Answer to Question 4 (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.)