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Find the distance of closest approach of a 5.5 MeV α-particle to a gold nucleus (Z = 79).
Given: K = 5.5 MeV; Z = 79; 1/4πε₀ = 9 × 10⁹
Use: K = 2Ze²/(4πε₀ r₀)
r₀ = 9 × 10⁹ × 2 × 79 × (1.6 × 10⁻¹⁹)² / (5.5 × 10⁶ × 1.6 × 10⁻¹⁹)
r₀ = 4.1 × 10⁻¹⁴ m
Answer: about 4 × 10⁻¹⁴ m
If the energy of the α-particle in the question above is doubled, what is the new distance of closest approach?
Given: K′ = 2K
Use: r₀ ∝ 1/K
r₀′ = 4.1 × 10⁻¹⁴ / 2 = 2.1 × 10⁻¹⁴ m
Answer: 2.1 × 10⁻¹⁴ m
The radius of the first orbit of hydrogen is 5.3 × 10⁻¹¹ m. Find the radii of the n = 2 and n = 3 orbits.
Given: r₁ = 5.3 × 10⁻¹¹ m
Use: rₙ = n² r₁
r₂ = 4 × 5.3 × 10⁻¹¹ = 2.12 × 10⁻¹⁰ m
r₃ = 9 × 5.3 × 10⁻¹¹ = 4.77 × 10⁻¹⁰ m
Answer: 2.12 × 10⁻¹⁰ m; 4.77 × 10⁻¹⁰ m
Find the energy of the n = 2, 3 and 4 levels of hydrogen.
Given: E₁ = −13.6 eV
Use: Eₙ = −13.6/n²
E₂ = −3.4 eV
E₃ = −1.51 eV
E₄ = −0.85 eV
Answer: −3.40, −1.51, −0.85 eV
The ground state energy of hydrogen is −13.6 eV. Find the kinetic and potential energies.
Given: E = −13.6 eV
Use: K = −E, U = 2E
K = +13.6 eV
U = −27.2 eV
Answer: 13.6 eV; −27.2 eV
Find the speed of the electron in the n = 1, 2 and 3 orbits of hydrogen.
Given: v₁ = 2.19 × 10⁶ m/s
Use: vₙ = v₁/n
v₁ = 2.19 × 10⁶ m/s
v₂ = 1.09 × 10⁶ m/s
v₃ = 7.3 × 10⁵ m/s
Answer: 2.19 × 10⁶; 1.09 × 10⁶; 7.3 × 10⁵ m/s
Find the orbital period of the electron in the n = 1, 2 and 3 orbits.
Given: r₁ = 5.3 × 10⁻¹¹ m
Use: T = 2πr/v
T₁ = 2π × 5.3 × 10⁻¹¹ / 2.19 × 10⁶ = 1.52 × 10⁻¹⁶ s
T₂ = 8 T₁ = 1.22 × 10⁻¹⁵ s
T₃ = 27 T₁ = 4.11 × 10⁻¹⁵ s
Answer: 1.52 × 10⁻¹⁶ s; 1.22 × 10⁻¹⁵ s; 4.11 × 10⁻¹⁵ s
Find the angular momentum of the electron in the n = 2 orbit.
Given: n = 2; h = 6.63 × 10⁻³⁴ J s
Use: L = nh/2π
L = 2 × 6.63 × 10⁻³⁴ / 6.28 = 2.11 × 10⁻³⁴ J s
Answer: 2.1 × 10⁻³⁴ J s
Find the ground state energy and the first orbit radius of Li²⁺ (Z = 3).
Given: Z = 3
Use: E = −13.6 Z²
E₁ = −13.6 × 9 = −122.4 eV
r₁ = 0.529/3 = 0.176 Å
Answer: −122.4 eV; 0.176 Å
Find the wavelength of the first line of the Balmer series.
Given: n₁ = 2; n₂ = 3; R = 1.097 × 10⁷ m⁻¹
Use: 1/λ = R(1/4 − 1/9)
1/λ = 1.097 × 10⁷ × 5/36 = 1.524 × 10⁶ m⁻¹
λ = 656 nm
Answer: 656 nm (red)
Find the longest and the shortest wavelength of the Lyman series.
Given: n₁ = 1
Use: 1/λ = R(1 − 1/n₂²)
Longest (n₂ = 2): 1/λ = R × 3/4, λ = 121.6 nm
Shortest (n₂ = ∞): λ = 1/R = 91.2 nm
Answer: 121.6 nm; 91.2 nm
Find the shortest wavelength of the Balmer series.
Given: n₁ = 2; n₂ = ∞
Use: 1/λ = R/4
λ = 4/R = 4/1.097 × 10⁷ = 364.6 nm
Answer: 364.6 nm
Two levels are 2.3 eV apart. Find the frequency of the radiation emitted in the transition.
Given: ΔE = 2.3 eV
Use: ν = ΔE/h
ν = 2.3 × 1.6 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ = 5.55 × 10¹⁴ Hz
Answer: 5.5 × 10¹⁴ Hz
A hydrogen atom in the ground state absorbs a photon and goes to n = 4. Find the wavelength and frequency of the photon.
Given: n: 1 → 4
Use: ΔE = E₄ − E₁
ΔE = −0.85 + 13.6 = 12.75 eV
λ = 1240/12.75 = 97.3 nm
ν = c/λ = 3.08 × 10¹⁵ Hz
Answer: 97 nm; 3.08 × 10¹⁵ Hz
How many different spectral lines can be emitted by hydrogen atoms excited to n = 4?
Given: n = 4
Use: N = n(n − 1)/2
N = 4 × 3/2 = 6
Answer: 6 lines
A 12.5 eV electron beam bombards gaseous hydrogen in the ground state. Which transitions can take place?
Given: E = 12.5 eV
Use: Energy needed for n = 3 is 12.09 eV, for n = 4 is 12.75 eV
Atoms can reach n = 3 (12.09 eV) but not n = 4
Lines: 3 → 1 (102.6 nm), 2 → 1 (121.6 nm), 3 → 2 (656 nm)
Answer: Two Lyman lines and one Balmer line
Find the de Broglie wavelength of the electron in the first Bohr orbit.
Given: r₁ = 5.3 × 10⁻¹¹ m
Use: 2πr₁ = λ
λ = 2π × 5.3 × 10⁻¹¹ = 3.3 × 10⁻¹⁰ m
Answer: 0.33 nm
Find the quantum number for the Earth's revolution (m = 6.0 × 10²⁴ kg, r = 1.5 × 10¹¹ m, v = 3 × 10⁴ m/s).
Given: m, r, v as given
Use: n = mvr/(h/2π)
mvr = 6 × 10²⁴ × 3 × 10⁴ × 1.5 × 10¹¹ = 2.7 × 10⁴⁰
n = 2.7 × 10⁴⁰ / 1.05 × 10⁻³⁴ = 2.6 × 10⁷⁴
Answer: 2.6 × 10⁷⁴
Find the frequency of revolution of the electron in the first orbit.
Given: T₁ = 1.52 × 10⁻¹⁶ s
Use: f = 1/T
f = 6.6 × 10¹⁵ Hz
Answer: 6.6 × 10¹⁵ Hz
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(a) no different from: both give an atom of size about 10⁻¹⁰ m.
(b) In the ground state of Thomson's model the electrons are in stable equilibrium, while in Rutherford's model electrons always experience a net force from the nucleus.
(c) A classical atom based on Rutherford's model is doomed to collapse (the revolving electron radiates and spirals in).
(d) An atom has a nearly continuous mass distribution in Thomson's model, but a highly non-uniform mass distribution in Rutherford's model.
(e) The positively charged part possesses most of the mass in both the models.
A hydrogen nucleus (a proton) is about 4 times lighter than the α-particle, and the charge is only +e, so the Coulomb force is weak.
The heavy α-particle would push the light proton aside, like a bowling ball hitting a ping-pong ball; it would continue almost undeflected. So no large-angle scattering and no back-scattering would be seen: the α-particles would pass nearly straight through the hydrogen sheet, which does not support Rutherford's conclusion about a heavy nucleus.
hν = ΔE = 2.3 eV = 2.3 × 1.6 × 10⁻¹⁹ = 3.68 × 10⁻¹⁹ J.
ν = 3.68 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ = 5.55 × 10¹⁴ Hz.
In a Bohr orbit the total energy E = −K, so the kinetic energy K = −E = +13.6 eV.
The potential energy U = 2E = −27.2 eV (so that K + U = −13.6 eV).
ΔE = E₄ − E₁ = −0.85 − (−13.6) = 12.75 eV = 12.75 × 1.6 × 10⁻¹⁹ = 2.04 × 10⁻¹⁸ J.
λ = hc/ΔE = 1240 eV nm / 12.75 eV = 97 nm.
ν = ΔE/h = 2.04 × 10⁻¹⁸ / 6.63 × 10⁻³⁴ = 3.08 × 10¹⁵ Hz.
(a) vₙ = e²/(2ε₀ n h) = 2.19 × 10⁶ / n m/s: n = 1: 2.19 × 10⁶ m/s; n = 2: 1.09 × 10⁶ m/s; n = 3: 7.3 × 10⁵ m/s.
(b) Using rₙ = 0.529 n² Å and T = 2πrₙ/vₙ: n = 1: 1.52 × 10⁻¹⁶ s; n = 2: 1.22 × 10⁻¹⁵ s; n = 3: 4.11 × 10⁻¹⁵ s. (T ∝ n³.)
rₙ = n² r₁. So r₂ = 4 × 5.3 × 10⁻¹¹ = 2.12 × 10⁻¹⁰ m, and r₃ = 9 × 5.3 × 10⁻¹¹ = 4.77 × 10⁻¹⁰ m.
The atoms are in the ground state (n = 1). The energy needed to reach n = 2 is 10.2 eV, to n = 3 is 12.09 eV and to n = 4 is 12.75 eV. A 12.5 eV electron can raise the atom to n = 3 but not to n = 4.
Possible transitions: 3 → 1, 2 → 1 and 3 → 2.
3 → 1: 12.09 eV, λ = 1240/12.09 = 102.6 nm (Lyman series). 2 → 1: 10.2 eV, λ = 121.6 nm (Lyman series). 3 → 2: 1.89 eV, λ = 656.3 nm (Balmer series, Hα).
Angular momentum mvr = nh/2π, so n = 2π mvr / h.
mvr = 6.0 × 10²⁴ × 3 × 10⁴ × 1.5 × 10¹¹ = 2.7 × 10⁴⁰ kg m²/s.
n = 2π × 2.7 × 10⁴⁰ / 6.63 × 10⁻³⁴ = 2.6 × 10⁷⁴. The quantum number is so large that the quantisation of the Earth's orbit is not noticed.