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Physics Paper 2, Nov/Dec. 2007  
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Main
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Question 15

(a) Explain the following terms:
(i) ground state energy;
(ii) excitation energy;
(iii) emission line spectra.

(b)  The surface of a metal of work function 3.20 x 10-19 J is illuminated with A radiation of wavelength 3.50 x 10-7 m.  Calculate the
(i)  frequency of the incident radiation;
(ii) energy of the incident radiation;
(iii) kinetic energy of the most energetic electrons.

                        [ c = 3.0 x 108 ms-1,  e = 1.6 x 10-19C,  h = 6.6 x 10-34 Js]
_____________________________________________________________________________________________________
Observation

This question on energy quantization attracted very few candidates.  The performance was also poor.

In part (a)(i), (ii) and (iii) the explanations of ‘ground state energy’, ‘excitation energy’ and ‘emission line spectra’ were poorly rendered as candidates did not show any acceptable trace of understanding of the subject matter.  Teachers need to develop greater effort for teaching of atomic and nuclear physics.
Part (b)(i0, (ii) and (iii), were numerical problems and was fairly well handled by most of the candidates.  They were able to recall the formular, substitute and evaluate correctly.           

The expected answers are:
(i) Ground state energy  - least energy of an atom
Valid additional information
     -  Atoms are most stable at this state

(ii) Excitation energy 
- energy required to raise the state of an atom from its ground   state to any other allowed state Valid additional information  e.g
-  excitation energy is quantized  

(iii) Emission line spectra consists of distinct and separate bright lines of

definite wavelengths/frequencies on a dark back ground 
Valid additional information e.g
- they result from excited atoms/molecules of (certain) gases
-  can be used to identify elements/compounds          
                                                     
(b)(i)    frequency  f  =   c/l    
                                                                        
               =   3  x 108
                   3.5  x 10-7                        
=  8.57  x 1014     Hz    

 (ii)  E  =  hf  
=  6.6  x  10-34  x  8.57   x   1014    
=  5.66   x  1019   J   

(iii)  Ke max    =  hf  -  Wo  
  =  5.66   x  10-19  -  3.20   x  10-19  
 =   2.46  x  10-19  J

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