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 15. 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 ? ‘ 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. 16.

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 1/2 mv2, exerted in the recoiling electron, is supplied – according to taw of conservation of energy – by an incident X-ray photon, h?.The energy of the scattered photon decreases, satisfying the following equation: h? ‘ = h? – 1/2 mv2 As the frequency of the scattering photon decreases, its wavelength, ? ‘=c/? ‘, is more than the wavelength, ? , 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 : ?? = h/mc(1-cos? ) where m is the electron mass, and ? is a scattering angle. 17.

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 18.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. 19. 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 20.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 = 6,63i?? 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: 21.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. 22. 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) 23. 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: h? = Wi – Wf (Wi -energy of the initial energetic state; Wf – of the final state). We know that ? =c/? (c-speed of light in vacuum, ? -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 24. 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=h? , and momentum, p=h/?. 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=h? /c=h/?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. 25. 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 ?=h/m? and h=6. 63i?? 10-34Js we calculate that wavelength is 6,3i?? 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 :)) 26. 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/2?. Therefore. The second of Bohr’s postulates results from the wave nature of matter. 27. 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.