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42.photo electric effect

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					      PHOTO ELECTRIC EFFECT AND WAVE PARTICLE QUALITY
                        CHAPTER 42
1.   1 = 400 nm to 2 = 780 nm
                 hc                           –34                     8
     E = h =               h = 6.63  10           j - s, c = 3  10 m/s, 1 = 400 nm, 2 = 780 nm
                  
            6.63  10 34  3  108       6.63  3
                                                    10 19 = 5  10 J
                                                                    –19
     E1 =                  9
                                      
                400  10                     4
           6.63  3
                     10 19 = 2.55  10 J
                                        –19
     E2 =
             7.8
                              –19           –19
     So, the range is 5  10 J to 2.55  10 J.
2.    = h/p
                      6.63  1034              –27            –27
      P = h/ =                J-S = 1.326  10 = 1.33  10 kg – m/s.
                   500  10 9
                             –9                        –9
3.   1 = 500 nm = 500  10 m, 2 = 700 nm = 700  10 m
     E1 – E2 = Energy absorbed by the atom in the process. = hc [1/1 – 1/2]
                                –19            –19
      6.63  3[1/5 – 1/7]  10 = 1.136  10 J
4.   P = 10 W  E in 1 sec = 10 J        % used to convert into photon = 60%
     Energy used = 6 J
                                                            6.63  10 34  3  108       6.633
     Energy used to take out 1 photon = hc/ =                             9
                                                                                                10 17
                                                         590  10                          590
                                     6            6  590                                    19
     No. of photons used =                                 1017  176.9  1017 = 1.77  10 
                              6.63  3            6.63  3
                                        10 17
                                590
                                      3      2                      power            3     2
5.   a) Here intensity = I = 1.4  10 /m            Intensity, I =        = 1.4  10 /m
                                                                     area
        Let no.of photons/sec emitted = n             Power = Energy emitted/sec = nhc/ = P
                         2
        No.of photons/m = nhc/ = intensity
            int ensity      1.9  103  5  10 9
        n=                                           3.5  1021
                  hc         6.63  10 34  3  108
     b) Consider no.of two parts at a distance r and r + dr from the source.
        The time interval ‘dt’ in which the photon travel from one point to another = dv/e = dt.
                                                                   p  dr
        In this time the total no.of photons emitted = N = n dt =  
                                                                   hc  C
         These points will be present between two spherical shells of radii ‘r’ and r+dr. It is the distance of the
      st
     1 point from the sources. No.of photons per volume in the shell
                        N      Pdr       1       p
        (r + r + dr) =                      
                      2r2dr     hc 2      2
                                        4r ch 4 hc 2r 2
                              11                          –9
        In the case = 1.5  10 m,  = 500 nm, = 500  10 m
            P                                           3      P     
                  1.4  103 ,  No.of photons/m =
         4r 2                                               4r 2 hc 2
                      3         500  10 9
        = 1.4  10                            1.2  1013
                      6.63  10 34  3  108
                                                 2
     c) No.of photons = (No.of photons/sec/m )  Area
                   21       2
        = (3.5  10 )  4r
                  21                     11 2           44
        = 3.5  10  4(3.14)(1.5  10 ) = 9.9  10 .

                                                                   42.1
                                                                              Photo Electric Effect and Wave Particle Quality
                    –9                                 19
6.    = 663  10 m,  = 60°, n = 1  10 ,  = h/p
                    –27                                                                                               60°
      P = p/ = 10
     Force exerted on the wall = n(mv cos  –(–mv cos )) = 2n mv cos .
                    19    –27             –8
        = 2  1  10  10  ½ = 1  10 N.
7.   Power = 10 W       P  Momentum
         h              h            P h
     =         or, P =          or,  
         p                          t t
         hc            E hc
     E=            or,         = Power (W)
                       t    t
     W = Pc/t      or, P/t = W/c = force.
     or Force      = 7/10 (absorbed) + 2  3/10 (reflected)
                      7 W          3 W        7   10          3   10
                   =         2                     2    
                     10 C         10 C       10 3  108      10 3  108
                              –8          –8
                   = 13/3  10 = 4.33  10 N.
8.   m = 20 g
     The weight of the mirror is balanced. Thus force exerted by the photons is equal to weight
          h              hc
     P=            E=        PC
                         
       E P
          C
        t   t
      Rate of change of momentum = Power/C
       30% of light passes through the lens.
       Thus it exerts force. 70% is reflected.
      Force exerted = 2(rate of change of momentum)
                = 2  Power/C
                        2  Power 
                   30%              mg
                             C    
                 20  10 3  10  3  108  10
      Power =                                   = 10 w = 100 MW.
                              23
9.   Power = 100 W
     Radius = 20 cm
     60% is converted to light = 60 w
                    power         60
     Now, Force =                       2  10 7 N .
                   velocity 3  108
                   force     2  107           1
     Pressure =                                     105
                   area 4  3.14  (0.2)2
                                            8  3.14
                                     –5                –7          –7         2
                   = 0.039  10           = 3.9  10        = 4  10    N/m .
10. We know,
    If a perfectly reflecting solid sphere of radius ‘r’ is kept in the path of a parallel beam of light of large
    aperture if intensity is I,
              r 2l
     Force =
               C
                  2                    8
     I = 0.5 W/m , r = 1 cm, C = 3  10 m/s
                  (1)2  0.5       3.14  0.5
     Force =             8
                                 
                    3  10     3  108
                           –8        –9
              = 0.523  10 = 5.2  10 N.

                                                                       42.2
                                                                                Photo Electric Effect and Wave Particle Quality
11. For a perfectly reflecting solid sphere of radius ‘r’ kept in the path of a parallel beam of light of large
                                                 r 2I
    aperture with intensity ‘I’, force exerted =
                                                  C
12. If the i undergoes an elastic collision with a photon. Then applying energy conservation to this collision.
                         2        2
    We get, hC/ + m0c = mc
    and applying conservation of momentum h/ = mv
                           m0
    Mass of e = m =
                        1 v2 / c2
    from above equation it can be easily shown that
    V=C          or      V=0
    both of these results have no physical meaning hence it is not possible for a photon to be completely
    absorbed by a free electron.
13. r = 1 m
                    kq2 kq2
     Energy =          
                     R   1
             kq2   hc                         hc
     Now,        =                 or  =
              1                              kq2
     For max ‘’, ‘q’ should be min,
                                 –19
     For minimum ‘e’ = 1.6  10 C
              hc                3
     Max  =       = 0.863  10 = 863 m.
              kq2
                                                      6.63  3  10 34  108           863
     For next smaller wavelength =                                                         = 215.74 m
                                                    9  109  (1.6  2)2  10 38        4
                                     –9
14.  = 350 nn = 350  10                 m
     = 1.9 eV
                            hC        6.63  10 34  3  108
     Max KE of electrons =                                   1.9
                                    350  109  1.6  10 19
                = 1.65 ev = 1.6 ev.
                  –19
15. W 0 = 2.5  10 J
    a) We know W 0 = h0
             W0     2.5  1019           14            14
          0 =                 = 3.77 10 Hz = 3.8  10 Hz
              h    6.63  1034
     b) eV0 = h – W 0
                h  W0 6.63  10 34  6  1014  2.5  10 19
          or, V0 =                                             = 0.91 V
                   e                 1.6  1019
                           –19
16.  = 4 eV = 4  1.6  10 J
    a) Threshold wavelength = 
        = hc/
            hC 6.63  10 34  3  108 6.63  3 10 27
      =      =                               9  3.1 107 m = 310 nm.
                    4  1.6  10 19      6.4  10
     b) Stopping potential is 2.5 V
      E =  + eV
                           –19         –19
      hc/ = 4  1.6  10 + 1.6  10  2.5
                 6.63  1034  3  108
                                         = 4 + 2.5
                       1.6  10 19
          6.63  3  1026                             –7
                   19
                                   = 1.9125  10            = 190 nm.
          1.6  10          6.5
                                                                         42.3
                                                                     Photo Electric Effect and Wave Particle Quality
17. Energy of photoelectron
               2     hc          4.14  10 15  3  108
     ½ mv =             hv 0 =                          2.5ev = 0.605 ev.
                                       4  10 7
                     P2       2
    We know KE =           P = 2m  KE.
                     2m
     2                –31                –19
    P = 2  9.1  10  0.605  1.6  10
                   –25
    P = 4.197  10 kg – m/s
                           –9
18.  = 400 nm = 400  10 m
    V0 = 1.1 V
        hc hc
                ev 0
            0
        6.63  10 34  3  108       6.63  10 34  3  108
                         9
                                                              1.6  10 19  1.1
               400  10                         0
                   19.89  10 26
     4.97 =                       1.76
                        0
        19.89  10 26
                      = 4.97 – 17.6 = 3.21
             0
           19.89  10 26             –7
     0 =                = 6.196  10 m = 620 nm.
               3.21
19. a) When  = 350, Vs = 1.45
       and when  = 400, Vs = 1
             hc
               = W + 1.45                      …(1)                                            Stopping
            350
                                                                                                potential
              hc
        and       =W+1                    …(2)
             400                                                                                            1/ 
        Subtracting (2) from (1) and solving to get the value of h we get
                    –15
        h = 4.2  10 ev-sec
                                         hc
    b) Now work function = w =              = ev - s
                                          
            1240
        =         1.45 = 2.15 ev.
             350
              hc                    hc
    c) w =         there cathod 
                                   w
         1240
        =        = 576.8 nm.
          2.15
                                         45
20. The electric field becomes 0 1.2  10 times per second.
                         1.2  1015           15
     Frequency =                   = 0.6  10
                              2
      h = 0 + kE
     h – 0 = KE
               6.63  10 34  0.6  1015
     KE =                           2
                     1.6  10 19
       = 0.482 ev = 0.48 ev.
                           7   –1
21. E = E0 sin[(1.57  10 m ) (x – ct)]
                   7
    W = 1.57  10  C

                                                             42.4
                                                               Photo Electric Effect and Wave Particle Quality

         1.57  107  3  108
     f=                      Hz            W 0 = 1.9 ev
                  2
    Now eV0 = h – W 0
                          1.57  3  1015
                          –15
         = 4.14  10                     – 1.9 ev
                                2
         = 3.105 – 1.9 = 1.205 ev
                  1.205  1.6  10 19
    So, V0 =                     = 1.205 V.
                  1.6  1019
                          15 –1            15 –1
22. E = 100 sin[(3  10 s )t] sin [6  10 s )t]
                              15 –1              15 –1
        = 100 ½ [cos[(9  10 s )t] – cos [3  10 s )t]
                      15            15
    The w are 9  10 and 3  10
    for largest K.E.
                  w max   9  1015
         fmax =         =
                   2        2
    E – 0 = K.E.
     hf – 0 = K.E.
         6.63  1034  9  1015
                            2  KE 
           2  1.6  10 19
     KE = 3.938 ev = 3.93 ev.
23. W 0 = hv – ev0
             5  10 3
                     1.6  10 19  2 (Given V0 = 2V, No. of photons = 8  10 , Power = 5 mW)
                                                                              15
         =
           8  1015
         = 6.25  10–19 – 3.2  10–19 = 3.05  10–19 J
             3.05  10 19
         =              = 1.906 eV.
          1.6  10 19
24. We have to take two cases :
    Case I … v0 = 1.656
                           14
                 = 5  10 Hz
    Case II… v0 = 0                                                                V (in volts)
                           14
                 = 1  10 Hz
                                                                                       2
    We know ;
                                                                                   1.656
    a) ev 0  h  w 0
                                    14
                                                                                       1
       1.656e = h  5  10 – w0          …(1)
                   14
       0 = 5h  10 – 5w0                 …(2)                                                     1    2    3    4    5
       1.656e = 4w0                                                                                   v(in 1014 Hz)
             1.656
     w0 =          ev = 0.414 ev
               4
    b) Putting value of w0 in equation (2)
                      14
     5w0 = 5h  10
                              14
     5  0.414 = 5  h  10
                       –15
     h = 4.414  10 ev-s
25. w0 = 0.6 ev
    For w0 to be min ‘’ becomes maximum.
             hc        hc   6.63  10 34  3  108
    w0 =        or  =    =                         
                      w0     0.6  1.6  10 19
                     –7
    = 20.71  10          m = 2071 nm

                                                        42.5
                                                                         Photo Electric Effect and Wave Particle Quality
26.  = 400 nm, P = 5 w
                       hc    1242 
    E of 1 photon =       =        ev
                            400 
                                 5                5  400
    No.of electrons =                      
                         Energy of 1 photon 1.6  10 19  1242
                                  6
    No.of electrons = 1 per 10 photon.
                                                        5  400
    No.of photoelectrons emitted =
                                     1.6  1242  1019  106
                                     5  400
                                                        1.6  1019 = 1.6  10 A = 1.6 A.
                                                                               –6
    Photo electric current =
                             1.6  1242  106  10 19
                         –7
27.  = 200 nm = 2  10 m
                         hc   6.63  10 34  3  108             –19
    E of one photon =       =                         = 9.945  10
                                    2  107
                            1 10 7                11
    No.of photons =                    19
                                             = 1  10    no.s
                        9.945  10
                                              1 1011
                                                    7
    Hence, No.of photo electrons =          = 1  10
                                      10 4
    Net amount of positive charge ‘q’ developed due to the outgoing electrons
               7           –19          –12
       = 1  10  1.6  10 = 1.6  10 C.
    Now potential developed at the centre as well as at the surface due to these charger
           Kq 9  109  1.6  10 12         –1
        =                           = 3  10 V = 0.3 V.
            r        4.8  10 2
28. 0 = 2.39 eV
    1 = 400 nm, 2 = 600 nm
    for B to the minimum energy should be maximum
      should be minimum.
                                                                                                          X   X   X
         hc                                                                                     10 cm
                                                                                                                           A
    E=       0 = 3.105 – 2.39 = 0.715 eV.                                                               X   X   X
          
    The presence of magnetic field will bend the beam there will be no current if
    the electron does not reach the other plates.
            mv
       r=
            qB
              2mE
     r=
              qB

                 2  9.1 10 31  1.6  10 19  0.715
     0.1 =
                         1.6  10 19  B
                       –5
     B = 2.85  10 T
29. Given : fringe width,
       y = 1.0 mm  2 = 2.0 mm, D = 0.24 mm, W 0 = 2.2 ev, D = 1.2 m                             B
                                                                                                 A
           D                                                                                    B
                                                                                                          S
        y=
            d                                                                                    A
                                                                                                 B
              yd 2  10 3  0.24  10 3         –7
    or,  =                              = 4  10 m                                                                  A
              D            1.2
           hc 4.14  10 15  3  108
        E=                            = 3.105 ev
                       4  10
    Stopping potential eV0 = 3.105 – 2.2 = 0.905 V
                                                                  42.6
                                                               Photo Electric Effect and Wave Particle Quality
30.  = 4.5 eV,  = 200 nm
                                              WC
     Stopping potential or energy = E –  =        
                                                
    Minimum 1.7 V is necessary to stop the electron
    The minimum K.E. = 2eV
       [Since the electric potential of 2 V is reqd. to accelerate the electron to reach the plates]
    the maximum K.E. = (2+1, 7)ev = 3.7 ev.
31. Given
              –9   –2
     = 1  10 cm , W 0 (Cs) = 1.9 eV, d = 20 cm = 0.20 m,  = 400 nm
    we know  Electric potential due to a charged plate = V = E  d
    Where E  elelctric field due to the charged plate =/E0
    d  Separation between the plates.
                    1 10 9  20
     V=      d                      = 22.598 V = 22.6
          E0      8.85  10 12  100
                      hc         4.14  10 15  3  108
     V0e = h – w0 =       w0                           1.9 
                                       4  10 7
        = 3.105 – 1.9 = 1.205 ev
    or, V0 = 1.205 V
    As V0 is much less than ‘V’
    Hence the minimum energy required to reach the charged plate must be = 22.6 eV
    For maximum KE, the V must be an accelerating one.
    Hence max KE = V0 + V = 1.205 + 22.6 = 23.8005 ev
32. Here electric field of metal plate = E = P/E0
          1 10 19
     =               = 113 v/m
       8.85  10 12
     accl. de =  = qE / m
             1.6  10 19  113                  12                                            Metal plate
         =                        = 19.87  10
                9.1 10 31
                                                                                 y = 20 cm
          2y      2  20  10 2            –7
     t=                         = 1.41  10 sec
          a      19.87  10 31
            hc
     K.E. =      w = 1.2 eV
             
                          –19
        = 1.2  1.6  10 J [because in previous problem i.e. in problem 31 : KE = 1.2 ev]
                 2KE       2  1.2  1.6  10 19         –6
      V=                             = 0.665  10
                 m      4.1 10 31
     Horizontal displacement = Vt  t
                   –6         –7
      = 0.655  10  1.4  10 = 0.092 m = 9.2 cm.
33. When  = 250 nm
                              hc 1240
     Energy of photon =               = 4.96 ev
                                  250
                hc
      K.E. =       w = 4.96 – 1.9 ev = 3.06 ev.
                 
     Velocity to be non positive for each photo electron
     The minimum value of velocity of plate should be = velocity of photo electron
      Velocity of photo electron =          2KE / m


                                                       42.7
                                                                   Photo Electric Effect and Wave Particle Quality

               3.06            3.06  1.6  10 19             6
        =                                           = 1.04  10 m/sec.
            9.1 10 31           9.1 10 31
34. Work function = , distance = d
    The particle will move in a circle
    When the stopping potential is equal to the potential due to the singly charged ion at that point.
              hc
       eV0 =     
               
            hc    1  ke  hc    1                                                                   ion
     V0 =               
                 e  2d       e                                                                     d

        Ke2 hc      hc Ke2      Ke2  2d
                      
         2d            2d        2d
              hc 2d               2hcd       8 hcd
     =                                   2 0        .
               2
            Ke  2d            1
                                       2d e  80 d
                              40 e2
35. a) When  = 400 nm
                               hc 1240
        Energy of photon =               = 3.1 eV
                                    400
       This energy given to electron
       But for the first collision energy lost = 3.1 ev  10% = 0.31 ev
       for second collision energy lost = 3.1 ev  10% = 0.31 ev
       Total energy lost the two collision = 0.31 + 0.31 = 0.62 ev
       K.E. of photon electron when it comes out of metal
       = hc/ – work function – Energy lost due to collision
       = 3.1 ev – 2.2 – 0.62 = 0.31 ev
                 rd
    b) For the 3 collision the energy lost = 0.31 ev
       Which just equative the KE lost in the 3rd collision electron. It just comes out of the metal
       Hence in the fourth collision electron becomes unable to come out of the metal
       Hence maximum number of collision = 4.
    
                                                          




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Description: HC VERMA SOLUTIONS PHYSICS