W AVE N ATURE OF M ATERIAL P ARTICLES
2.8 HEISENBERG’S UNCERTAINTY PRINCIPLE OR THE PRINCIPLE OF INDETERMINACYINDETERMINACY
The dual nature of matter and radiation requires profound changes in our concepts built on the basis of common sense and everyday experience. The formulation of classical mechanics implies that the position and momentum of a particle are assumed to have well defined values and can be determined simultaneously with perfect accuracy. But the wave particle duality compels us to abandon the idea of simultaneous determination of position and momentum with perfect accuracy. In 1927 Werner Heisenberg, a German physicist, enunciated that it is impossible to determine both position and momentum simultaneously with perfect accuracy. If ,x is the uncertainty in position and ,px is the uncertainty in the corresponding momentum then
Dpx. Dx ≥h ...(2.8.1)
Similarly if DE is the uncertainty in energy and ,t is uncertainty in time then
DE . Dt ≥h ...(2.8.2)
From Eqn. (2.8.1) it is evident that if we try to measure the position of particle with utmost accuracy i.e., Dx ® 0, the corresponding uncertainty in momentum becomes very large i.e., Dpx ® ¥ and vice versa.
Let us illustrate the above assertion. Consider a particle having well-defined momentum px (= hk). Such a particle has well defined k or l and is represented by a sinusoidal (monochromatic wave). A monochromatic wave has no beginning and end i.e., it is infinitely long; its amplitude is constant for all values space coordinates x and therefore the particle may be anywhere between x = – ¥ to + ¥. Thus the position of the particle is completely uncertain (Dx ® ¥).
Now consider a particle having well-defined position (Dx ® 0). A wave packet having very small extension in space describes such a particle. Fourier’s transform of this wave packet shows that it is formed by superposition of a very large number of waves having continuous distribution of k or l within a large range of Dk. Thus the uncertainty in k or p is very large (Dk ® ¥).
Fig. 2.8.1 A particle with well-defined momentum p is described by a sinusoidal wave extending from x = – ¥ to + ¥. Here Dp ® 0 but Dx ® ¥
Thus a particle with relatively small uncertainty in momentum has large uncertainty in position.
A sinusoidal wave has well defined frequency and so is its energy (E = hw). A particle described this wave also has well-defined energy E and therefore DE = 0. In order to see the constancy of amplitude of such a sinusoidal wave, which exists from t = – ¥ to + ¥, we have to look for a very long time. Therefore, the uncertainty in time is infinite (Dt ® ¥).
Fig. 2.8.2 Some wave packets and their Fourier’s transforms, Dx Dk≅1
Consider a particle, which is described by a wave packet as shown in the Fig. (2.8.2). The Fourier’s transform of the wave packet is also shown adjacent to it. Let Dx be the spread of the wave packet in space and Dk the spread in propagation constant. It can be shown by standard mathematical technique that
Dx Dk ³ 1 ...(2.8.3)
Since p = hk and Dp = h Dk we have
Dpx Dx ³h ...(2.8.4)
which is the uncertainty relation.
It should be carefully noted that the uncertainties in measurement of position and momentum are not because of inadequacies in our measuring instruments. Even with ideal instrument we can never in principle do better. This principle is the fundamental law of nature. The indeterminism is inherent in the very structure of matter. The momentum and position don’t assume well-defined values simultaneously.
Notice that it is the smallness of Planck’s constant that makes the uncertainty principle insignificant in macroscopic world. In microscopic world the consequences of uncertainty principle cannot be ignored.
S OLVED E XAMPLES
Ex. 1. Show that the wavelength of electron accelerated through a potential difference V is given by
( )
h 12.3
Å.
2m eV V volt
= =
λ
Sol. The kinetic energy of electron T =
2
2 p
m = eV and p= 2 Tm = 2meV
Therefore 2 T 2 eV
h h h
p m m
λ = = =
Substituting m = 9.1 × 10–31 kg, e = 1.6 × 10–19 C, h = 6.6 × 10–34 Js, we have
12.3 10 12.3
10 m
V V
λ = × − = Å
For other charged particles appropriate values of m and charge q should be substituted in the above equation.
Ex. 2. Obtain expression for the wavelength of a particle moving with relativistic speed.
Sol. The relativistic momentum of a particle
0
2 2
1 /
= −
p m v
v c
\
( ) ( )
( )
1/ 2 1/ 2
2 2 2 2
0 0
1 / 1 /
/
h v c v c
h h
p m v m c v c
− −
λ = = =
The momentum p of a relativistic particle can also be expressed as follows.
E 2 = p2c2 + (m0c2)2 = (T + m0c2)2
T T( 2m c0 2)
p c
= +
Hence l = h/p
= T T( +hc2m c0 2) = 2mh0T 1 T/2+ 1 m c0 2
=
1/ 2 2
0 0
1 T
2 T 2
h
m m c
−
+
If the particle under consideration is an electron accelerated through a potential difference of V volt, its de Broglie wavelength is given by
l =
1/ 2 2
0 0
1 eV
2 V 2
h
m e m c
−
+
Ex. 3. Find the de Broglie wavelength of (i) electron moving with velocity 1000 m/s (ii) an object of mass 100 gram moving with the same velocity.
Sol. (i) de Broglie wavelength of electron
34 31
6.63 10 Js (9.1 10 kg)(1000 m/s) h
mv
−
−
λ = = × =
× 7285 × 10–10 m
= 7285 Å (ii) de Broglie wavelength of object
34
6.63 10 Js 36
8.63 10 m (0.1 kg)(1000 m/s)
h mv
− −
λ = = × = ×
Owing to extremely short wavelength of the object its wave behavior cannot be demonstrated.
Ex. 4. Find the de Broglie wavelength of electron, proton and a-particle all having the same kinetic energy of 100 eV.
Sol. For electron
34
31 19
6.63 10 Js
2 T 2 (9.1 10 kg)(100 1.6 10 J) h
m
−
− −
λ = = ×
× × × ×
= 1.23 × 10 –10 m = 1.23 Å For proton m = 1.67 × 10–27 kg,
\ l = 0.028 Å
For a-particle m = 4 × 1.67 × 10– 27 kg
\ l = 0.014 Å
Ex. 5. At what kinetic energy would an electron have the wavelength equal to that of yellow spectral line of sodium, l = 5896 Å ?
Sol.Since
2 T h λ = m
\
2
T 2
2 h
= m λ
Substituting h = 6.63 × 10–34 Js, m = 9.1 × 10–31 kg, l = 5896 × 10–10 m, we have T = 6.93 × 10–25 J = 4.3 × 10–6 eV
Ex. 6. What is the wavelength of thermal neutron at 300 K?
Sol. Kinetic energy of thermal neutron E =
23
3 3 1.38 10 J/K 300K 21
T 6.2 10 J
2k 2
− −
× × ×
= = ×
\ 34
27 21
6.63 10 Js
1.45 2 E 2 1.67 10 kg 6.2 10 J
h m
−
− −
λ = = × =
× × × × Å
Ex. 7. Find the de Broglie wavelength of hydrogen molecules, which corresponds to their most probable speed at room temperature 27°C.
Sol. The most probable speed of hydrogen molecule at temperature T is
2 Tk v= m
Momentum of molecule p = mv = 2mkT
\
2 T
h h
p mk
λ = =
10
27 23
6.63 10 Js
2 (3.34 10 kg)(1.38 10 J/K)(300K)
−
− −
= ×
× × ×
= 1.26 × 10–10 m = 1.26 Å.
Ex. 8. At what value of kinetic energy is the de Broglie wavelength of an electron equal to its Compton wavelength?
Sol. Energy of electron E = T + m0c2 = ( )pc 2+(m c0 2 2)
\
2 2
T 2m c0 T
p c
= +
de Broglie wavelength of electron
2 2
T 2 0 T
h hc
p m c
λ = = + Given that
0
λ = λ =c h m c
2 2
0 T 2 0 T
h hc
m c m c
= +
2 2
0 0
T= −m c ± 2m c (– sign is meaningless) T=m c0 2( 2− =1) (0.51 MeV)(0.414)=0.21 MeV.
Ex. 9. Find the de Broglie wavelength of relativistic electrons reaching the anticathode of an X-ray tube if the short wavelength limit of continuous X-ray spectrum is equal to 0.10 Å.
Sol. Short wavelength limit of X-ray spectrum
0
0
eV eV
hc hc
λ = ∴ =
λ Kinetic energy of electron T = eV = hc/l0 Momentum of electron
2
T(T 2m c0 )
p c
= +
de Broglie wavelength of electron
2
T(T 2 0 )
h hc
p m c
λ = =
+
0 2
2 0 0
0
0 0
(2 )
2 1
hc
m c hc hc
m c hc
λ = = λ
+ + λ
λ λ
For electron m0 c2 = 0.51 MeV, hc = 0.0124MeV Å.
0.10 A
0.033A.
1.02 0.10 1 0.0124
λ = =
+ ×
o o
Ex. 10. Find the de Broglie wavelength of electron traveling along the first Bohr orbit in hydrogen atom.
Sol. The angular momentum of electron in the first Bohr orbit 2
mvr=nh π
\ 2 mv h
= r π
de Broglie wavelength of electron h 2
mv r
λ = = π =2 × 3.14 × 0.53 Å = 3.3 Å.
Ex. 11. Describe the Bohr’s quantum condition in terms of de Broglie wave.
Sol. A stationary Bohr orbit must accommodate whole number of de Broglie wavelengths. If r is the radius of electron orbit then
2pr = nl ...(1)
According to de Broglie hypothesis l = h
mv ...(2)
Eliminating l from these equations we have 2 n h
r m v π = or
2 mvr=nh
π which is the Bohr’s quantum condition.
Ex. 12. An object has a speed of 10000 m/s accurate to 0.01%. With what fundamental accuracy can we locate its position if the object is (a) a bullet of mass of 0.05kg (b) an electron?
Sol. Momentum of bullet p = mv = (0.05 kg)(1000 m/s) = 50 kg m/s Uncertainty in momentum Dp = 50 × 0.0001= 5 × 10–3 kg m/s Minimum uncertainty in position
34
31 3
1.054 10 Js
2.1 10 m 5 10 kg m/s
x p
− −
−
∆ = = × = ×
∆ ×
h Momentum of electron
p = mv = (9.1 × 10–31 kg)(1000 m/s) = 9.1 × 10–28 kg m/s Uncertainty in momentum
Dp = 9.1 × 10–28 × 0.0001= 9.1 × 10–32 kg m/s Uncertainty in position
34 32
1.054 10 Js
0.115 m 9.1 10 kg m/s
x p
−
−
∆ = = × =
∆ ×
h
The uncertainty in bullet’s position is so small that it is far beyond the possibility of measurement.
Thus, we see that for macroscopic objects like bullet, the uncertainty principle practically sets no limits to the measurement of conjugate dynamic variables position and momentum. For electron, the uncertainty in its position is very large, nearly 107 times the dimensions of atom. Thus for microscopic objects such as electrons, the uncertainty in their position is significant and cannot be overlooked.
Ex. 13. The position and momentum of 1 keV electrons are measured simultaneously. If its position is located within 1Å, what is the percentage uncertainty in its momentum? Is this consistent with the binding energy of electrons in atoms?
Sol. The uncertainty in position of electron
34
24 10
1.054 10 Js
1.054 10 kg m/s
10 m
p x
− −
−
∆ ≥ = × = ×
∆h
The momentum of electron inside the atom is at least equal to p = 1.054 × 10–24 kg m/s. The corresponding kinetic energy is
− −
−
= = × = × =
× ×
2 24 2
17 31
(1.054 10 kg m/s)
T 0.061 10 J 3.8 eV
2 2 9.1 10 kg
p m
The ionization potential of atoms is of this order and hence the uncertainty in momentum is consistence with the binding energy of electrons in atoms.
Ex. 14. Imagine an electron to be somewhere in the nucleus whose dimension is 10–14 m. What is the uncertainty in momentum? Is this consistent with the binding energy of nuclear constituents?
Sol. If an electron were in the nucleus, its momentum would be uncertain by amount Dp given by
34
20 14
1.054 10 Js
1.054 10 kg m/s 10 m
p x
− −
−
∆ ≥ = × = ×
∆ h
The momentum itself must be at least equal to p = 1.54 × 10–20 kg m/s. The corresponding kinetic energy of electron is many times greater than the rest energy m0c2 of electron and therefore the kinetic energy of electron may be taken equal to pc.
T = pc = (1.054 × 10–20 kg m/s)(3 × 108 m/s) = 3.3 × 10–12 J
= 20 MeV
Experiments show that energy of electrons in nuclear disintegration (b decay) is very much less than 20 MeV. Hence the uncertainty principle rules out the possibility of electrons being a nuclear constituent.
Ex. 15. Consider a proton or neutron to be inside the nucleus. What is the uncertainty in momentum of electron? Is this consistent with the binding energy of nuclear constituents?
Sol. If a proton or neutron were inside the nucleus, the uncertainty in momentum would be
34
20 14
1.054 10 Js
1.054 10 kg m/s 10 m
p x
− −
−
∆ ≥ = × = ×
∆ h
The corresponding kinetic energy T << m0c2 . Hence
2 20 2
14 27
(1.054 10 kg m/s)
T 3.6 10 J 0.23MeV
2 2 1.67 10 kg
p m
− −
−
= = × = × =
× ×
The binding energies of nuclei are of this order.
Ex. 16. The energy of a harmonic oscillator is given by
2
1 2
2 2
= p +
E kx
m
where p is momentum and k is force constant. Using uncertainty principle, show that the minimum energy of the oscillator is DM0, where ω0= k/m.
Sol. According to uncertainty principle, ∆ ≥
∆h
p x. The momentum of oscillator is at least equal to p where p=h/x . The energy of oscillator may be written as
2
2 2
E 1
2 2kx
= mxh +
...(1)
For minimum energy
2
2 3
E 0 kx x
x mx mk
∂ = = − + ⇒ =
∂
h h
Substituting the value of x2 in (1), we get E=h k m/ = ωh 0.
Ex. 17. A nucleus exists in excited state about 10–12 sec. What is uncertainty in energy of the gamma ray photon emitted by the nucleus?
Sol. The minimum uncertainty in energy is at least equal to DE, where DE.Dt = h
Therefore
DE =
34
22 12
1.054 10 Js
1.054 10 J.
10 s t
− −
= × = ×
∆ h
Ex. 18. The average excited atom has a life-time of about 10–8 sec. During this period it emits a photon. What is the minimum uncertainty in the frequency of photon?
Sol. According to uncertainty principle DEDt ³h hDwDt ³ h
Minimum uncertainty in frequency of photon
1 8
10 rad/s.
∆ω = t =
∆
Ex. 19. Making use of uncertainty principle give an estimate of radius and binding energy of electron in hydrogen atom in the ground state.
Sol. Energy of electron
2 2
0
E 2 4
p e
m r
= + −
πε ...(1)
Assuming that the uncertainties in momentum and energy are equal to the momentum and energy themselves, we can write the uncertainty principle as
pr=h ...(2)
Eliminating p from these equations, we get
2 2
2 0
E 2 4
e mr r
= −
πε
h ...(3)
In the ground state, energy is minimum. Therefore,
2 2
3 2
0
E 2
0
2 4
e
r mr r
∂ = − − =
∂ πε
h
This gives
2
10
0 2
4 0.529 10 m
r
me
= πε h = × −
...(4) This value of r, that is ground state radius of hydrogen atom, is called Bohr radius (a0).
Substituting the value of r in (3) we get ground state energy of atom
2 4
0 2
1 1
E .
2 4
me
= − πε h ...(5)
Q UESTIONS AND P ROBLEMS
1. Give reasoning that led de Broglie to speculate the wave behavior of material particles. Derive de Broglie relation.
2. Show that the wavelength of an electron beam accelerated through a potential difference of V volt is 12.3 A.
V(volt)
λ = o
.
3. Describe Davisson-Germer experiment and interpret its results.
4. State and explain Heisenberg uncertainty principle. Use this principle to show that (i) electrons cannot reside inside the nucleus (ii) radius of the first orbit of hydrogen atom is
2
0 2
4 .
r
= πε meh
5. Show that the principle of indeterminacy can be expressed as ∆ ∆θ ≥L. h where DL is uncertainty in angular momentum and Dq is uncertainty in angular position of the particle under investigation.
6. (a) An electron beam of energy 100 eV is passed through a circular hole of radius 5 ´ 10–4 cm. What is the uncertainty introduced in the angle of emergence? (b) A lead ball of mass 200 gram is passed through a hole of radius 25 cm, calculate the uncertainty in the angle of emergence. The velocity of ball is 20 m/s.
7. According to uncertainty principle. A particle of momentum p cannot be confined by a central force to a circle of radius r less than h/p. Assume that pr = h, show that the total energy of electron in hydrogen atom is
2 2
2 0
E .
2 4
e mr r
= −
πε
h Show that for minimum value E, r has the value
2 0
0 2
4 0.53
r me
= πε h =
Å Bohr radius.
8. The accuracy in measurement of wavelength of photon is one part per million (i.e., Dl/l = 10–6). What is the minimum uncertainty Dx in a simultaneous measurement of the position of the photon in the case of (a) a photon with l = 6000 Å and (b) an x-ray photon with l = 5 Å.
9. The atoms in a solid posses a certain minimum zero-point energy even at 0 K. Using uncertainty principle, explain this statement.
S CHRệDINGER E QUATION