1) The document discusses the concept of de Broglie wavelength and uses examples to calculate the wavelength of a golf ball and electron moving at given velocities. 2) It then discusses several problems involving calculating kinetic energy, momentum, and phase/group velocities of particles like protons and electrons using their de Broglie wavelengths and relativistic equations. 3) The document also summarizes the Davisson-Germer experiment that provided direct evidence of the wave nature of electrons by observing diffraction peaks matching the wavelengths calculated from de Broglie's equation.

1. Quantum Mechanics Tutorial l
Engineering Physics
Indian Institute of Information Technology, Allahabad

2. Let us consider a problem which involves the concept of de Broglie wavelength
 Find the wavelength of (a) a 46-g golf ball with a velocity 30 m/s and, (b) an electron with a velocity of
𝟏𝟏𝟏𝟏𝟕𝟕 m/s.
(a) Momentum of golf ball p = mv
Since we know that p = h/λ, therefore λ = h/mv
So now, λ =
ℎ
𝑚𝑚𝑚𝑚
=
6.63×10−34 𝐽𝐽.𝑠𝑠
(0.046 𝑘𝑘𝑘𝑘)(30 𝑚𝑚/𝑠𝑠)
= 4.8× 10−34 m
• The wavelength of the golf ball is so small compared with its dimension. Thus, we would not expect to find
its wave nature.
(b) Mass of electron m = 9.1× 10−31 kg
Now λ =
ℎ
𝑚𝑚𝑚𝑚
=
6.63×10−34 𝐽𝐽.𝑠𝑠
(9.1×10−31 𝑘𝑘𝑘𝑘)(107 𝑚𝑚/𝑠𝑠)
= 5.3× 10−11 m
• The dimension of atoms are comparable with above finding (e. g. the radius of hydrogen atom is ~5.3 ×
10−11 𝑚𝑚). It is not surprising that the wave nature of moving electron is the key to understand atomic
structure and behaviour.

3. Relativistic energy in terms of momentum
The famous Einstein mass energy equivalence relation is
And the relativistic momentum is
Using above two expressions:
Therefore,

4. Let’s look at another problem
 A proton has de Broglie wavelength 1 fm. Calculate the kinetic energy of the proton.
Rest energy of proton 𝐸𝐸𝑜𝑜 = 𝑚𝑚𝑜𝑜𝑐𝑐2 = 0.938 GeV
Where m0 is the rest mass of the proton having value 1.672× 10−27 kg
𝑝𝑝𝑝𝑝 =
ℎ𝑐𝑐
λ
=
[ 6.63×10−34 /1.6×10−19) 𝑒𝑒𝑒𝑒.𝑠𝑠](3×108 𝑚𝑚/𝑠𝑠)
1×10−15 𝑚𝑚
= 1.241× 109 eV = 1.241 GeV
Since 𝑝𝑝𝑝𝑝 > 𝐸𝐸𝑜𝑜 a relativistic calculation is required.
The expression for total energy ( including both kinetic energy and rest energy) of proton is given as-
E = 𝐸𝐸𝑜𝑜
2 + 𝑝𝑝2𝑐𝑐2 = (0.938 𝐺𝐺𝐺𝐺𝐺𝐺)2+ 1.241 𝐺𝐺𝐺𝐺𝐺𝐺 2 = 1.555GeV
Hence,
Kinetic Energy = 𝐸𝐸 − 𝐸𝐸𝑜𝑜 = (1.555-0.938) GeV = 617 MeV

5.  An electron has a de Broglie wavelength of 2× 𝟏𝟏𝟏𝟏−𝟏𝟏𝟏𝟏 m. Find its (a) kinetic energy (b) phase and
group velocity of its de Broglie wave.
(a) Rest energy of electron 𝐸𝐸𝑜𝑜 = 𝑚𝑚𝑜𝑜𝑐𝑐2 = 511 𝑘𝑘𝑘𝑘𝑘𝑘 (Rest mass of electron 𝑚𝑚𝑜𝑜 = 9.1 × 10−31 kg)
Now to calculate pc: 𝑝𝑝𝑝𝑝 =
ℎ𝑐𝑐
λ
=
(4.136×10−15 𝑒𝑒𝑒𝑒.𝑠𝑠)(3×108 𝑚𝑚/𝑠𝑠)
2×10−12 𝑚𝑚
= 620 𝑘𝑘𝑘𝑘𝑘𝑘
Since 𝑝𝑝𝑝𝑝 > 𝐸𝐸o, so we will use the relativistic approach.
And kinetic energy of electron using relativistic approach is given by:
𝐾𝐾𝐾𝐾 = 𝐸𝐸 − 𝐸𝐸𝑜𝑜 = 𝐸𝐸𝑜𝑜
2 + (𝑝𝑝𝑝𝑝)2− 𝐸𝐸o = (511 𝑘𝑘𝑘𝑘𝑘𝑘)2+ (620 𝑘𝑘𝑘𝑘𝑘𝑘)2 − 511 𝑘𝑘𝑘𝑘𝑘𝑘 = 292 𝑘𝑘𝑘𝑘𝑘𝑘
(b) The energy of electron with velocity v is of the fallowing form
𝐸𝐸 = 𝛾𝛾𝐸𝐸o where 𝛾𝛾 =
1
1−
𝑣𝑣2
𝑐𝑐2
𝐸𝐸 =
𝐸𝐸𝑜𝑜
1 − 𝑣𝑣2/𝑐𝑐2
Gives 𝑣𝑣 = 𝑐𝑐 1 −
𝐸𝐸0
2
𝐸𝐸2
= c 1 −
511 𝑘𝑘𝑘𝑘𝑘𝑘
803 𝑘𝑘𝑘𝑘𝑘𝑘
2
= 0.771 𝑐𝑐
Hence the phase and group velocities are respectively
vp =
𝑐𝑐2
𝑣𝑣
=
𝑐𝑐2
0.771𝑐𝑐
= 1.30 c And vg = 0.771c

6.  The Davisson – Germer experiment: An experiment that confirms the existence of de
Broglie waves
n = 1, θ = 650 (highest intensity observed with a 54 V) and d = 0.091 nm (spacing of crystalline planes of
nickel)
The Bragg equation for maxima in the diffraction pattern
nλ = 2d sinθ = 2(0.091 nm)(sin650) = 0.165 nm
Now we use de Broglie’s formula to find expected wavelength of the electrons i.e. 𝜆𝜆 =
ℎ
𝛾𝛾𝑚𝑚𝑚𝑚
Kinetic energy of electron KE = eV = 54 eV
We also know that K =
1
2
m𝑣𝑣2=
𝑝𝑝2
2𝑚𝑚 Gives p = 2𝑚𝑚𝑚𝑚𝑚𝑚
since KE < 0.51 MeV (rest energy of electron). So we can let 𝛾𝛾 = 1
𝜆𝜆 =
ℎ
2𝑚𝑚𝑚𝑚𝑚𝑚
=
6.63×10−34 𝐽𝐽.𝑠𝑠
2 9.1×10−31 𝑘𝑘𝑘𝑘 54𝑒𝑒𝑒𝑒 1.6×10−19 𝐽𝐽
𝑒𝑒𝑒𝑒
= 0.166 nm
Which agrees well with the observed wavelength of 0.165 nm. The Davisson - Germer experiments thus
directly verifies de Broglie hypothesis of the wave nature of moving bodies.
Measured by XRD

7.  In the Davisson-Germer Experiment, if the electron beam was accelerated by 100 volts instead of
54 volts, at which scattering angle would they have found a peak in the intensity?
First determine the wavelength of the electron. The kinetic energy of the electron accelerated
By 100 volts is 𝐾𝐾 = 100 𝑒𝑒𝑒𝑒.
Therefore, 𝜆𝜆 =
ℎ
2𝑚𝑚𝑚𝑚
=
6.63 × 10−34 𝐽𝐽 ⋅ 𝑠𝑠
2 × (9.1 × 10−31𝑘𝑘𝑘𝑘) × (100 𝑒𝑒𝑒𝑒)
= 0.123 𝑛𝑛𝑛𝑛
The spacing between two crystalline planes in nickel is 𝑑𝑑 = 0.091 𝑛𝑛𝑛𝑛.
From the Bragg diffraction formula,
𝜃𝜃 = sin−1
𝜆𝜆
2𝑑𝑑
= sin−1
0.123
2 × 0.091
= sin−1(0.676) = 42.53°
Therefore, the scattering angle is
𝜙𝜙 = 180° − 2𝜃𝜃 ≈ 95°