Nanoelectronics Workshop 1
17:00-18:00 on Friday, 20/01/2017 (V 123)
To be handed in by 12:00 on Friday, 03/02/2017 to the General Office
Note: 12.5% of Final Mark (50% from Workshops and 50% from Final Examination)
Question 1. Calculate the de Broglie wavelength for the following cases:
(1) A human of mass 60 kg walking at 1.0 m/s
(2) A bullet of mass 10 g flying at 103 m/s
(3) A free electron in a metal of speed 108 m/s
(4) A Helium atom at 1 K
Answer:
De Broglie wavelength is defined as λ =
€
€
€
h
h
(Planck constant: h = 6.625 ´ 10 -34 [J×s])
=
p mv
(1) λ =
6.625 × 10−34€[ J ⋅ s]
h
=
= 1.104  × 10−35 [ m] ≈ 1.1 × 10−35 [ m]
mv 60 [ kg] × 1.0 [ m/s]
(2) λ =
6.625 × 10−34 [ J ⋅ s]
h
=
= 6.625 × 10−35 [ m] ≈ 6.6 × 10−35 [ m]
mv 10 × 10−3 [ kg] × 10 3 [ m/s]
(3) λ =
6.625 × 10−34 [ J ⋅ s]
h
=
= 7.273 × 10−12 [ m] ≈ 7.3 × 10−12 [ m]
mv 9.10938 × 10 -31 [ kg] × 10 8 [ m/s]
(4) A kinetic energy of a Helium atom: E =
1
1
2
mv 2 =
( mv) , which leads mv = 2mE . At temperature
2
2m
3
T, the kinetic energy equals to k BT (Boltzmann constant: kB = €1.38065 ´ 10 -23 [kg×m 2/s 2×K]). For a
€ 2
Helium atom (m = 6.64 ´ 10 -27 [kg]), the de Broglie wavelength is calculated to be
λ=
h €
=
mv
h
=
6.625 × 10 −34 [ J⋅ s]
3
3 × 6.64 × 10 −27 [kg] × 1.38065 × 10 −23 [kg⋅ m2 /s2 ⋅ K] × 1 [K]
2m × k BT
2
−9
= 1.263 × 10 [m] ≈ 1.3 × 10 −9 [m]
€
Question 2. (1) In order to observe a crystal, of which atomic separation is 1 Å, estimate energy of the
following waves; (a) X-ray, (b) electron-beam and (c) neutron-beam.
(2) In order to investigate atomic core structures of 10-15 m, estimate energy of the following waves; (a)
g-ray, (b) electron-beam and (c) proton-beam.
Note you do not need to consider the theory of relativity for the above calculations.
Answer:
For observation, de Broglie wavelength needs to be in the same order with sample structures.
(1) (a) X-ray energy is calculated to be
E=
=
−34
[ J ⋅s] × 2.9979 × 10 8 [ m/s] = 1.988× 10−15 J
hc 6.625× 10
=
[]
λ
10 -10 [ m]
1.988× 10−15 [ J ]
1.6022× 10−19 [ J/eV]
= 1.240× 10 4 [ eV] ≈ 1.2 × 10 4 [ eV]
€
h
1
1
2
(b) A kinetic energy: E = mv 2 =
( mv) , which leads mv = 2mE . Hence, λ = =
mv
2
2m
Electron-beam energy (electron mass: 9.1094 ´ 10 -31 [kg]) is calculated to be
€
2
€
−34
€
2
6.625×
10
J
⋅s
[ ]
h
−17
E=
=
=
2.409×
10
[J]
2mλ2 2 × 9.1094 × 10 -31 [ kg] × 10 -10 [ m] 2
(
)
(
=
2.409× 10−17 [ J ]
1.6022× 10−19 [ J/eV]
)
= 1.504× 10 2 [ eV] ≈ 1.5× 10 2 [ eV]
(c)€Similarly, neutron-beam energy (neutron mass: 1.6749 ´ 10 -27 [kg]) is
(
)
2
6.625× 10−34 [ J ⋅s]
h2
E=
=
2mλ2 2 × 1.6749 × 10 -27 [ kg] × 10 -10 [ m]
(
=
1.310× 10−20 [ J ]
1.6022× 10−19 [ J/eV]
(2)€(a) g-ray energy is
)
2
= 1.310× 10−20 [ J ]
= 8.178× 10−2 [ eV] ≈ 8.2 × 10−2 [ eV]
h
.
2mE
E=
=
−34
[ J ⋅s] × 2.9979 × 10 8 [ m/s] = 1.988× 10−10 J
hc 6.625× 10
=
[]
λ
10 -15 [ m]
1.988× 10−10 [ J ]
1.6022× 10−19 [ J/eV]
= 1.240× 10 9 [ eV] ≈ 1.2 × 10 9 [ eV]
(b)€Electron-beam energy (electron mass: 9.1094 ´ 10 -31 [kg]) is calculated to be
(
)
2
6.625× 10−34 [ J ⋅s]
h2
E=
=
2mλ2 2 × 9.1094 × 10 -31 [ kg] × 10 -15 [ m]
(
=
2.409× 10−7 [ J ]
1.6022× 10−19 [ J/eV]
)
2
= 2.409× 10−7 [ J ]
= 1.504× 1012 [ eV] ≈ 1.5× 1012 [ eV]
N.B. For precise calculations, the theory of relativity needs to be taken into account, resulting
E€=
=
h 2c 2
+ m 2c 4 − mc 2
λ2
(6.625 × 10
−34
[ J ⋅ s])
(10
2
(
)
× 2.9979 × 10 8 [ m/s]
[ m])
-15
(
2
− 9.1094 × 10 -31 [ kg] × 2.9979 × 10 8 [ m/s]
)
2
(
2
) (
2
= 1.985 × 10−10 [ J ]
=
€
1.985 × 10−10 [ J ]
1.6022 × 10
−19
[ J/eV]
= 1.239 × 10 9 [ eV] ≈ 1.2 × 10 9 [ eV]
(c) Similar to (b), proton-beam energy (proton mass: 1.67262 ´ 10 -27 [kg]) is calculated to be
2
(6.625 × 10−34 [J⋅ s])
h2
E=
=
= 1.314 × 10 −10 [ J]
2mλ2 2 × 1.67262 × 10 -27 [kg] × (10 -15 [m]) 2
=
1.314 × 10 −10 [ J]
= 8.20 × 10 8 [eV] ≈ 8.2 × 10 8 [eV]
1.6022 × 10 −19 [ J/eV]
N.B. For precise calculations,
€
)
+ 9.1094 × 10 -31 [ kg] × 2.9979 × 10 8 [ m/s]
4
E=
h 2c 2
+ m 2c 4 − mc 2
λ2
(6.625× 10
−34
=
[ J ⋅s])
(10
2
(
)
× 2.9979 × 10 8 [ m/s]
-15
[ m])
(
2
− 1.67262 × 10 -27 [ kg] × 2.9979 × 10 8 [ m/s]
)
2
(
2
= 9.876× 10−11 [ J ]
=
€
9.876× 10−11 [ J ]
1.6022× 10
−19
[ J/eV]
2
) (
)
+ 1.67262 × 10 -27 [ kg] × 2.9979 × 10 8 [ m/s]
= 6.164× 10 8 [ eV] ≈ 6.2 × 10 8 [ eV]
4
Question 3. In a Rutherford Hydrogen model, derive quantum energy levels by assuming angular
momentum of an electron is quantised as integer multiples of the Planck constant ( n; n = 1,2,3, ).
Answer:
€
Since angular momentum of an electron is quantised as integer multiples of the Planck constant,
mvr = n
( n = 1,2,3,) ,
where m : the electron mass, v : electron speed and r : radius of an electron orbit.
The
€ total energy of an electron in the Rutherford Hydrogen model is obtained from
E = ( kinetic energy) + ( potential energy)
=
,
1
1 q2
mv 2 −
2
4 πε 0 r
where the potential energy is derived from the Coulomb’s law with using permittivity: e0 and the electron
€
charge:
q.
Along the radius direction, the balance of force is
m
v2
q2
=
r
4 πε 0 r 2
(rotational motion: F = m
v2
)
r
By comparing the above two equations, (kinetic energy) = 2 ´ (potential energy).
€
1 q2
1
∴E = −
= − mv 2
2 4 πε 0 r
2
∴mv =
€
q2
4 πε 0 rv
q2
Therefore, mvr =
. By comparing with mvr = n ,
4 πε 0 v
€
v=
€
q2
4 πε 0 n
€
By substituting this speed into the total energy,
€
2
1 2
1 & q2 )
E = − mv = − m⋅ (
+
2
2 ' 4 πε 0 n *
=−
€
N.B. From mvr =
mq 4
32π 2ε 0 2 n 2  2
q2
4 πε 0 2 2
4 πε 0 2
n
and mvr = n , r =
.
Here,
is the Bohr radius of ~ 0.53 Å.
4 πε 0 v
mq 2
mq 2
€
€
€
€