Answer to Question 4

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(a)  Use:

 

 

 

 

    Temperature = 1000 + 273 = 1273 K (Who put 1000 into their equation?)

 

     ln (Probability)  = -((1.602 10-19 J eV-1 3.45 eV) (1.38 10-23 J K-1 1273 K)) = -(5.52 10-19 J 1.757 10-20 J) = -31.4

 

     Probability = e-31.4 = 2.26 10-9 (which is about 2 10-14  QED)

 

 

 

 

(b)  Use:

 

     J = 1.2 106 A m-2 K-2 (1273 K)2 2.26 10-14 = 0.0439 A m-2 = 0.044 A m-2 (2 s.f.)

 

 

 

 

(c) Area = p (1.0 10-3)2 = 3.14 10-6 m2.

 

 

     Current = current density area = 0.0439 A m-2 3.14 10-6 m2 = 1.38 10-9 A = 0.14 mA

 

 

 

(d) If the temperature is room temperature, the probability of the emission is very low since the result of f/kT is high.  As probability is e to the power of a large number, the probability becomes very low.

 

    The current density is the product of the Richardson Constant, the square of the temperature, and the  probability.  The square of a low temperature makes the current density extremely low.

 

 

    (The figures (not needed) are:  Probability = e-136 = 8.6 10-60.  J = 1.2 10-48 A m-2, not very high.)