NQA7 kho tài liệu bách khoa

NUMERICALS-ANSWERS QUESTIONS What is the de Broglie wavelength of a kg object moving with a speed of m/s? (AISSCE 1992) Calculate the frequency associated with a photon of energy 3.3 × 10–10 J (AISSCE 1990) Calculate the de Broglie wavelength of an electron beam accelerated through a potential difference of 60 V (AISSCE Delhi 1993) Work function of Na is 2.3eV Does sodium show photoelectric emission for light of wavelength 6800 Å? Light of wavelength 3500 Å is incident on two metals A and B Which metal will yield photoelectrons if their work functions are 4.2 eV and 1.9 eV respectively (AISSCE 1991) Calculate the maximum kinetic energy of photoelectron (in eV) emitted on shining light of wavelength 6.2 × 10–6 m on a metal surface The work function of the metal is 0.1 eV (h = 6.6 × 10–34 Js) (AISSCE 1992) For a photosensitive surface, the work function is 3.3 × 10–19 J Find its threshold frequency Take h = 6.6 × 10–34 Js Obtain the energy in joules acquired by an electron beam when accelerated through a potential difference of 2000 V (AISSCE 1994) Calculate the threshold frequency of photon for photoelectric emission from a metal of work (AISSCE Delhi 1992) function 0.05 eV (h = 6.6 × 10–34 Js) 10 Calculate the photon energy in electron-volt for radiation of wavelength m (AISSCE Delhi 1990 C) 11 What is the energy associated in joule with a photon of wavelength 4000 Å (AISSCE Delhi 1994) S Chand & Company Limited 12 Calculate the velocity of an electron accelerated through a potential of 100 volt Given e = 1.6 × 10–19 C and m = 9.1 × 10–31 kg 13 An electron, moving with a speed of 2.5 × 107 m/s, is deflected by an electric field of 1600 V/m perpendicular to its path If the radius of the electron trajectory is 2.3 m, calculate the elm for the electron? (AISSCE Delhi 1991) 14 What is the value of stopping potential between the cathode and the anode of a photo cell, if the maximum kinetic energy of electrons emitted is 5eV? (AISSCE 1993) 15 When light of wavelength 400 nm is incident on the cathode of a photocell, the stopping potential recorded is 6V If the wavelength of the incident light is increased to 600 nm, calculate the new stopping potential (AISSCE 2000) 16 If the maximum speed of photoelectrons is 10 m/s, what is the frequency of the incident radiation on potassium metal (work function = 2.26 eV) 17 The work function of a photosensitive material is 2.5 eV What must be the threshold wavelength of incident light to eject photoelectrons from it? 18 Calculate the de Broglie wavelength for an electron beam whose energy is 400 eV 19 Find the frequency of light which ejects electrons from a metal surface fully stopped by a retarding potential of 3V The photoelectric effect begins in this metal at frequency of × 104 Hz Find the work function for this metal 20 Calculate the momentum and the de Broglie wavelength associated with an election of kinetic energy 54 eV 21 In an experiment on photoelectric effect, the slope of the cut off voltage versus frequency of incident light graph is found to be 4.12 × 10–15 Vs Given e = 1.60 × 10–19 C, estimate the value of Planck’s constant 22 Find the typical de Broglie wavelength associated with a He atom in helium gas at temperature 27oC and pressure atm S Chand & Company Limited 23 The photoelectric cut-off voltage in a certain experiment is 1.5 V What is the maximum kinetic energy of photoelectrons emitted? 24 Light of wavelength 5000 Å falls on a metal surface of work function 1.9 eV Find (a) the energy of a photon in eV, (b) the maximum kinetic energy of photoelectrons, and (c) the stoping potential 25 Some values of work function are given below Na : 1.92 eV, K : 2.15 eV Mo : 4.17 eV, Ni : 5.0 eV Which of these metals will not give photoelectric emission for radiation of wavelength 3300 Ao 26 Two metals X and Y have work functions eV and eV respectively Which metal will emit electrons, when irradiated with light of wavelength 400 nm and why ? (AISSCE 1999) ANSWERS λ = = h mv 6 × − × = 1 × − m From the formula E = hν, we have ν = E/h S Chand & Company Limited = 3.3 × 10 −20 6.6 × 10 −34 = × 1013 Hz λ= 12.27 == o A V 12.27 60 o = 1.583 A o λ = 6800 A = 6800 × 10–10 m Energy of incident photon = = 6.6 × 10 −34 × × 108 6800 × 10 −10 = 6.6 × × 10 −18 68 × 1.6 × 10 −19 hc λ J eV = 1.83 eV Since the energy of incident photon is less than the work function of Na, photoelectric emission is not possible with the given light o λ = 3500 A So, energy of incident photon, E = hc λ S Chand & Company Limited = 6.6 × 10 −34 × × 108 3500 × 10 −10 × 1.6 × 10 −19 eV = 3.53 eV Since 1.9 eV < E < 4.2 eV, only metal B will yield photoelectrons λ = 6.2 × 10–6 m, W = 0.1 eV Maximum kinetic energy of photo electron is given by Emax = hν – W = hc –W λ  6.6 × 10 −34 × × 108  – 0.1 eV = −6 −19  6.2 × 10 × 1.6 × 10   66 ×  = − 0.1 = 0.2 − 0.1 = 0.1 eV 62 16 ×   Threshold frequency is ν0 = = W h 3.3 × 10 −19 6.6 × 10 −34 = 5.0 × 1014 Hz S Chand & Company Limited V = 2000 V Energy acquired E = eV = 1.6 × 10–19 × 2000 = 3.2 × 10–16 J From the formula W = h ν0, we have W h Here, W = 0.05 eV = 0.05 × 1.6 × 10–19 J Therefore, ν0 = ν0 = 10 E= 0.05 × 1.6 × 10−19 6.6 × 10−34 = 1.21 × 1012 Hz hc λ 6.6 × 10−34 × × 108 J 6.6 × × 10−26 = eV = 1.24 × 10−6 eV 1.6 × 10−19 = o 11 λ = 4000 A = × 10–7 m E= hc 6.6 × 10 −34 × × 108 = λ × 10 −7 = × 10–19 J 12 From the equation m v2 = eV S Chand & Company Limited v= = eV m × 1.6 × 10 −19 × 100 −31 × 10 = 5.9 × 106 m/s = 32 × 1013 13 Force due to the electric field will provide the required centripetal force So m v2 eE = r e v = or m rE (2.5 × 107 ) 25 × 25 = = × 1011 2.3 × 1600 23 × 16 = 1.7 × 1011 C/kg 14 Maximum kinetic energy Emax = eV0, where V0 is the stopping potential So, E × 1.6 × 10−19 V0 = max = e 1.6 × 10−19 = 5.0 V 15 Using Einstein’s photoelectric equation eV1 = hν1 – W eV2 = hν2 – W S Chand & Company Limited h (ν1 − ν ) e hc  λ − λ1  =   e  λ1λ  V1 − V2 =  6.6 × 10−34 × × 108  200 × 10−9  −19 −18  1.6 × 10  600 × 400 × 10  =1V V2 = V1 − = − = V = 16 W = 2.26 eV = 2.26 × 1.6 × 10–19 J = 3.61 × 10–19 J or = hν − W mvmax 2 +W hν = m vmax = × × 10−31 × (106 ) + 3.61 × 10−19 = 4.55 × 10−19 + 3.61 × 10−19 = 8.16 × 10−19 or ν = 8.16 × 10−19 6.6 × 10−34 = 1.23 × 1015 Hz S Chand & Company Limited 17 Work function W = 2.5 eV = 2.5 × 1.6 × 10–19 = × 10–19 J Now W = hc λ0 λ0 = hc 6.62 × 10 −34 × × 108 = W × 10 −19 or = 5.0 × 10 −7 m λ= 18 h = p h 2mE E = 400 eV = 400 × 1.6 × 10–19 J = 6.4 × 10–17 J So, λ= 6.6 × 10 −34 × × 10 −31 × 6.4 × 10 −17 = 0.62 × 10 − 10 m 19 Work function W = hν0 = 6.63 × 10–34 × × 1014 = 39.78 × 10–20 J S Chand & Company Limited Now, or eV0 = hν – W ν= = eV0 + W h (1.6 × 10 −19 × 3) + (39.78 × 10−20 ) 6.6 × 10 −34 = 13.2 × 1014 Hz 20 E = 54 eV = 54 × 1.6 × 10–19 J = 86.4 × 10–19 J p = mE momentum = × 9.1 × 10 −31 × 86.4 × 10 −19 = 3.96 × 10–24 kg m/s = 4.0 × 10–24 kg m/s de Broglie wavelength λ = = h p 6.6 × 10 −34 3.96 × 10 −24 = 1.66 × 10 −10 m o = 1.66 A S Chand & Company Limited 21 Planck’s Constant h = (Slope of cut-off voltage versus frequency graph) × (charge of an electron) = 4.12 × 10–15 × 1.60 × 10–19 = 6.59 × 10–34 Js 22 de Broglie wavelength λ = h = h = h mE m (3 / 2) kT mkT where T is the absolute temperature and k is the Boltzmann constant : T = 27 + 273 = 300 K k = 1.38 × 10–23 J/K Mass of one helium atom m= ⇒ M × 10−3 = = × 10−26 kg N × 1023 6.6 × 10−34 λ= (3 × × 10−26 × 1.38 × 10 −23 × 300) = 0.73 × 10 −10 m o = 0.73 A 23 We have Emax = eV0 = 1.6 × 10–19 × 1.5 = 2.4 × 10–19 J S Chand & Company Limited 24 Here o λ = 5000 A = 5000 × 10–10 m W = 1.9 eV = 1.9 × 1.6 × 10–19 J (a) Energy of a photon E = = = hc λ 6.6 × 10 − 34 × × 10 5000 × 10 −10 J 6.6 × 10 −34 × 10 × 5000 × 10 −10 × 1.6 × 10 −19 eV 2.48 eV = 2.5 eV (b) Emax = hν – W = 2.48 – 1.90 = 0.58 eV (c) Emax = eV0 E or V0 = max e 0.58 × 1.6 × 10−19 = 1.6 × 10 −19 = 0.58 V S Chand & Company Limited 25 Incident photon energy is E= hc 6.6 × 10 −34 × × 108 = J λ 3300 × 10−10 = 6.6 × × 10 −26 3300 × 10−10 × 1.6 × 10−19 eV = 3.76 eV For photoelectric emission, E should be equal to or greater than the work function So Mo and Ni will not give photoelectric emission 26 Photon energy is hc E= × eV λ e 6.6 × 10−34 × × 108 = = eV 400 × 10−9 × 1.6 × 10−19 Since E > eV, metal X will give photoelectric emission S Chand & Company Limited