Photoelectric Effect Questions and Answers - Physics Form 4 Topical Revision

Questions

1. Light of frequency 5.5x 10¹⁴ Hz is made to strike a surface whose work function is 2.5ev. Show that photoelectric effect will not take place. H= 6.6 X 10³⁴Js
2. Photoelectrons emitted by illuminating a given metallic surface constitute a “photocurrent”. What is the effect of increasing the intensity of the illumination of the magnitude of the photocurrent?
3. The diagram below shows a photocell in action
   
   
   1. The photocell is either evacuated or filled with an inert gas at low pressure. Give one reason for this
   2. What is the function of the resistor R in the circuit?
   3. State one reason for using a particular radiation such as ultraviolet for a given photocell.
   4. Explain how the set-up shown in the diagram may be used as an automatic switching device for a burglar alarm.
4. A monochromatic beam of radiation is directed on a clean metal surface so as to produce photoelectrons. Give a reason why some of the ejected photoelectrons have more kinetic energy than others.
5.  
   1. Describe with the aid of a labelled diagram an experimental set-up for observing the photoelectric effect.
   2. The table shows the relationship between the wavelength of a radiation falling on a surface and the energy, k of the emitted electrons.
      

      λ (m) × 10-7	1.0	1.5	1.0	0.5
      K(J) × 10-19	10	13	20	40

      1. Plot a graph of energy k(Y-axis) against the frequency, f, of the incident light.
      2. Determine the work function of the surface used (h=6.663 x 10^-34Js)
6. Name a device used to convert light energy directly into electric energy.
7. Electrons emitted from a metal when light of a certain frequency is shone on the metal are found to have a maximum energy of 8.0 x 10^-19 J. If the work function of the metal is 3.2 x 10^-19 J, determine the wavelength of the light used.
8. The figure below shows ultra violet light striking a polished zinc plate placed on a negatively charged gold leaf electroscope.
   
   Explain the following observation
   1. The leaf of the electroscope falls.
   2. When the same experiment was repeated with a positively charged electroscope, the leaf did not fall.
9. The work function of a certain material is 3.2 ev. Determine the threshold frequency for the material. (1 electron volt (eV) = 1.6 x 10^-19 and planks constant H= 6.62 x 10^-34Js)
10. State what is meant by the term accommodation as applied to the human eye. (1 mk)
11. The graph in the figure below shows the variation of photoelectric current with applied voltage when a surface was illuminated with light of a certain frequency.
    
    Use this information in the figure to answer questions a and b.
    1. On the same axes, sketch the graph of when light of higher intensity but same frequency is used to illuminate the surface. (1 mk)
    2. Explain your answer in a above . (1 mk)
12. Calculate the energy of a photon of red light and ultra-violet light 
    (λ_R = 7.0 x 10-⁷m: λ_v = 4.0 x 10^-7m)
13. The wavelength of light from a sodium lamp is 5.9 x 10^-7m. A 200W sodium vapour has an efficiency of 40%. Calculate:
    1. The energy of one quantum of sodium light.
    2. The number of quanta emitted in one second
14. The threshold frequency for potassium is 5.37 x 10¹⁴ Hz. When the surface of potassium is illuminated by another radiation, photoelectrons are emitted with a speed of 7.9 x 10⁵ m/s
    Calculate:
    1. The work function for potassium
    2. The k.e of the photoelectrons
    3. The frequency of the second source
15. Explain the term “work function”
16. A metal has a work function of 2eV. Calculate the threshold wavelength of the metal given that e= 1.6 x 10-¹⁹C and h= 6.63 x 10^-19C and Me = 9x 10^-31kg.

Answers

1. Energy (incoming) = hf = 6.6 x 10^-34 x 5.5 x 10¹⁴
   = 3.63 x 10^-19 J
   = 3.63 x 10^-19 = 2.27 eV
       1.6 x 10^-19 
   This energy is less than work function hence no photoelectric emission.
2. Higher photocurrents or more photoelectrons produced.
3. 1. Reduce/ prevent collisions of electrons with air molecules and hence increase current
   2. Control/ limit current, lowers current
   3. The energy of the radiation must be greater than the work function of the emitting surface.
   4. Current flows when uv falls on the cathode; interruption of the uv beam cuts off the circuit: use with relay to switch on a second circuit with alarm.
4. Electrons ejected from inside the metal lose more energy on the way out while those on the surface require very little work function to be removed.
5.  
   1.  
      
   2. Work function is given by = hf_o
      fo is the x- intercept in the graph fo (from graph)
      = 1.2 x 10¹⁵
      Φ= 6. 63 x 10^-34 js x 1.2 x 10¹⁵
      = 7.95 x 10^-19
6. Solar cell (photovoltaic); Photocell/ photo electric cell
7. hf = hf_o + ½ mv²
   Energy of photon (hf) = w.f + k.e
                                    = 3.2 × 10^-19 + 8.0 × 10^-19
                                     = 11.2 × 10^-19
   f = c 
        λ
   ^hc/_λ = 11.2 × 10^-19
   λ = 3.0 x 10⁸ x 6.64 x 10^-34
               11.2 x 10^-19
   = 1.76 x 10^-7m
8.  
   1. Photo – electric effect takes place releasing the extra electrons.
   2. The electrons released are attracted back by the positive charge.
9. hf₀ = 3.2 x 1.6 x 10^-19 J
             6. 62 x 10^-34
   f₀ = 7.76 x 10¹⁴Hz
10.  The process of the eye lens being adjusted to focus objects at various distances
11.  
    1.  
       
    2. - The higher the intensity
       - Implies greater number of electrons and hence
       - Higher saturation current
12.  ER = 2.83 x 10^-19J;       4.95 x 10^-19J
13.  
    1. 3.37 x 10^-19J
    2. 2.37 x 10^-20J
14.  
    1. 3.56 x 10^-19J
    2. 2.84 x 10 ^-19J
    3. 9.7 x 10¹⁴Hz
15. This is the minimum amount of work required to free an electron from a metal surface
16. No = hf₀
    f₀ = w₀
           h
    = 2.0 x 1.6 x 10^-19 Hz
         6.63 x 10^-34
    = 4.85 x 10¹⁴ Hz
    λ_o = ^V/_f
    = 3.0 x 10⁸ ms^-1 Hz
       4.85 x 10¹⁴s^-1
    = 6.2 x 10^-7m
17. eV = ½ meV²
    V² = 2ev
            Me
    
    = 1.3 x 10⁷ ms^-1