PHYSICS NOTES ONLINE: WAVE NATURE OF MATTER

WAVE NATURE OF MATTER

Wave nature of matter means that matter behaves like wave and exhibits the characteristic or properties of waves. Wave nature of matter implies that matter undergo reflection, interference, refraction, diffraction and other properties of waves.

EXAMPLES OF MATTER (PARTICLES) THAT HAVE WAVE NATURE

( i) electron (ii ) x – ryas ( iii ) photons ( iv ) light energy

Albert Einstein in his explanation of photoelectric effect assumed that the energy of light beam travel through space in packets or bundle called photons.

Louis de Broglie, a French scientist suggested that light energy behaves like a wave because it exhibits reflection, refraction and diffraction, which are features of waves. He later stated that the wavelength of the wave along with particles were given as ʎ = h /ƿ = h / mv ( ƿ = mv ). Where ʎ is wavelength, h is Planck constant = 6.63 x 10⁻³⁴ Js,
v is velocity and m is mass of electron = 9.1x10 ⁻³¹ kg.
He said that electron has wave particles , x – rays.

Compton, with experiment showed that the conservation of momentum and energy were obeyed when x –rays were incidented on an electron+

Drag’s Law of diffraction of x-rays

Drag law of diffraction of x- rays state that when a monochromatic x-ray impinging upon the atom in a crystal lattice, each atom acts as a source of scattered radiation of the same wavelength, the crystal act as a series of parallel reflecting planes.

Therefore, x –rays would be reflected by ordinary laws of reflection and the planes will reflect x – rays at all angles.

DUALITY OF MATTER

Wave duality of matter simply refers to as the behavior of matter, like a wave and like a particle. The wave particle duality of matter is a general nature of matter. Matter has the properties of wave and the properties of particles.

FORMULA FOR SOLVING PROBLEM IN WAVE-PARTICLE OF MATTER

Since matter has the properties of wave and particle, the Einstein energy equation for particle and light are combined to obtain the formula for solving problem in wave-particle nature of matter. Thus:

Particle enegy equation is E = mc² … equation 1, and

wave/ light energy equation is E = hf … equation 2.

Let us substitute equation 1 for equation 2. Therefore,

E = mc² = hf → mc² = hf. Recall that f = v / λ. Note that c = v. then

mc² = hf → mc² = h * v / λ → mc = h / λ.

Where c = v is the velocity, h is planck constant, λ is wavelength and m is mass.

WORKED EXAMPLES

*What is the wavelength of an electron moving at the speed of 10⁶ m/s if the mass of the electron is

1 x 10 ³⁰ kg, e = 1.6 x 10 ̄ ¹⁹C, c = 8.0 x 10⁸ m/square second, h = 6.6 x 10 ̄ ³⁴ Js

Solution:

Given from the question: Me =1 x 10 ̄ ³⁰ kg, e = 1.6 x 10 ̄ ¹⁹C,
Speed c of light = 8.0 x 10⁸ m/sq. second

Speed of electron c = 10⁶ m/s.

Formula: mc = h / λ. Let us substitute for the variables, then

1 x 10 ̄ ³⁰ * 10⁶ = 6.6 x 10 ̄ ³⁴/ λ. Let us make λ the subject of the formula. Then

λ = 6.6 x 10 ̄ ³⁴ / 1 x 10 ̄ ³⁰ * 10⁶ → λ = 6.6 x 10 ̄ ³⁴ / 1 x 10 ̄ ²⁴

λ = 6.6 x 10 ̄ ¹⁰ m

*what is the energy of an electron with a wavelength of 10⁻¹² m. calculate the potential difference of the electron before it could be accelerated to acquire that velocity?

e = 1.6 x 10 ̄ ¹⁹C, Me = 9.1 x 10 ̄ ³¹kg, h = 6.6 x 10 ̄ ³⁴ Js,

Solution:

Extract data given in the question:

λ = 10⁻¹² m, e = 1.6 x 10 ̄ ¹⁹C, Me = 9.1 x 10 ̄ ³¹kg, h = 6.6 x 10 ̄ ³⁴ Js.

Formula for calculation: mc² = hf. Where f = v / λ, and c = v.
Therefore,

mc² = h * v / λ. → mc = h / λ.

Note, we are to calculate the potential difference of the electron. The formula that contains the potential difference is the electron volt formula which is eV. Recall that electron volt is defined as energy of electron when the electron is under the influence of a p.d of 1volt.

Therefore, eV = K. E → eV = ½mv².

From the question, v is not given. So we must first calculate v using the formula, mc² = hf, before we calculate the potential difference.

Therefore,

Mc² = hf → mc² = h * v / λ. (f = v / λ). Let us substitute for the variables,

mc² = h * v / λ. → Mc = h / λ.

9.1 x 10 ̄ ³¹ * c = 6.6 x 10 ̄ ³⁴ / 10⁻¹². (Let us make c the subject of the formula).

C = 6.6 x 10 ̄ ³⁴ / 10⁻¹² * 9.1 x 10 ̄ ³¹ → C = 6.6 x 10 ̄ ³⁴ / 9.1 x 10 ̄ ⁴³

C = 0.7253 x 10 ⁹ m/s.

Now that we have calculate velocity of the electron, we will now be able to calculate the potential difference using the formula,

eV = K.E → eV = ½mv². Let us substitute for the variables.

1.6 x 10 ̄ ¹⁹ * V = ½ * 9.1 x 10 ̄ ³¹ *(0.7253 x 10 ⁹)². Let us make V the subject of the formula

V = (½ * 9.1 x 10 ̄ ³¹ *(0.7253 x 10 ⁹)²) / 1.6 x 10 ̄ ¹⁹. →

V = ( 4.55 x 10 ̄ ³¹ * ( 0.5260 x 10¹⁸ )) / 1.6 x 10 ̄ ¹⁹
V = 2.393 x 10 ̄ ¹³ / 1.6 x 10 ̄ ¹⁹

Potential difference, V = 1.495 x 10 ⁶ volt

UNCERTAINTY PRINCIPLE

Heisenberg’s uncertainty principle states that there is always an uncertainty in the determination or measurement of the position and momentum of an electron at the same time. There must be a deviation in the actual value compared with the true value.

In other words, he states that it is practically impossible to obtain accurate result in determining or measuring the position and momentum of an electron at the same time.

FORMULAE

H e assumed that if change in distance and velocity are the u8ncertainty of a distance and velocity o fan atomic scale. Then,

∆x . ∆v > h / m. where m is mass of particle. Therefore

∆p . ∆x > h ( m∆v = ∆p ) or ∆E . ∆x ≥ h, ∆p. ∆x ≥ h, ∆. ∆t ≥ h

Where p is momentum; ∆p is change in momentum, x is distance; ∆x is change in distance, E is energy; ∆E is change in energy, t is time; ∆t is change in energy.

WORKED EXAMPLES

EXERCI:SE

*Calculate the frequency of photon whose energy is required to eject a surface electron with a kinetic energy of 1.97 x10¹⁶ eV if the work function of the metal is 1.33 x 10 ̄ ¹⁶ eV (, h = 6.6 x 10 ̄ ³⁴ Js, 1eV = 1.6 x 10 ̄ ¹⁹ J )

*in a photoelectric effect, no electron is ejected from the metal surface until threshold frequency is reached. Explain what happened to the energy of the light before the emission of electron begins. State one factor that may affect the numbers of emitted electrons. ( SSCE).

*calculate the energy of phoy\ton of red light of wavelength 6000 Ȧ ( I A = 10 ̄ ¹⁰m).

*Ultraviolet light of wavelength 2000 Ȧ is incidented on a potassium surface. What is the maximum K.E of the photoelectron ejected? What retarding potential difference will be required to stop the emission of electrons? Assume that the photo electric threshold wavelength of the potassium is 4400 Ȧ and

1Ȧ = 10 ̄ ¹⁰m.

*Calculate the energy in joule of ultraviolet light of wavelength 3 x 10 ̄ ⁶m.

*with the aid of a diagram, describe the production of x-rays. State five properties of x – rays and four uses to which x – rays are put in practically life.

*by means of a label diagram, explain the mode of operation of x – rays tube. State the energy transformation which take place during the operation.. Explain the term hardness and intensity as applied to x – rays. State three uses of x –rays. State two hazard of overexposure to x – rays in a radiological lab, outline two safety precautions.

*what is the energy of a photon of light of wavelength 4500 Ȧ, in joule and in eV (1Ȧ = 10 ̄ ¹⁰m).

What is the work function of a metal whose photo electric threshold wavelength is 6.8 x 10 ̄ ⁴Ȧ?

*define photo electric effect, threshold frequency, work function.

The threshold frequency for a metal 5.5 x 10¹⁴ Hz. What is the work function for the metal? If a light of wavelength 4.5 x 10⁻⁸ m shines on the metal, calculate the maximum K.E of the emitted photoelectrons in joules and in electron volt. (1eV = 1.6 x 10 ̄ ¹⁹ J).

Describe Frank hertz experiment and the conclusions that can be drawn from the results of the experiment.

Explain what is meant by photo electric emission. Draw a label diagram showing the structure of a simple type of photo cell and explain its mode of operation. State four application of photo electric emission.

*Explain the term photoelectric effect. Draw a photo cell with its associated electric circuit and label its parts.

What factor determines the current produced by a photo cell, the maximum K.E of the photoelectrons parts?. State one difference and similarity between photoemission and evaporation. Name two other methods a beam of free electrons may be produced apart from photoemission. State two application ¹of photoelectric effect. A light wavelength 5.0 x 10⁻⁶m is incident on metal resulting in photoemission of electrons. If the work function of the metal is 3.04 x 10⁻¹⁹ j, calculate the: (i.) frequency of the light (ii) energy of the incident photon (iii) max. K.e of the photoelectrons (speed of light = 3.0 x 10⁸m/s, h = 6.6 x 10 ̄ ³⁴ Js).

*State two: i) properties of x – rays. Ii) Reasons to show that x – rays are waves. (iii) Uses of x – rays other than those in medicine. (iv) Hazards of x – rays.

The potential difference between the cathode and target of an x-ray tube is 5.0 x 10⁴V and the current in the tube is 2.0 x 10⁻²A. Given that only 1% of the total energy supplied is emitted as x- radiation, determine: (i) max. Frequency of the radiation. (ii) Rate at which heat is removed from the target in other to keep it at a steady temperature (h = 6.6 x 10 ̄ ³⁴ Js), electronic charge e = 1.6 x 10 ⁻¹⁹C ).

*The uncertainty in determining the duration during which an electron remains in a particular energy level before returning to the ground state is 2.0 x 10 ⁻⁹s. calculate the uncertainty in determining its energy at that level. ( pai = 22/7, h = 6.6 x 10 ̄ ³⁴ Js. SSCE,2006.

*The mass and wavelength of a moving electron are 9. X 10⁻³¹ kg and 1.0 x 10 ̄ ¹⁰ respectively. Calculate the K.E of the electron. h = 6.6 x 10 ̄ ³⁴ Js.

*State the Heisenberg’s uncertainty principle. Mention two phenomena that can only be explained in terms of the particulate nature of light. SSCE,2004.

*Explain how neon signs can be produced. State two factors on which the colour of light from a fluorescent tube depend. SSCE, 2002.

*Name two types of gas which can be used in gas discharge tube device.