Energy Levels and the Bohr Model of the Atom Problems and Solutions 2

 Problem#1

In a set of experiments on a hypothetical one-electron atom, you measure the wavelengths of the photons emitted from transitions ending in the ground state (n = 1) as shown in the energy-level diagram in Fig. E39.31. You also observe that it takes 17.50 eV to ionize this atom. (a) What is the energy of the atom in each of the levels (n = 1, n = 2, etc.) shown in the figure? (b) If an electron made a transition from the n = 4 to the n = 2 level, what wavelength of light would it emit?

Answer:

The wavelength of the photon is related to the transition energy E_i – E_f of the atom by

E_i – E_f = hc/λ where hc = 1.240 x 10^-6 eV.m

(a) The minimum energy to ionize an atom is when the upper state in the transition has E = 0,

So E₁ = –17.50 eV.

For n = 5 → n = 1, λ = 73.86 nm and

E₅ – E₄ = (1.240 x 10^-6 eV.m)/(73.86 x 10^-9 m) = 16.79 eV

E₅ = –17.50 eV + 16.79 eV = –0.71 eV

For n = 4 → n = 1, λ = 75.63 nm and E₄ = –1.10 eV

For n = 3 → n = 1, λ = 79.76 nm and E₃ = –1.95 eV

For n = 2 → n = 1, λ = 94.54 nm and E₂ = –4.38 eV

(b) E_i – E_f = E₄ – E₂ = –1.10 eV – (–4.38 eV) = 3.28eV and

λ = hc/(E_i – E_f) = (1.240 x 10^-6 eV.m)/(3.28eV) = 378 nm

The n = 4 → n = 2 transition energy is smaller than the n = 4 → n = 1 transition energy so the wavelength is longer. In fact, this wavelength is longer than for any transition that ends in the n = 1.

Problem#2

Find the longest and shortest wavelengths in the Lyman and Paschen series for hydrogen. In what region of the electromagnetic spectrum does each series lie?

Answer:

For the Lyman series the final state is n = 1 and the wavelengths are given by

1/λ = R(1/1² – 1/n²), n = 2,3,…. For the Paschen series the final state is n = 3 and the wavelengths are gives by 1/λ = R(1/3² – 1/n²), n = 4,5,…. R = 1.097  x 10⁷ m^-1. The longest wavelength is for the smallest n and the shortest wavelengt is for n → ∞.

Lyman:

Longest: 1/λ = R(1/1² – 1/2²) = 3R/4 → λ = 4/(3 x 1.097 x 10⁷ m^-1) = 121.5 nm

Shortest: 1/λ = R(1/1² – 1/∞²) = R → λ = 1/(1.097 x 10⁷ m^-1) = 91.16 nm

Paschen:

Longest: 1/λ = R(1/3² – 1/4²) = 7R/144 → λ = 144/(7 x 1.097 x 10⁷ m^-1) = 1875 nm

Shortest: 1/λ = R(1/3² – 1/∞²) = R/9 → λ = 9/(1.097 x 10⁷ m^-1) = 820 nm

The Lyman series is in the ultraviolet. The Paschen series is in the infrared.

Problem#3 

(a) An atom initially in an energy level with E = –6.52 eV absorbs a photon that has wavelength 860 nm. What is the internal energy of the atom after it absorbs the photon? (b) An atom initially in an energy level with E = –2.68 eV emits a photon that has wavelength 420 nm. What is the internal energy of the atom after it emits the photon?

Answer:

Apply conservation of energy to the system of atom and photon.

The energy of a photon is E_γ = hc/λ.

(a) E_γ = hc/λ = (6.63 x 10^-34 J.s)(3.00 x 10⁸ m/s)/(8.60 x 10^-7 m) = 2.31 x 10^-19 J = 1.44 eV.

So the internal energy of the atom increases by 1.44 eV to E = –6.52 eV + 1.44 eV = –5.08 eV.

(b) E_γ = hc/λ = (6.63 x 10^-34 J.s)(3.00 x 10⁸ m/s)/(4.20 x 10^-7 m) = 4.74 x 10^-19 J = 2.96 eV.

So the final internal energy of the atom decreases to E = –2.68 eV – 2.96 eV = –5.64 eV.

When an atom absorbs a photon the energy of the atom increases. When an atom emits a photon the energy of the atom decreases.

Problem#4

Use Balmer’s formula to calculate (a) the wavelength, (b) the frequency, and (c) the photon energy for the H_γ line of the Balmer series for hydrogen.

Answer:

Balmer’s formula is 1/λ = R(1/2² – 1/n²).

For the H_γ spectral line n = 5. Once we have λ, calculate f from f = c/λ and E.

(a) 1/λ = R(1/2² – 1/n²) = R(1/2² – 1/5²) = 21R/100

Thus λ = 100/21R = 100/(21 x 1.097 x 10⁷ m^-1) = 4.341 x 10^-7 m = 434.1 nm

(b) f = c/λ = (2.998 x 10⁸ m/s)/(4.341 x 10^-7 m) = 6.906 x 10¹⁴ Hz

(c) E = hf = (6.63 x 10^-34 J.s)(6.906 x 10¹⁴ Hz) = 4.576 x 10^-19 J = 2.856 eV

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