given Data:
- Potential difference, \( V_0 = 40 \, \text{kV} = 40000 \, \text{V} \)
To Determine:
- Shortest wavelength, \( \lambda_{\text{min}} = ? \)
Calculations:
\[ \lambda_{\text{min}} = \frac{h c}{V_0 e} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{40000 \times 1.6 \times 10^{-19}} = 3.11 \times 10^{-11} \, \text{m} \]
Problem 20.10: The orbital electron of a hydrogen atom moves with a speed of \( 5.456 \times 10^5 \, \text{ms}^{-1} \). (a) Find the value of the quantum number n associated with this electron. (b) Calculate the radius of this orbit. (c) Find the energy of the electron in this orbit.
Given Data:
- Speed of electron, \( v_n = 5.456 \times 10^5 \, \text{ms}^{-1} \)
To Determine:
- (a) Quantum number, \( n = ? \)
- (b) Orbital radius, \( r_n = ? \)
- (c) Energy of electron, \( E_n = ? \)
Calculations:
(a) Quantum number:
\[ v_n = \frac{2 \pi K e^2}{n h} \Rightarrow n = \frac{2 \pi K e^2}{h v_n} \]
\[ n = \frac{2 \times 3.14 \times 9 \times 10^9 \times (1.6 \times 10^{-19})^2}{6.63 \times 10^{-34} \times 5.456 \times 10^5} = 4 \]
(b) Orbital radius:
\[ r_n = \frac{n^2 h^2}{4 \pi^2 K e^2 m} = \frac{(4)^2 (6.63 \times 10^{-34})^2}{4 \times (3.14)^2 \times 9 \times 10^9 \times (1.6 \times 10^{-19})^2 \times 9.1 \times 10^{-31}} = 8.46 \times 10^{-10} \, \text{m} \]
(c) Energy of electron:
\[ E_n = -\frac{13.6}{n^2} = -\frac{13.6}{4^2} = -0.85 \, \text{eV} \]
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