^{-34}Js.

^{14}Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons. (b) stopping potential and (c) maximum speed of the emitted photoelectrons. Given ; h = 6.63⨯ 10

^{-34}Js, 1 eV = 1.6 ⨯ 10

^{-19}J; c = 3⨯ 10

^{8}m/s.

^{-19}C.

*have*to travel in order to have the same momentum a s that of the photon, h = 6.63⨯10

^{-34}Js, 1 a. m. u. = 1.66 ⨯ 10

^{-27}kg.

^{3}W/m

^{2}. How many photons (nearly) per square metre are incident in the earth per second ? Assume that the photons in the sunlight have an average wavelength of 550 nm.

^{-15}Vs. Given e = 1.6 ⨯ 10

^{-19}C, estimate the value of Planck’s constant.

Sol. (a) Energy associated per photon with the sodium light is,

^{14}Hz. If light of frequency 8.2 ⨯ 10

^{14}Hz is incident on the metal, predict the cut off voltage for photoelectric emission, Given plank’s constant h = 6.62 ⨯ 10

^{-34}Js.

^{-19}C, c = 3 ⨯ 10

^{8}m/s; h = 6.62 ⨯ 10

^{-34}Js.

As energy of incident photon E < hence photoelectric emission will take place.

^{14}Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 ⨯ 10

^{5}ms

^{-1 }are ejected from the surface. What is the threshold frequency for photoelectrons of electrons ? h = 6.63 ⨯ 10

^{-34}Js, m

_{e}= 9.1 ⨯ 10

^{-31}kg.

^{-34}Js, 1 eV = 1.6 ⨯ 10

^{-19}J.

^{-34}Js, m

_{e}= 9 ⨯ 10

^{-31}kg ; e = 1.6 ⨯ 10

^{-19}C.

^{-34}Js, m

_{e}= 9 ⨯ 10

^{-31}kg ; 1 eV = 1.6 ⨯ 10

^{-19}J.

h = 6a.6 × 10^{-34} Js, m_{e} = 9 × 10^{-31} kg ; m_{n } = 1.66 × 10^{-27 } kg.

^{-1}. (b) a ball of mass 0.060 kg moving at a speed of 1.0 ms

^{-1 }and (c) a dust particle of mass 1.0 ⨯ 10

^{-9 }kg drifting with a speed of 2.2 ms

^{-1}? h = 6.63⨯ 10

^{-34}Js.

^{-34}Js, m

_{e}= 9 ⨯ 10

^{-31}kg.

^{-10}m ? (b) Also find the di-Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of

Sol. The momentum of a electromagnetic wave of frequency v wavelength λ is given by,

Thus wavelength of electromagnetic radiation is equal to the de-Broglie wavelength.

Q.19. What is the de-Broglie wavelength of a nitrogen molecule in air at 300 K ? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u.). h =

= 6.63 × 10^{-3}Js ; k = 1.38 × 10^{-23} JK^{-1} ; 1u = 1.66 × 10^{-27} kg.

Q.20. (a) Estimate the speed with which electrons emitted from a heated cathode of an evacuated tube impinge on the anode maintained at a potential difference of 500 V with respect to the cathode. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its e/m is given to be 1.76 ⨯ 10^{11} C kg^{-1}.

(b) Use the same formula you employ in (a) to obtain electron speed for an anode potential of 10 MY. Do you see what is wrong ? In what way is the formula to be modified ?

Q.21. (a) A monoenergetic electron beam with electron speed of 5.20 ⨯ 10^{6} ms^{-1 }is subjected to a magnetic field of 1.30 ⨯ 10^{-4 }T, normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 ⨯ 10^{11} C. kg^{-1}.

(b) Is the formula you employ in (a) valid for calculating radius of the path of 20 MeV electron beam ? If not, in what way is it modified !

Which is greater than the velocity of light.

Therefore, the formula r = mv/eb is not valid for calculating the radius of the path of 20 MeV electron beam because electron with such a high energy has velocity in the relativistic domain (i.e. comparable with the velocity of light). For this we use relativistic formula as follows.

^{-2 }mm of Hg). A magnetic field of 2.83 ⨯ 10

^{-4}T cures the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electron and emitting light by electron capture ; this method is known as ‘fine beam tube’ method). Determine e/m from the data.

Sol.Here, V = 100 V, B = 2.83 × 10^{-4} T, r = 12.0 cm = 12.0 × 10^{-2} m

When electrons are accelerated through V volt, the gain in K.E. of the electron in given by

Q.23. (a) An X-ray tube produces a continuous spectrum of radiation with its hort wavelength end at 0.45 Å. What is the maximum energy of a photon in the radiation ?

(b) From your answer to (a), give what order of accelerating voltage (for electrons) is required in such a tube. Given, h = 6.63 ⨯ 10^{-34} Js.

(b) In X- ray tube, accelerating voltage provides the energy to the electrons which on striking the anticathode produce X-rays. For getting X-ray photons of 27.6 keV, it is required the incident electrons must possess kinetic energy of atleast 27.6 keV. Therefore, accelerating voltage of the order of 30 keV should be applied across the X-ray tube to get the required X-ray photons.

^{9}eV)

Q.25. Estimating the following two number should be interesting. The first number will tell you why radio engineers do not need to worry much about ‘photons’. The second number tells you why our eye can never ‘count photon’ even in barely detectable light.

(a) The number of photons emitted per second by a MW transmitter of 10 kW power emitting radio waves of length 500 m.

(b) The number of photon entering the pupil of our eye per second corresponding to the minimum intensity of white light that we human can perceive( ~10 ^{-10}Wm^{-2}). ( Take the area of the pupil to be about 0.4 cm^{2}, and the average frequency of white light to be about 6⨯ 10^{4} Hz.

We see that the energy of a radio photon is exceedingly small and the number of photons emitted per second in a radio beam is enormously large. There is, therefore, negligible error involved in ignoring the existence of a minimum quantum of energy (photon) and treating the total energy of a radio wave is continuous.

Though this number is not large as in part (a) above, it is large enough for us never to ‘sense’ of ‘count’ the individual photons by our eye.

^{-5 }Wm

^{-2}). ( red light of wave length 6328 Å produced by a He-Ne laser ?

Since,v_{r} < v_{0 }therefore, the photocell will not respond to this red light, howsoever strong its intensity may be.

^{-9}m) from a neon lamp irradiates a photo-sensitive material made of calcium of calcium or tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photocell. Predict the new stopping voltage.

Q.28. A mercury lamp is a convenient source for studying dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used :

Λ_{1} = 3650Å,λ_{2} = 4047 Å.λ_{3} = 4358 A^{0} , λ_{4} = 5461 A^{0} , λ^{5} = 6907 A^{0}

The stopping voltages, respectively were measured to be :

V_{01 }= 1.28 V, V_{02} = 0.95 V, V_{03} = 0.74 V, V_{04} = 0.16 V, V_{05} = 0 V.

(a) Determine the value of Planck’s constant h.

(b) Estimate the threshold frequency and work function for the material.

Hence a graph between frequency v and stooping potential v_{0}, must be a straight line.

On plotting a graph between given frequencies v and corresponding values of stopping potentials V_{0}, we get a straight line with first four points_{. }This straight line is cutting v-axis at 5.0 × 10^{14 }Hz. The fifth point comes in the range of v < V_{0} which shown there is no photoelectric emission and hence no stopping potential is required to stop the current. From graph the slope of this straight line.

^{0}from a He-Cd laser placed 1 m away from the photocell ? What happens if the laser is brought nearer and placed 50 cm away ?

This energy of the incident radiation is greater then the work function of Na and K but less than those of MO and Ni, so photoelectric emission will occur only in Na and K metals and not in MO and Ni. If the laser is brought closer, the intensity of incident radiation increases. This does not affect the result regarding MO and Ni metals, while photo electric emission from Na and K will increase in proportion to intensity.

^{-5}Wm

^{-2}falls on a sodium photocell of surface area 2 cm

^{2}. Assuming that the top 5 layers of sodium absorb the incident energy, estimate, the time required for photo-electric emission in the wave picture of radiation. The work function of the metal is given to be about 2 eV. What is the implication of your answer ? Effective atomic area = 10

^{-20}m

^{2}.

Sol. Here, I = 10^{-5} Wm^{-2} , A = 2 × 10^{-4} m^{2} ,n = 5, t = ? Ø_{0} =2eV = 2 × 1.6 × 10^{-19} J

Sodium has one conduction electron per atom and, effective atomic area = 10^{-20} m^{2}

Since the time for photelectric emission is very large so wave picture cannot be applied to this experiment.

_{e}= 9.11⨯ 10

^{-31}kg.

Thus for the same wave length, a photon has much greater kinetic energy than an electron.

Q.32. (a) Obtain the de-Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in previous problem 31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable ? Explain.

Given m_{n} = 1.675 × 10^{-27} kg.

(b) Obtain the de-Broglie wavelength associated with thermal neutrons at room temperature (27^{0} C). hence explain why a fast neurton beam needs to be thermalized with the environment before it can be used for neutron diffraction experiments.

The interatomic spacing ~1 Å (= 10^{-10}m) isabout a hundred times greater than this wave length. Therefore, a neutron beam of energy 150 eV is not suitable for diffraction experiment.

Since this wavelength is comparable to interatomic spacing (~ 1Å) in a crystal, therefore, thermal neutrons are suitable probe for diffraction experiments : so a high energy neutron beam should be fust thermalized before using it for diffraction.

Since resolving power (R.P.) is inversely proportional to wave length, therefore R.P. of an electron microscope is about 10^{5} times that of an optical microscope. In practice, differences in other (geometrical) factors can change their comparison somewhat.

^{-15}m or less. This structure was first probed in early 1970’s using high energy electron beam produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams (Rest mass energy of electron= 0.511 MeV).

Thus the electron energies from the accelerator must have been of a few BeV.

^{0}C) and 1 atm pressure ; and compare it with the mean separation between two atoms under these conditions.

^{0}C and compare it with the mean separation between two electrons in a metal which is given to be about 2 ⨯ 10

^{-10}m.

Q.37. answer the following questions : (a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2.3)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 pressure ? (d) Every if incident definite work function. Why do all photoelectrons not come out with the same energy distribution of photoelectrons ?

(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations : E = hv, p = h/λ

Speed v λ) has no physical significance. Why ?

Sol. (a) The quarks having fractional charges are thought to be confined within a proton and a neutron. These quarks are bound by forces. These force become stronger when the quarks are tried to be pulled apart. That is why, the quarks always remain together. It is due to this reason that though fractional charges does exist in nature but the observable charges are always integral multiple of charge of electron.

(b) The motion of electron in the electric and magnetic field is related with the basic equations

All these equation involve e and m together, i.e., there is no equation in which e or m occurring alone. As a result of it, we study e/m of electron and do not talk of e and m separately far an electron.

(c) At ordinary pressures a few positive ions and electrons produced by the gas molecules by energetic rays (like X-ray, Y-ray, cosmic rays etc. coming from outer space and entering the earth’s atmosphere are not able to reach their respective electrodes, even at high voltages, due to their frequent collisions with gas molecules and recombinations. That is why the gases at ordinary pressures are insultors.

At low pressures, the density of the gas decreases, the mean free path of the gas molecules become large. Now under the effect of external high voltage, the ions acquire sufficient energy before they collide with molecules causing further ionization. Due to it, the number of ions in the gas increases and it becomes a conductor.

(d) By work function of a metal, we mean the minimum energy required for the electron in the highest lever of conduction band to get out of the metal. Since all the electrons in the metal do not belong to that level but they occupy a continuous band of levels, therefore, for the given incident radiation, electrons knocked off from different levels come out with different energies.

(e) Debroglie wavelength associated with the moving particle is λ

For the relations of E and p, we note that λ is physically significant but v has direct physical significance.