UNIT VII DUAL NATURE OF MATTER AND RADIATIONS VERY SHORT ANSWER TYPE QUESTIONS:-

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1 UNIT VII DUAL NATURE OF MATTER AND RADIATIONS (4marks) VERY SHORT ANSWER TYPE QUESTIONS:-. An electron and photon hae same waelength. Which one of the two has more energy? Relatiistic energy of a particle, E (m c 4 +p c )½ Hence the electron has more energy than photon. 3. If waelength of electromagnetic waes are doubled what will happen to energy of photon? E hνy KF E HQHUJ\ RI SURWRQ UHGXFHV WR KDOI 4. Alkali metals are most suitable for photoelectric emission. Why? Alkali metals hae too low work functions. Een isible light can eject electrons from them. 5. Out of microwaes, UV, IR which radiation will be most effecting for emission of electrons from a metallic surface? UV are most effectie since they hae highest frequency hence more energetic. 7. If the intensity of incident radiation on a metal is doubled what happens to the K.E of electrons emitted? K.E of photons remains unaffected since they do not depend 8. What is the alue of stopping potential between the cathode and anode of photocell? If the max K.E of electrons emitted is 5eV? Stopping potential V K max /e 5e/e 5 V 9. It is easier to remoe an electron from sodium than from copper, which has a higher alue of threshold waelength? w hν hc/λ λ / w Since sodium has lower work functions than copper it is easier for electron ejection. As it is lower work function, higher waelength.. An electron and proton possessing same K.E. Which one will hae greater waelength? / m (m )/m p /m

2 λ e > λ p electrons hae greater De broglie waelength than proton.. In Daisson Germer experiment if the angle of diffraction is 5 find Glancing angle? θ 9 - φ/ 9 5/ What is the effect on the elocity photo electrons, if the waelength of incident light is decreased? KE of photoelectrons is gien by Einstein s photoelectric equation. E k / m h - w 9. As waelength decreases elocity increases. 4. The stopping potential for some material is. V. What is the energy range of the emitted photoelectrons? The range of energies of the emitted photoelectrons will be from to. ev. 5.The intensity of incident radiations in a photoelectric experiment is doubled. How the stopping is potential affected? The stopping potential will remain unaffected because it does not depend on the intensity of the incident light. 6. If the intensity of the incident radiation on a metal is doubled, what happens to the kinetic energy of the emitted photoelectrons? There is no change in the kinetic energy of the emitted electrons. This K.E. is independent of the intensity of the incident radiation as long as its frequency remains the same. 7. The frequency (υ) of incident radiation is greater than threshold frequency (υ q ) in a photocell. How will the stopping potential ary if frequency (υ) is increased, keeping other factors constant? When, (υ >υ o ) stopping potential will be increased. 8.What is the energy and waelength of a thermal neutron?

3 Kinetic energy of a thermal neutron f/ k B T. Since degree of freedom of a thermal neutron is three. K.E 3/ k B T 6.6 x - J Wae length e h/(mk.e) / e h/(3m k B T) /.47nm. 9. An X ray tube produces a continuous spectrum of radiation with its shots waelength end.45 A.What is the maximum energy of a photon in the radiation in electron olt?(ii) From your answer to (i) what order of accelerating oltage (for electron) is required in such a tube?. ( K KF [ 3 ev (ii)in this case energy of X-rays photon is 7.6 KeV, the striking electron must be of energy higher than 7.6 KeV. Therefore an accelerating oltage of the order of 3 KV is required.. It is difficult to remoe free electrons from metal X as compare to metal Y. What you infer? Work function of metal X is higher than metal Y.. A particle behaes like a wae. What determine the waelength of the wae? Momentum of the moing particle.. Draw a graph giing ariation of maximum kinetic energy of photoelectrons with frequency of incident radiations. What is the slope of this graph? The equation of this straight line is K.E. /m max h (υ - υ o ) 3. The waelength of radiations incident on a material is decreased. Does the maximum elocity of photoelectrons increase or decrease? A decrease in waelength implies an increase in frequency. Since an increase in frequency increases the maximum K.E. of the emitted photoelectrons, the maximum elocity would increase. 4. For photoelectric effect in sodium, the figure shows the plot of cut off oltage ersus frequency of incident radiation. Calculate the threshold frequency (υ ) the work function for sodium (ii)the work function (W) is related to the threshold frequency (υ ) by the relation. W o hυ o 6.6 x -34 x 4.5 x 4 J 9.7 xl-j.97 xl-i9j

4 -9.97 x ev.856ev Suppose the photoelectric effect occurs in a gaseous target rather than a solid. Will photoelectrons be produced at all frequencies of the incident photons? No; we are likely to get photoemission only for those frequencies whose photons hae an energy equal to or more than the (minimum) ionization energy for the gas concerned. 6. Yellow light does not eject photoelectrons from a gien photosensitie surface, whereas green light does. What shall be situation in case of red and iolet light? We will not get any photoemission with red light since its frequency is less than that of yellow light. We will, howeer, get photoemission with iolet light since its frequency is more than that of green light. 7. By what factor does the maximum elocity of the emitted photoelectrons change when the waelength "the incident radiation is increased four times? (Gien that the initial frequency used is fie times the threshold frequency) factor of four. Now, When the waelength is increased four times, the frequency goes down by a 5ν ν 5 ν 4 ν m h ν ν ν ν 4ν ν 4 ( ν ν ) and m h( ν ν ) 6 Thus, the maximum elocity goes down by a factor of A cesium photocell, with a steady potential difference of 9 olt across it, is illuminated by a small bright light placed one metre away. The number of electrons that cross the photocell is n. What will be the number of electrons crossing the photocell when the same light is placed half metre away? When the light is brought to a distance of.5 m, the intensity of the light falling on the photocell goes up four times. Since the number of photoelectrons emitted is

5 directly proportional to the intensity of incident light, the new number of photoelectrons emitted would become 4n. 9. The sun rays are focused on a metal surface, and it produces a current. The lens forming the image is then replaced by another lens of the same diameter but only half in focal length. What will be the effect on the photoelectric current? Hence, the intensity of light falling on the metal surface gets reduced in case (b) and, we, therefore, get a reduced photoelectric current. 3. To work functions e and 5e for two metals x and y respectiely. Which metal will emit electrons, when it is irradiated with light and wae length 4nm and why? 4 x -9 m4 x -7 m E h c/ (6.6 x -34 x 3 x 8 )/ (4 x -7 ) 4.98 x -9 j E (4.98 x -9 )/ (.6 x -9 ) 3e Hence, metal x will emit electrons. 3. A photon and an electron hae same de-broglie waelength. Which has greater total energy. Explain? For a photon E hc/ For an electron h/m or mh/ E mc (h/ ) x c E /E c/> Therefore,E >E.thus, electron has total energy greater then that of photon. 3. The de-broglie wae length of a photon is same as the wae length of electron. Show that K.E. RI D SKRWRQ LV PF /h times K.E. of electron. Where m is mass of electron, c is elocity of light. ph e h/m K.E. of photon E ph hνhc/ K.E. of electrons E/m / m [h/m ] h /m

6 E ph /E e (hc/ ) x m /h mc /h E ph E e (mc /h) 33. How may photons are required for emission of one photo electron if frequency of incident radiation is (i) (ii) less than threshold frequency. More than threshold frequency. (i)no photo electron will be emitted and photons are absorbed by electrons. ii)one photon will emit one photo electron. 34. State the dependence of work function on the kinetic energy of electrons emitted in a photocell. If the intensity of incident radiation is doubled, what changes occur in the stopping potential and the photoelectric current? According to Einstein theory of photoelectric effect, kinetic energy of emitted electron is Ek mvmax hν W Greater the work function of the metal, lesser the kinetic energy of the photoelectron. on doubling the intensity of the incident radiation stopping potential remains the same, whereas photoelectric current is doubled. 35. Using Daisson and Germer Experiment to establish the existence of de Broglie waes. The experimental set up consists of an electron gun connected to a low tension battery. The electrons emitted by the gun are accelerated to a desired elocity by applying a suitable potential difference V using a high tension battery. Value of V can be aried. Using Bragg's law of diffraction of light through crystal lattices, the waelength of the wae showing diffraction peak at 5 was calculated as.65 A.

7 The predicted de Broglie waelength of electrons accelerated through 54 V is h h λ p me h.7 mev V A.7 54 A.66 A The excellent agreement between theoretical and experimental alues of de Broglie waelength confirms the wae nature of electrons and hence the existence of the de Broglie waes. 36. Through what potential difference an electron be accelerated so that it may hae de broglie waelength.5ö" Let potential through which electron is accelerated. Therefore, Energy ½ m ev! KP( / h/(me) / > h PH here, h Js m kg e.6-9 C - m hence V 37. An e - and photon hae the same energy of ev. Which has greater associated waelength? Here, e h/(mee) / > E e h P e..() Now, E p KF p > E p h c..() p as E e E p E ev..(3) Diiding equation () by () and using equation (3) we get E h c p h P e

8 > E mc e p e p (E/mc ) / as E ev > mc H9 therefore E << mc LH e p Therefore the waelength associated with a photon is greater than an electron for the same energy. 38. the frequency of the light falling on the metal is doubled, what will be the effect on photocurrent & the maximum Kinetic Energy? The photo current does not depend on the frequency of incident radiation as: E k h W If the frequency is doubled E / k h - W > E / k /E k (h W)/(h W) (h W + W)/(h W) + W/(h W) > i.e. maximum KE will increase slightly more than double. 39.e work function of Li and Cu are.3ev & 4eV. Which of these metals will be useful for the photoelectric cells working with the isible light? The threshold waelength of a metal is: o hc/w 475 / (W in ev) Ö > Li o 375 /.3 Ö Ö and Cu 375 / 4 Ö Ö The waelength 538Ö OLHV LQ WKH YLVLEOH UHJLRQ KHQFH µ/l ZLOO EH XVHIXO IRU WKH photoelectric cell. 39. An e - has a speed of 5 6 m / s in a magnetic field of -4 T. What is the acceleration of e - if e / m.76 C / kg. BeV m /r as the force due to magnetic field proides the necessary centripetal force. > /r Be / m as acceleration /r > a B( e / m ) > a /r > a ms Assume that the potential difference between cathode and anode is the same as that between two deflecting plates. If this potential difference is doubled, calculate by what factor the magnetic field should changed to keep the electron beam undeflected? e / m E /B But E / d e / m /B d /B d or, B. ( e / m )d or B. > B. ; if is doubled, then B / /B () / or B / () / B Thus, the magnetic field should be increased by a factor of () /

Chapter-11 DUAL NATURE OF MATTER AND RADIATION

Chapter-11 DUAL NATURE OF MATTER AND RADIATION Work function (j o ): The minimum energy required for an electron to escape from the surface of a metal i.e. The energy required for free electrons to escape