Answer the following questions: (a) Quarks inside protons and neutrons are thought to carry fractional charges [( + \frac{2}{3} e ; - \frac{1}{3}e )]. Why do they not show up in Millikan’s oil-drop experiment? (b) What is so special about the combination e/m? Why do we not simply talk of e and m separately? (c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures? (d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?

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(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations:

E = h\upsilon , p = \frac{h}{\lambda}

But while the value of \lambda is physically significant, the value of \upsilon (and therefore, the value of the phase speed \upsilon \lambda ) has no physical significance. Why?

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~~N ." $ itOL'it At .A

... it up. It was after the discovery of electricity that the history of•Particle Physics as ... This forced the oil droplet, which had been electrified by friction in th e ... that is, its atomic number times the average mass of the proton and neutron. ... e truly fundamental particles with B 1/3 and fractional charges, so that baryons woul d.

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Zero-point energy - Wikipedia

Zero-point energy (ZPE) is the lowest possible energy that a quantum mechanical system may have. Unlike in classical mechanics, quantum systems constantly ...

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Millikan’s oil-drop experiment


Two up quarks and one down quark are found in each proton. One up quark and two down quarks are found in neutrons.


                               u = + \frac{2}{3}e and d = - \frac{1}{3}

Quarks are carrying fractional charges inside protons and neutrons. This is because nuclear power rises enormously if it is pulled apart. As a result, fractional charges can occur in nature; observable charges are still an integral multiple of the electrical charge.


Step 1: Recall the electric and magnetic field basic relations  


                  After being accelerated by 1 volt of electricity, an electron gains energy.

                                       eV = \frac{1}{2} mv^2 .................(1)


              When electron passes through magnetic field region

                 Force acting on electron F = evB

                Centripetal force F = \frac{mv^2}{r}


                         evB = \frac{mv^2}{r} ...............................(2)

                Where, B - magnetic field, v - velocity, e -  electron charge, m - mass, r - radius and V - potential


Step 2:  Set up an equation for velocity

                From equation (1)

                       v^2 = \frac{2eV}{m}

                       v = \sqrt{2V \frac{e}{m}}

                 From equation (2)

                       eB = \frac{mv}{r}

                       v = Br \frac{e}{m}

              It can be concluded from these relations that the dynamics of an electron is calculated not by e and m separately, but

              by the ratio e/m.


Atom 1:                                                                                             Atom 2:


Because of collisions and recombination with other gas molecules, ions of gases have no chance of meeting their respective electrons at atmospheric pressure. As a consequence, at atmospheric pressure, gases serve as insulators. Electrons have a chance of touching their respective electrodes and creating a current at low pressures. As a result, at these pressures, they conduct electricity.


The minimum energy needed for a conduction electron to leave the metal surface is known as the work function of the metal. The energy levels of all electrons in an atom are different. When a photon-emitting ray strikes a metal surface, electrons emerge from various levels at various energies. As a result, the energy distributions of the emitted electrons vary.


Within the additive constant, a particle's absolute energy value is arbitrary. Therefore, while the wavelength ( \lambda ) associated with an electron is significant, the frequency ( \upsilon ) associated with an electron has no direct physical significance. As a result, the product \upsilon \lambda (phase speed) has no physical meaning.

The speed of a group is expressed as follows

                 v_G = \frac{dv}{dK}

                 v_G = \frac{dv}{d \frac{1}{\lambda}}   

                 v_G = \frac{dE}{dp}

                 v_G = \frac{\frac{p^2}{2m}}{dp}

                 v_G = \frac{p}{m}

          This quantity has a physical meaning