The Physics
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Opus in profectus

# Matter Waves

## Practice

### practice problem 1

The Davisson-Germer experiment that first demonstrated the wave nature of matter used electrons accelerated to 54 V.
1. Determine the energy of the electrons in…
1. electronvolts
2. joules
2. How fast were the electrons moving in this experiment?
3. What momentum did these electrons have?
4. Determine the wavelength of these electrons.
5. What form of electromagnetic radiation has the same wavelength as the value you calculated in part d.?

#### solution

1. When electrons are accelerated through a potential difference of 54 V they acquire 54 eV of energy. To convert this to joules, multiply by the elementary charge.

(54 eV)(1.60 × 10−19 C) = 8.64 × 10−18 J

2. Rearrange the kinetic energy equation to compute the speed of the electrons.

v =
 2K m

v =  2(8.64 × 10−18 J)
9.11 × 10−31 kg
v = 4,360,000 m/s

3. Use the definition of momentum to compute the momentum of an electron.

 p = mvp = (9.11 × 10−31 kg)(4,360,000 m/s)p = 3.97 × 10−24 kgm/s
4. Use de Broglie's equation to find the wavelength of an electron.

λ =
 h p

λ =  6.63 × 10−34 Js
3.97 × 10−24 kgm/s
λ = 1.67 × 10−10 m

5. This is comparable to an x-ray photon.

### practice problem 2

Repeat the Davisson-Germer experiment using a pitching machine that launches 145 g baseballs with a speed of 100 m/s at two doors separated by one meter. Determine…
1. the momentum of the baseballs
2. the de Broglie wavelength of the baseballs
3. the distance in light years to the "screen" that would result in "bright" interference fringes separated by a distance greater than the diameter of one baseball (7.4 cm)

#### solution

1. Use the definition of momentum to compute the momentum of a baseball.

 p = mvp = (0.145 kg)(100 m/s)p = 14.5 kgm/s
2. Use de Broglie's equation to find the wavelength of a baseball.

λ =
 h p

λ =  6.63 × 10−34 Js
14.5 kgm/s
λ = 4.57 × 10−35 m

3. There's an equation for determining the separation between interference fringes. Actually there are two — one that always works and one that is approximately correct when angles are small. With this ridiculously small wavelength, the small angle approximation equation is the one to use.

 λ ≈ x d L

where…

 λ = wavelength d = separation between sources x = separation between fringes L = distance to the screen

Substitute

 4.57 × 10−35 m ≈ 0.074 m 1 m L

Solve

L = 1.62 × 1033 m

Convert to light years by dividing by the conversion factor (9.46 × 1015 m).

L = 1.71 × 1017 light years

The observable universe is about 93 billion light years in diameter or 1011 to the nearest order of magnitude. Since the observable universe is a million (106) times too small and the observable universe is the biggest thing we have access to, the wave nature of baseballs will probably never be observed.

### practice problem 3

As a first approximation, an electron in a hydrogen atom can be said to orbit the nucleus at a distance of 0.0529 nm. Determine the following quantities for an electron that just fits in the circumference of this orbit…
1. its de Broglie wavelength in meters
2. its momentum in kgm/s
3. its kinetic energy in electronvolts
4. its electric potential energy in electronvolts (watch the sign)
5. its total energy in electronvolts

#### solution

1. In the Bohr model of the atom, a ground state electron is the simplest possible standing wave — one wavelength. The electron exists as a standing wave wrapped around the nucleus — the circumference.

 λ = 2πrλ = 2π(0.0529 × 10−9 m)λ = 3.32 × 10−10 m
2. Use de Broglie's equation to compute the electron's momentum.

p =
 h λ

p =  6.63 × 10−34 Js
3.14 × 10−9 m
p = 1.99 × 10−24 kgm/s

3. There's an equation that connects kinetic energy to momentum. Divide by the elementary charge at the end to convert from joules to electronvolts.

K =
 p2 2m

K =  (1.99 × 10−24 m)2
2(9.11 × 10−31 kg)
K =  2.18 × 10−18 J

K = 13.6 eV

4. A hydrogen atom consists of a lone electron orbiting a lone proton. Use the equation for the electric potential surrounding a point elementary charge. The electric potential we compute in volts is then the electric potential energy of the electron in electronvolts. Since the electron is trapped in the electric potential well of the proton, it has a negative potential energy.

V =
 kq r

V =  (8.99 × 109 Nm2/C2)(1.60 × 10−19 C)
0.0529 × 10−9 m
V =  27.2 V

U = −27.2 eV

5. Combine kinetic and potential energies to get total energy. This is the ground state energy of an electron in a hydrogen atom. More on this in the next section of this book.

 E = K + UE = (13.6 eV) + (−27.2 eV)E = −13.6 eV

### practice problem 4

Write something completely different.