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Problem Set 3 : Bohr’s Atom
Solution
- A photon is emitted as an atom makes a transition from n = 4 to n = 2 level. What is the frequency, wavelength and energy of the emitted photon? Answer 1 we are given that, Initial orbit of electron = ni = 4 Final orbit of electron = nf = 2. Also we know that, Speed of electromagnetic waves = c = 3 × 108 m/s Planck’s constant = h = 6. 63 × 10 −^34 Js. We want to calculate the frequency, wavelength and energy of the emitted photon, when the atom makes a transition from ni = 4 to nf = 2. Frequency of the emitted photon can be calculated by using the equation, E = hf, ⇒ f = Eh. Let’s at first calculate E by using the following relation, E = − 13. 6
[ 1
n^2 f^ −^
n^2 i
]
eV
= − 13. 6
[ 1
22 −^
]
eV
= − 13. 6
[ 1
4 −^
]
eV = − 13. 6 × 16 3 eV = 2.55 eV. Thus frequency of photon will be, f = Eh = (^6). 632 ×.55 eV 10 − (^34) Js =^2.^55 ×^1.^6 ×^10
− 19 J
- 63 × 10 −^34 Js = 6. 15 × 1014 Hz.
Wavelength of the emitted photon can be calculated by using the following equation. c = f λ ⇒ λ = (^) fc = 3 ×^10
(^8) m/s
15 × 1014 Hz = 4. 875 × 10 −^7 m = 488 nm.
For the Balmer series i.e., the atomic transitions where final state of the electron is n = 2, what is the longest and shortest wavelength possible? Is any of the frequency of Lyman series, which corresponds to transitions where electron ends up in n = 1 level, in the visible region? Answer 2 The atomic transitions where final state of the electron is n = 2, atoms emit a series of lines in the visible part of the spectrum. This series is called the Balmer Series. Balmer examined the four visible lines in the spectrum of the hydrogen atom. Wave length of this series can be calculated by using the equation, 1 λ =^ R
[ 1
22 −^
n^2
]
, where n = 3, 4 , 5 , .... R is the Rydberg constant, whose value is 1. 097 × 107 m−^1. The number n is just an integer; the above formula gives the longest wavelength, when n = 3, and gives each of the shorter wavelengths as n increases. From Balmer’s equation, it looks like when n gets bigger, the lines should start getting really close together. That’s exactly right; as n gets larger, 1 over n squared gets smaller, so there’s less and less difference between the consecutive lines. We can see that the series has a limit, that is, as n gets larger and larger, the wavelength gets closer and closer to one particular value. If n is infinity, then 1 over n squared is 0, and the corresponding wavelength is shortest. The individual lines in the Balmer series are given the names Alpha, Beta, Gamma, and Delta, and each corresponds to a ni value of 3, 4, 5, and 6 respectively as shown in the figure below.
δ γ β α
6 25 2 4 2 n=3^2
Thus the shortest wavelength in the Balmer series is 365 nm.
In order to see that whether any of the frequency of Lymen series is in the visible region, let’s calculate the largest and smallest frequencies of Laymen series. the wavelength of the Laymen series is given by, 1 λ =^ R
[
1 − (^) n^12
]
, where n = 2, 3 , 4 , 5 , ....
The longest wavelength corresponds to the smallest energy difference between energy levels, which in this case will be between n = 1 and n = 2.
Wavelength for transition from n = 2 1 λ =^ R
[
1 − (^) n^12
]
= 1. 097 × 107 m−^1 ×
[
]
= 1. 097 × 107 m−^1 ×
[
]
= 1. 097 × 107 m−^1 × (^34) = 8. 23 × 106 m−^1 ⇒ λ = 1. 22 × 10 −^7 m = 122 nm.
Thus the longest wavelength in the Laymen series is 122 nm, and corresponding fre- quency will be,
f = (^) λc = 3 ×^10
(^8) m/s
- 22 × 10 −^7 m = 24.^7 ×^10
(^14) Hz.
The shortest wavelength corresponds to the largest energy difference between energy levels, which in this case will be between n = 1 and n = ∞.
Wavelength for transition from n = ∞ 1 λ =^ R
[
1 − (^) n^12
]
= 1. 097 × 107 m−^1 ×
[
1 − ∞^12
]
= 1. 097 × 107 m−^1 ×
[
1 − ∞^1
]
= 1. 097 × 107 m−^1 × 1 = 1. 097 × 107 m−^1 ⇒ λ = 9. 12 × 10 −^8 m = 91.2 nm.
Thus the shortest wavelength in the Laymen series is 91.2 nm, and corresponding frequency will be, f = (^) λc = 3 ×^10
(^8) m/s
- 12 × 10 −^8 m = 32.^9 ×^10
(^14) Hz.
From above we can see that frequency range of Laymen series is 24. 7 × 1014 Hz to
9 × 1014. As frequency range of visible light is from 4 × 1014 to 8 × 1014 , therefore no frequency of Laymen series is in the visible region.
A Hydrogen atom initially in its ground state i.e., n = 1 level, absorbs a photon and ends up in n = 4 level. What must have been the frequency of the photon? Now the electron makes spontaneous emission and comes back to the ground state. What are the possible frequencies of the photons emitted during this process. Answer 3 We are given that, Initial state of electron = ni = 1 final state of electron = nf = 4. We want to calculate the frequency of photon absorbed by the electron to make this transition. As we know that energy of photon absorbed is given by, E = − 13. 6
n^2 f^ −^
n^2 i
eV
hf = − 13. 6
n^2 f^ −^
n^2 i
eV
f = −^13 h.^6
n^2 f^ −^
n^2 i
eV
= − (^4). 14 ×^13 10.^6 − (^15) eVs
n^2 f^ −^
n^2 i
eV
= − 3. 28 × 1015
n^2 f^ −^
n^2 i
Hz
f 1 → 4 = − 3. 28 × 1015
42 −^
Hz
= − 3. 28 × 1015
16 −^1
Hz = − 3. 28 × 1015 × − 1615 Hz = +3. 08 × 1015 Hz,
From n = 4 to n = 2 f 4 → 2 = − 3. 28 × 1015
22 −^
Hz = − 6. 15 × 1014 Hz. From n = 2 to n = 1
f 2 → 1 = − 3. 28 × 1015
12 −^
Hz = − 2. 46 × 1015 Hz. From n = 4 to n = 3
f 4 → 3 = − 3. 28 × 1015
32 −^
Hz = − 1. 6 × 1014 Hz.
From n = 3 to n = 1 f 3 → 1 = − 3. 28 × 1015
12 −^
Hz = − 2. 92 × 1015 Hz.
From n = 3 to n = 2 f 3 → 2 = − 3. 28 × 1015
22 −^
Hz = − 4. 57 × 101 Hz.
Negative sign with each frequency represents that photon has been emitted.
- Non-relativistic Nature of Electron Inside Bohr’s Atom
Using Bohrs quantization rule, derive a formula for electrons speed in the quantized Bohrs orbits. By putting in the values of constants, explicitly derive the value of electrons speed in n = 1 orbit. What fraction of speed of light it is? Is it justified to treat electrons motion as non-relativistic in Hydrogen atom as we have been doing? What happens to the speed in higher levels? Answer 4 The basic feature of quantum mechanics that is incorporated in the Bohr Model the energy of the particles in the Bohr atom is restricted to certain discrete values. One
says that the energy is quantized. This means that only certain orbits with certain radii are allowed; orbits in between simply don’t exist.
In the Bohr model, the wavelength associated with the electron is given by the De- Broglie relationship,
λ = (^) mv h ,
and the standing wave condition that circumference = whole number of wavelengths. In the hydrogenic case, the number n is the principal quantum number.
2 πr = nλ.
These can be combined to get an expression for the angular momentum of the electron in orbit. (Note that this assumes a circular orbit, a generally unwarranted assumption.) 2 πr n =^
h mv mvr = nh 2 π mvr = nℏ r = (^) mv nℏ. (1)
The electron is held in a circular orbit by electrostatic attraction. The centripetal force is equal to the Coulomb force. mv^2 r =^
Zke^2 r^2 r = Zke
2 mv^2.^ (2) Compare equation (??) and (??) we get, nℏ mv =^
Zke^2 mv^2 v = Zke
2 nℏ.^ (3) where
Z = Atomic number for Hydrogen = 1 k = Coulomb’s constant = 9 × 109 Nm^2 /C^2 e = Charge on an electron = 1. 6 × 10 −^19 C ℏ = Reduced Planck’s constant = 1. 05 × 10 −^34 J.s
the center of mass and may be considered as stationary, while electron moving around it. Reduced mass for such a system can be calculated by using the equation,
Reduced mass = μ = (^) mme e+m mpp ,
where mp >> me, ⇒ me + mp ≈ mp. Thus reduced mass will become,
μ = (^) mme +em mpp μ ≈ m mempp μ ≈ me.
From above we can see that the reduced mass is approximately equal to the mass of electron, that is why in our formulas we put mass of electron instead of reduced mass.
Now let’s consider the positronium. Positronium is made up of a positron (anti particle of electron) and an electron. Since electron and positron both have the same masses the center of mass will be at the center of the line joining the two particles. In this case the reduced mass is given by,
μ = (^) (m(me−^ )(me+^ ) e−^ ) + (me+^ ) μ = (m2(e−m^ )(em+ (^) )e+^ ) μ = m 2 e −= m 2 e.
Bohr radius is given by,
r = n
kmee^2. For positronium Bohr radius will be,
r = n
kμe^2 = 2 n
kmee^2 , Substitute values of constants, we get,
r = 2 ×^ (1.^05 ×^10
− (^34) J.s) 2 9 × 109 Nm^2 /C^2 × 9. 1 × 10 −^31 kg × (1. 6 × 10 −^19 C)^2 = 1. 05 × 10 −^10 m = 1.05˚A.
Numerical value for Bohr’s radius for hydrogen atom is 0.529 ˚A. Therefore Bohr’s radius for positronium is increased by 2 times. As we know that Rydberg constant in case of hydrogen atom is given by,
R = mee
4 8 ε^20 h^2. For positronium Rydberg constant will be,
R ′^ = μe
4 8 ε^20 h^2 R ′^ = mee
4 2 × 8 ε^20 h^2 = R 2.
Since Rydberg constant for positronium is reduced to half, formula for Balmer series for positronium is given by, 1 λ =^
R
[ 1
22 −^
n^2
]
, where n = 3, 4 , 5 , .... f c =^
R
[ 1
22 −^
n^2
]
f = cR 2
[ 1
22 −^
n^2
]
For n=
f 1 ′ =^3 ×^10
(^8) m/s × 1. 097 × 107 m− 1 2
[ 1
22 −^
]
=^3 ×^10
(^8) m/s × 1. 097 × 107 m− 1 2 ×^
= 2. 28 × 1014 Hz.
For n=
f 2 ′ =^3 ×^10
(^8) m/s × 1. 097 × 107 m− 1 2
[ 1
22 −^
]
=^3 ×^10
(^8) m/s × 1. 097 × 107 m− 1 2 ×^
= 3. 08 × 1014 Hz.
For n=
f 3 ′ =^3 ×^10
(^8) m/s × 1. 097 × 107 m− 1 2
[ 1
22 −^
]
=^3 ×^10
(^8) m/s × 1. 097 × 107 m− 1 2 ×^
= 3. 46 × 1014 Hz.
(c) Explain in detail what would one observe (about the radiation) if an electron were revolving around a proton in a macroscopic sized radius and how it matches with Maxwells predictions. Do we need to use quantum theory for motion of particles in macroscopic orbits? (d) Find the frequency of photons emitted by an electron moving around a proton in a radius of 1 cm.
Answer 6
(a) As we have already discussed in lecture 9 that total energy of an electron in nth orbit is given by, En = K.E. + P.E. =^12 mv^2 − Ze
2 kr (4) Coulomb’s Force = Centripetal Force Ze^2 4 πε 0 r^2 =^
mv^2 r ⇒ Ze
2 4 πε 0 r =^ mv
2 Ze^2 8 πε 0 r =^
mv^2
Thus equation (??) can be written as, En = Ze
2 8 πε 0 r −^
Ze^2 4 πε 0 r = − Ze
2 8 πε 0 r , where r is the Bohr radius that is given by, r = ε^0 n
(^2) h 2 πZmee^2. Therefore total energy of nth orbit will become, En = − Ze
2 8 πε 0 ×^
πZmee^2 ε 0 n^2 h^2 = − meZ
(^2) e 4 8 n^2 ε^20 h^2. Similarly total energy for (n − 1)th orbit will be, E(n−1) = − meZ
(^2) e 4 8(n − 1)^2 ε^20 h^2.
Energy of photon emitted as the electron makes a transition from an orbit with quantum number n to an orbit of quantum number (n − 1) is given by, En − En− 1 = meZ
(^2) e 4 8(n − 1)^2 ε^20 h^2 −^
meZ^2 e^4 8 n^2 ε^20 h^2 ∆E = meZ
(^2) e 4 8 ε^20 h^2
[ 1
(n − 1)^2 −^
n^2
]
hf = mee
4 8 ε^20 h^2
[ 1
(n − 1)^2 −^
n^2
]
since Z = 1
f = mee
4 8 ε^20 h^3
[ 1
(n − 1)^2 −^
n^2
]
= mee
4 8 ε^20 h^3
[ (^2) n − 1 n^2 (n − 1)^2
]
Hence proved.
(b) According to Bohr’s quantization,
mevr = nℏ me × rω × r = nℏ since v = rω, Also ω = 2πf , ℏ = h/ 2 π and Bohr radius r = ε 0 n^2 h^2 /πZmee^2 , therefore me ×
( (^) ε 0 n^2 h^2 πZmee^2
× 2 πf = n × 2 hπ
f = Z
(^2) mee 4 4 ε^20 n^3 h^3 = mee
4 4 ε^20 n^3 h^3 since^ Z^ = 1. (c) One would observe a continuous radiation because the levels would be placed so close, that distinguishing of each packet of radiation from another would be impossible. No, we do not need quantum theory. The electron would still emit discrete photons but they are not detectable due to closely packed energy levels as n → ∞.
(d) Now we are given that,
Radius of electron’s orbit = r = 1 cm = 10−^2 m. We want to calculate the frequency of the photon emitted. As we have already calculated in part (b), frequency of emitted photon is given by, f = mee
4 4 ε^20 n^3 h^3.
Answer 7
(a) Consider 0K temperature (just to eliminate any discrepancies about value of temperature). At any given time (under intensity of light), there are 15 electrons in conduction level. This means that there is a dynamic equilibrium established of let’s say 20 electrons going into conduction level and 5 dropping back to valence band. Probability of an electron absorbing a photon and going to conduction band is same as that of it emitting a photon and returning to valence band from conduction band. Thus when we double the intensity there is a probability of electrons jumping back to valence band while initially there was not as there were no electrons in conduction band. So doubling intensity would result a decrease of electrons in conduction band. (b) When we manage to send 50 electrons in conduction shell, now no more electrons can jump to conduction band as that many would jump back to valence band. Hence number of electrons from valence to conduction shell is equal to number of electrons coming from conduction to valence shell.
- Laser Characteristics
Recall our laser set-up from the class with energy levels 1, 2 and 3. Level 2 is the highest and Level 3 has energy in between levels 1 and 2. Level 2 is very short lived and spontaneously goes down to level 3 with much more probability than going to level 1. Level 3 on the other hand is long lived and obviously electron there can only go to level 1 by spontaneous emission. Photons of energy E 12 = E 2 − E 1 are used to pump electrons and achieve population inversion between level 1 and 3. For each of the following changes, describe what would change in output laser charac- teristics in terms of its frequency and power and what changes would be required in the light that is being used for pumping in terms of its frequency and power.
(a) Both levels 2 and 3 are raised by same fixed amount with respect to level 1. All probabilities remaining same. (b) Level 2 remains same while level 3 is brought closer to the level 1, all probabilities remaining same.
(c) Level 3 remains same while level 2 is raised up with respect to level 1, all proba- bilities remaining same. (d) All levels stay at the same level but the probability of level 2 going to level 3 by spontaneous emission is reduced (but still higher) compared to it going to level
(e) All else remains same but the probability of level 2 going to 1 by spontaneous emission becomes more than it going to level 3. (f) All else remains same but life time of level 3 is reduced.
Answer 8
(a) The photons of energy E 2 − E 1 = E 12 would be required to achieve population inversion. As difference increased, thus pumping photon frequency and hence power would be increased. Also as E 31 = E 3 − E 1 is increased, the output laser photons frequency and hence power is increased.
(b) The frequency and power of input light remains the same as E 12 is unchanged. The frequency and hence power of output laser light is decreased.
(c) As E 2 − E 1 = E 12 has increased, thus pumping photon’s frequency and hence power will increase. The frequency and hence power of output laser light will
energy is converted to give off a photon. What is the frequency of this photon, in terms of ω and h? (c) Now we send this newly created photon back to height h. In order to avoid the creation of energy, this photon must lose energy so that at height h it just have enough energy to create the pair of same mass again, without having any extra energy. What must be this energy loss for this to happen? What is the change in frequency in climbing up the height h? Does it agree with the formula we derived in class?
Answer 9
(a) As the photon has to be created two particles of equal mass and they have no energy due to motion implying their velocity is zero. Thus according to law of conservation of energy,
Ephoton = E+E− ℏω = m 0 c^2 + m 0 c^2 ℏω = 2m 0 c^2 ⇒ m 0 = 2 ℏcω 2 ,
where m 0 is the mass of each particle. (b) As they fall down they would gain an energy mgh, where m = m 0 =rest mass. As they were at rest, the velocity with which they fall is not relativistic.
E 1 = m 0 c^2 + mgh Etotal = 2E 1 = 2(m 0 c^2 + mgh),
is the total energy of the particle downstairs. Now at the time of annihilation,
according to law of conservation of energy Ephoton = 2m 0 c^2 + 2m 0 gh ℏω′^ = 2(c^2 + gh)m 0 = 2(c^2 + gh) 2 ℏcω 2 ω′^ = ω
1 + ghc 2
2 πf ′^ = .ω
1 + ghc 2
f ′^ = 2 ωπ
1 + ghc 2
Thus frequency of this photon is higher, because due to the attraction of gravity, energy of photon increase. (c) To create pair of same mass again, the photon must have the same energy as the old photon and as it goes back to height h. Thus energy loss will be, Energy loss = ℏω′^ − ℏω = ℏω
1 + ghc 2
− ℏω = ℏω + ℏωghc 2 − ℏω = ℏω ghc 2. Change in frequency while climbing up the height will be, ∆f = f ′^ − f = ω
′ 2 π −^
ω 2 π = (^21) π · ω
1 + ghc 2
− 2 ωπ
= 2 ωπ + 2 ωghπc 2 − 2 ωπ = 2 ωghπc 2 = f ghc 2 Yes it again agrees that formula we derived in class.
- Gravitational Red Shift
