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Physics Paper 2, Nov/Dec. 2009  
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Main
General Comments
Weakness/Remedies
Strength












Question 15
Part II:      Candidates were required to answer any five questions from this part.
  1. (a) (i)   A radioactive substance emits α, β and γ radiations.  Arrange them in order of decreasing;
                (A)   penetrating power;
    (B)   ionizing power.

         (ii)   With the aid of a labelled diagram, explain how the radiations in (i) above are affected
    by a magnetic field placed perpendicularly to the source.

    (b)         State one factor each that is responsible for the:
                (i)         occurrence of photoemission;
                (ii)        number of electrons emitted during photoemission;
                (iii)       energy of photoelectrons.

    (c)        A photon has a frequency of 5.02 x 1014 Hz.
                Calculate the energy of the photon in:
                (i)         joules;
                (ii)        electron volts.
                            [h = 6.6 x 10-34 Js;  1eV = 1.6 x 10-19 J]

     

_____________________________________________________________________________________________________
Observation

Part (a) was well attempted.  However, many of the candidates drew the expected diagram carelessly and so lost valuable marks.  One such example is given below:

Part (b) was fairly well attempted.
Part (c) was well attempted by the responding candidates.

The expected answers are:
(a)      (i)          (A)       gamma rays → beta rays →  alpha particles.             

                        (B)       alpha particles → beta rays → gamma rays.     
              
            Accept:    γ → β →  α   and
                            α → β  → γ
       (ii)

β ray is strongly deflected  (because it is negatively charged.                                    

    1. α particle is deflected but not as strong as the  β  - beta (because it is massive)
    2. γ rays is not affected (because it has neither  charge nor mass).                     

 (i)  -     The energy/frequency of incident radiation must be greater than thework function/threshold frequency of the metal       
(ii)  -   the intensity of the radiation 
(iii)- Energy/frequency of incident radiation/ work function of the metal irradiated    
         
     (c)  E = hf                                                                                                                    
=  6.6 x 10-34 x 5.02 x 1014                                                                                  

=  3.31  x 10-19 J 
1 e V = 1.6 X 10-19 J
∴  3.31 x 10-19 J  =  2.07  e V

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