Problem#1
The silicon–silicon single bond that forms the basis of the mythical silicon-based creature the Horta has a bond strength of 3.80 eV. What wavelength of photon would you need in a (mythical) phasor disintegration gun to destroy the Horta?
Answer:
The minimum energy the photon would need is the 3.84 eV bond strength.
The photon energy E = hf = hc/λ must equal the bond strength
hc/λ = 3.80 eV, so
λ = hc/3.80 eV
λ = (4.136 x 10-15 eV.s)(3.00 x 108 m/s)/(3.80 eV) = 327 NM
Any photon having a shorter wavelength would also spell doom for the Horta!
Problem#2
A hydrogen atom is in a state with energy –1.51 eV. In the Bohr model, what is the angular momentum of the electron in the atom, with respect to an axis at the nucleus?
Answer:
Use the energy to calculate n for this state. Then use the Bohr equation, Eq. (39.6), to calculate L.
En = –(13.6 eV)/n2, so this state has
n2 = –(13.6 eV)/(–1.51 eV)
n = 3
In the Bohr model, L = nℏ so far this state L = 3ℏ = 3.16 x 10-34 kg.m2/s
We will find in Section 41.1 that the modern quantum mechanical description gives a different result.
Problem#3
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 the photon.
Answer:
For a hydrogen atom En = –13.6 eV/n2.
∆E = hc/λ, where ∆E is the magnitude of the energy change for the atom and λ is the wavelength of the photon that is absorbed or emitted.
∆E = E4 – E1 = –13.6 eV(1/42 – 1/12) = +12.75 eV.
λ = hc/∆E = (4.136 x 10-15 eV.s)(3.00 x 108 m/s)/(12.75 eV) = 97.3 nm
f = c/λ = 3.08 x 1015 Hz.
This photon is in the ultraviolet region of the electromagnetic spectrum.
Problem#3
A triply ionized beryllium ion, Be3+ (a beryllium atom with three electrons removed), behaves very much like a hydrogen atom except that the nuclear charge is four times as great. (a) What is the ground-level energy of Be3+?. How does this compare to the groundlevel energy of the hydrogen atom? (b) What is the ionization energy of Be3+?. How does this compare to the ionization energy of the hydrogen atom? (c) For the hydrogen atom, the wavelength of the photon emitted in the n = 2 to n = 1 transition is 122 nm (see Example 39.6). What is the wavelength of the photon emitted when a ion undergoes this transition? (d) For a given value of n, how does the radius of an orbit in compare to that for hydrogen?
Answer:
The force between the electron and the nucleus in Be3+ is F = kZe2/r2, where Z = 4 is the nuclear charge. All the equations for the hydrogen atom apply to Be3+ if we replace e2 by Ze2.
(a) En = –me4(Ze2)2/(8ε02n2h2) = Z2{–me4(Ze2)2/(8ε02n2h2)}
En = Z2(–13.6eV/n2) for Be3+
The ground-level energy of Be3+ is E1 = 16(–13.6eV/12) = –218 eV
(b) The ionization energy is the energy difference between the n → ∞ level energy and the
n = 1 level energy.
The n → ∞ level energy is zero, so the ionization energy of Be3+ is 218 eV.
This is 16 times the ionization energy of hydrogen.
(c) 1/λ = R(1/n12 – 1/n22) just as for hydrogen but now R has a different value
RH = me4/(8ε02h3c) = 1.097 x 107 m-1 for hydrogen becomes
RBe = Z2me4/(8ε02h3c) = 16(1.097 x 107 m-1) = 1.755 x 108 m-1 for Be3+
For n = 2 to n = 1,
1/λ = RBe(1/12 – 1/22) = 3RBe/4
λ = 4/(3RBe) = 4/(3 x 1.755 x 108 m-1) = 7.60 x 10-9 m = 7.60 nm.
This wavelength is smaller by a factor of 16 compared to the wavelength for the corresponding transition in the hydrogen atom.
(d) rn = ε0n2h2/(πme2) (hydrogen)
rn = ε0n2h2/(πmZe2) for Be3+
For a given n the orbit radius for Be3+ is smaller by a factor of Z = 4 compared to the
corresponding radius for hydrogen.
Post a Comment for "Energy Levels and the Bohr Model of the Atom Problems and Solutions"