- How to determine the work function?
Work function, Φ, is the part of energy given to a single electron by a light quantum which collides with a metal surface. This energy is needed to pull a surface electron out of the material. The work function depends on the particular metal (table5.1)
- Find the relation between stopping potential and light frequency.
Stopping potential is the negative potential of the anode needed to stop photoelectrons. When it is reached even the fastest photoelectrons are not able to reach it, which means that ν0 can be expressed by the photoelectron’s ejected energy using the following equation:
So using the photoelectric equation we obtain:
Vse = hν – Φ
Thus, we come to the following conclusion: if Einstein’s photoelectric equation is correct, the experimentally determined stopping potential should be a linear function light frequency.
- How does the photoelectric effect allow us to calculate Planck’s constant?
If conclusion form question 9 is correct, then the slop of stopping potential as linear function of the light frequency should be Planck’s constant. This let us to determinate the linear slop :
∆Vs/∆ν = h/e
From this equation we can calculated the Planck’s constant, which should be 6,63×10-34Js.
II. PHOTONS AND ELECTRONS
- Draw a system producing X-rays.
Fig.5.7 – present-day oil-cooled X-ray tube consists of two electrodes placed in a vacuum gas bulb. Electrons are emitted from an incandescent cathode and accelerated to high speed by a potential difference existing between the cathode and the anode. Then, electrons are stopped while striking the anode target. On penetrating the target, the electrons are decelerated by Columbic force acting between an electron and the nucleus of a target atom → Fig. 5.8
The electrons which changes its velocity produce electromagnetic radiation, which should have a continuous spectrum (according to Maxwell theory). Continuous means that every wavelength, with no exception, should be present in this spectrum. → question no 11.
An X-ray tube is not a only source of X radiation; it will occur every time when fast electrons collide with matter. Generally, X-rays are produced during deceleration of electrons obtained from an accelerator or emitted by radioactive emitted by radioactive nuclei.
- Show how X-ray spectra change with the potential difference between a cathode and a target.
(Fig.5.9)It shows that:
-
a broad band of X-ray radiation is interrupted at a certain wavelength, λmin, which depends on the potential difference between the cathode an the target
- at higher voltages, certain characteristic spectral lines occur
-
the relation between wavelength and potential difference which is λmin ~ 1/V
→ question no 12
-
Why is there λmin in the X-ray spectrum.
The appearance of λmin in continuous is X-ray spectra is fully understandable if we consider the quantum nature of X-rays.
The majority of electrons colliding with the target lose their energy by multiple collisions, which cause an increase of the target’s temperature. However, some of the bombing electrons (about one ore two percent) lose most of their kinetic energy in a single collision. Such an electron is suddenly decelerated, and an impulse of electromagnetic radiation is generated; thus, an electron produced an X-ray photon. If an electron is accelerated by the potential difference V and loses its total energy in a single collision, its kinetic energy should be equal to the energy of X-ray photon generated. Thus,
and
- How do you explain the mechanism of characteristic X-ray generation ?
The X-ray generation phenomena can be consider as a reverse to the photoelectric effect.
When the photoelectric effect occurs, photon is absorbed and its energy and momentum are transmitted to an electron. (the part of energy transmitted to the atom’s nucleus is usually ignored). In the process of X-ray generation a photon is emitted. The photon’s energy and momentum are due to an interaction between a rushing electron and an atom.
→ mechanism of photon emission → Bohr’s atom model
Considering the electrons-target collision, we can imagine another form of energy exchange. The bombarding electron can transmit its energy to electrons of a target atom. The amount of energy of accelerated electrons are usually sufficient to remove an electron from outer as well as from inter shell and the resulting atomic structure is not stable. An electron from adjacent L shell will jump into the vacant space emitting an X-ray photon. When we assume that an electron is knocked out of the K shell and replaced by an electron from L shell, the jumping process will be continued until outermost shell is reached. During such emission processesm the energy if the successive photons decreases. The radiation of K series has high energy and is usually called “hard” radiation, while the L series belongs to the “soft”, less energetic X-rays.
Spectra of above described transition differ from one target material to another, as all materials have their own specific structure of atoms, number of protons in the nucleus (and the energy of the inner-shell electrons depends mainlyon the number of protons in the nucleus. This is why spectra lines are called characteristic X-ray.
- Explain the essence of Compton’s effect.
The Compton experiment:
(Fig.5.12) X-rays limited to a narrow beam of rays struck a face of a small graphite target and scatter in various directions. An X-ray spectrograph measured the wavelength of scattered radiation for fixed angle θ (measurements are repeated for different angles). Only in case of the angle of θ = 0 dec. the maximum of spectrum appears at exactly the same wavelength λ as the maximum of incident radiation. In other cases the intensity of the radiation at λ’ is more then the peak of X-rays scattered at θ = 0 dec. When the angle increases, the wavelength shift ∆λ = λ – λ’, increases.
The Compton effect provides incontrovertible argument confirming the particle-like aspect of electromagnetic radiation. (Compton's experiment became the ultimate observation that convinced all physicists that light can behave as a stream of particles whose energy is proportional to the frequency)
→ wave-photon duality:
- accting like a waves – X-rays may be scattered at different angles and produce diffraction patterns.
- acting like a particles – X-ray photons may collide with atoms (resulting in a photoelectric effect which is more complex than the photoelectric effect)
Difference between X-ray photoelectric effect and photoelectric effect produced by light:
- the energy of X-rays phptons is essentially higher as commpared with light photons; high enough to eject electrons from inner shells
- the process of ejecting inner-shell photoelectrons is connected with generation of new X-ray photons – here energy necessary to eject electrons from an inner shell is provided by external X-ray photons instead of accelerated external electrons.
Explanation of Compton’s effect→ question no 16
- Why cannot the Compton’s effect be explained by the electromagnetic wave theory?
Because, contrary to the wave theory, the spectra of scattered X-rays have additional maxima at wavelengths λ’ longer that the spectra of the incident radiation.
Classical electromagnetic theory:
When the energy of X-ray is higher then 0,1 MeV, the photons usually interacting with electrons lose only a small part of their energy and change their direction (the X-ray beam is scattered) because while interacting with an electron, the oscillating electric field of the X-ray wave should force the electron’s oscillation. Since an oscillating change is a source of electromagnetic waves, the electron itself should produce electromagnetic waves. The electron should radiate in all directions X-rays of the same wave frequency as the incident wave.
- What facts confirm that Compton’s effect can be explained by treating light a beam of photons?
Compton consider the photon-electron collision as a collision between two perfect elastic balls. (Fig.5.13.) Assuming that, before the collision with a photon, the electron is at rest, the energy equal to ½ mv2, exerted in the recoiling electron, is supplied – according to taw of conservation of energy – by an incident X-ray photon, hν. The energy of the scattered photon decreases, satisfying the following equation:
hν’ = hν – ½ mv2
As the frequency of the scattering photon decreases, its wavelength, λ’=c/ν’, is more than the wavelength, λ , of the incident photon.
T attain a quantitative agreement with experiment, Compton derived the following equation using the laws of conservation of energy and momentum :
∆λ = h/mc(1-cosθ)
where m is the electron mass, and θ is a scattering angle.
- Why is Compton’s effect not observed for light photons?
Because energy of light photons is much smaller than then energy of X-rays photons I SUPPOSED that it is to small to observe this effect → a light photon when interact with the electron, imparts its total energy to a single electron.
III. THE BOHR MODEL OF THE ATOM OF HYDROGEN
- Explain the contradiction between the classical theory of electromagnetism and Bohr’s model of the hydrogen atom.
Bohr proposed the planetary model of the atom of hydrogen assuming that:
- the electron moves in a circular orbit of radius r around a fixed nucleus
- the mass, m, of the electron does not charge during the motion (Fig.5.14)
According to Bohr’s assumptions, an electron moves with a constant orbital speed, v. Such movement takes place when the inward centripetal force, Fc=mv2/r, keeps the electron in its orbit. We see from this equation that the charge is accelerate (v2/r). Form the principles of classical electrodynamics we know that any rotating charge is accelerated, and when electric charge is accelerated it should continuously emit electromagnetic radiation. As a consequence electron lose its energy so it should spiral into the nucleus. So we see that the atom with an electron which rotates in a stable circular orbit around the nucleus CANNOT EXIST.
- Discuss the first of Bohr’s postulates
Bohr postulated the existence of certain stationary orbits. According to his first postulate, if an electron rotates in one of these circular orbits it does not emit electromagnetic radiation. This postulate directly contradicts the predictions of classical electromagnetism → question no 18
- Find an expression for the orbital radius of the hydrogen atom.
According to the second of Bohr’s postulates (the quantum hypothesis) the angular momentum of an electron in its stable orbit is quantized so an electron cannot rotate in any orbit of any radius, but only in certain define and discrete orbits. Bohr expressed the quantization condition as follows:
(n=1,2,3….)
angle between vector of velocity and of radius is equal 90dec so sin(v,r) = 1
where L is angular momentum, m is the electron mass, v is an electron’s speed in a circular orbit of radius r. The integer n is called the principal quantum number (describe the energy of the shell); h = 6,63×10-34 Js (Planck’s constant)
We know that there are two forces acting on the electron :
Electrostatic force :
And centripetal force :
Fc=mv2/r
We know that so we compare this two equations and we obtain :
Now, we divide the this equation by quantisation condition in order to obtain equation of velocity:
from quantisation condition we know that:
so substituting v we obtain the expression for radius, rn, for each quantized orbit:
- Determine the energy of the orbital radius of the hydrogen atom.
We know from mechanics that mechanic energy is equal to the sum of potential and kinetic energies.
Potential energy we find from definition of potential energy:
When we taking to account the distance ∞ the Ep tense to 0
Kinetic energy can be calculated from the equation : , knowing that
Wk = Ek = mv2/2, kinetic energy is
So mechanic energy is :
We see that mechanic energy depends on kinetic energy only.
rn is given by the equation : (question 20) so we can write the mechanic energy as :
So energy of the lowest energy state – ground state - is equal to -13,6 eV. The energy states for n>1 are called exited states. The transition from ground state to exited state of an atom is called an absorption of energy.
- Explain Bohr’s second postulate.
Considering question 19 we can see that quantization of a mechanical quantity such as angular momentum, L, leads to the quantization of the other mechanical proprieties of the atom, such as the velocity or radius. Finally, an electron’s mechanical energy, being a sum of its potential and kinetic energy, is also quantized. It is a general idea of Bohr’s second postulate which assume that angular momentum of an electron is quantized. (but this postulate is satisfied only if electron orbits are discrete)
- How is the hydrogen spectrum explained in Bohr model?
In order to explain emission of hydrogen spectrum Bohr introduced a third postulate into his theory. An atom emits light ONLY when it undergoes a transition from one stationary state t another of lower energy. Bohr said that the frequency of emitted light is not determined by the frequency of electron revolution but by the difference in energy between the initial and final energetic states. According to the law of conservation of energy, the quanta of emitted energy can be expressed as:
hν = Wi – Wf
(Wi –energy of the initial energetic state; Wf – of the final state). We know that ν=c/λ (c-speed of light in vacuum, λ –wavelength), so
→question no 21
this formula allows us to calculate a set of wavelengths, which should be obtained if we consider a series of electron jumps from its various initial energetic states to a selected final state (ni>nf).
Exited states are not stable and after a short time an electron will jump to the ground state, emitting an additional photon.
We should not that any act of light emission necessarily involves an electron’s excited state. An electron’s excitation can by achieve by various means, such as an absorbtion of radiation, atomic collision or electron discharge.
The hydrogen spectrum following from the Bohr model agrees exactly with experimentally observations. Unfortunately, Bohr’s theory does not account for the spectra of helium and more complex atoms. Even for the hydrogen spectrum, the Bohr model does not allow us to predict the intensity of spectra lines.
IV. WAVE-PARTICLE DUALITY
- Formulate the de Broglie hypothesis.
The de Broglie hypothesis says that all moving particles have wave-like proprieties. Wave-photon duality is the most characteristic attribute of electromagnetic radiation.
- as a particle, a photon exists only when it is in movement, as its rest mass equals zero.
- While motion, photon carries energy, E=hν, and momentum, p=h/λ.
-
For photon which have no rest mass E=pc → Einstein’s relativistic energy-momentum relation
So the momentum of photon is given by: p=E/c=hν/c=h/λ
According to he Broglie’s hypothesis the wave-like properties of moving particles can be expressed as matter waves associated with the moving particles. The wavelength of matter wave and the momentum of a particle are related by
The wavelength defined by this equation is called the de Broglie wavelength.
- Prove that matter wave is not measurable for macroscopic objects.
The de Broglie wavelength for a macroscopic objects, such as bullet, is too small to be measurable. Let’s assume that 15-g bullet is moving with the speed of v = 700 m/s
Knowing that λ=h/mν and h=6.63×10-34Js we calculate that wavelength is 6,3×10-35m. So we see that is very small value. But it is possible to determinate de Broglie wave for such small particles as electrons (what is logical because small number divided by small number it always gives some bigger value ☺)
- What is the relation between de Broglie’s hypothesis and second of Bohr’s postulates?
The metter-wave hypothesis is very useful if we want to justify the second of Bohr’s postulates: that the angular momentum of an electron in its stationary orbit is quantized. If the wave-particle dualityia a fundamental feature of any moving particle, an electron moving in its stationary orbit is also associated whit its wave. Thus, we may equivalently represent an electron by a matter wave. That it should be a standing wave results from the fact, that we consider a stationary orbit, so that n electron’s energy is localized in this orbit. If a standing wave forms the circumference of an electron’s circular orbit with radius e, full n de Broglie’s wavelength are found within this circumference. Thus,
and, hence, mvr=nh/2π. Therefore. The second of Bohr’s postulates results from the wave nature of matter.
- Draw and explain the G.P. Thomson experiment of electron diffraction.
(Fig.5.17) An electron gun is a source of electrons of known velocity which struck a thin metal film. Electrons diffracted and formed diffraction rings on a photographic screen. The dark and white rings are the result of interference among electron waves diffracted by atoms of the metal foil. It is a prove of wave nature of mater.
