Problem Solutions (Part 2)

(1) Photo-electrons with minimum velocity imply that K _min   applies here  so indicates the wavelength of incident light (l = 0.50 mm) is at the cutoff wavelength so: l = l _c     Then the cutoff or threshold frequency is:

 

f _c = c /l _c =    (3 x 10⁸ ms^-1)/ (500 x  10^-9 m) =  6 x 10 ¹⁴ s^-1

 

Then work function would be: f   = hf _c  =    

(6.62 x 10 ^-34 Js)   ( 6 x 10 ¹⁴ s^-1)  

 

=   3.97  x 10 ^-19 J  =    (3.97  x 10 ^-19 J )/ 1.60  x 10 ^-19 J /eV) = 2.48 eV

 

(2) Stopping potential V_s = 0.54 V and l = 750 nm

 

f=  c/ l =   (3 x 10⁸ ms^-1)/ (750 x  10^-9 m) =  4 x 10 ¹⁴  s^-1

 

The work function is obtained from: eV_s =  hf  -  f

 

So, transposing:   f  =  hf -    eV_s     =  

 

  (6.62 x 10 ^-34 Js)  (4 x 10 ¹⁴ /s) -   (1.6 x 10 ^-19 C) 0.54 J/C]

 

f  =   2.64 x 10 ^-19 J  -   0.86 x  10 ^-19 J    = 1.78 x  10 ^-19 J   

 

In eV:  f  =    1.78 x  10 ^-19 J / (1.6 x 10 ^-19 J/ eV) =  1.11 eV     

 

 

(3) K _max  =   0.57 eV   and photo-electrons dislodged from a metal surface by incident radiation with l = 500 nm.

 

The incident energy E = hf = h c/l  = 

  (6.62 x 10 ^-34 Js) (3 x 10⁸ ms^-1)/ (500 x  10^-9 m)

 

 h c/ l = 3.97  x 10^-19 J

 

K _max  =   0.57 eV (1.6 x 10 ^-19 J/ eV) = 0.91 x 10 ^-19 J

 

Therefore:

K _max  =  0.91 x 10 ^-19 J

 

 = [3.97  x 10^-19 J  -   f ]

 

So, the work function is: f  =  [3.97  x 10^-19 J    -  0.91 x 10 ^-19 J]  =  3.06 x 10 ^-19 J

 

To get in eV:   (3.06 x 10 ^-19 J)/ (1.6 x 10 ^-19 J/ eV) =  1.91 eV

 

The stopping potential in volts:

 

V_s =   hf/ e   -  f/e    where the slope h/e = 4.13 x 10^-15 Js/C

 

The frequency f = c/l  =    (3 x 10⁸ ms^-1)/ (500 x  10^-9 m) = 

6 x 10 ¹⁴ c/s

 

Then:

 

V_s = (4.13 x 10^-15 J s/C )(6 x 10 ¹⁴ s^-1) - 

(3.06 x 10 ^-19 J)/ (1.6 x 10 ^-19 C)

 

V_s   = 2.47 V – 1.91 V =  0.56 V