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+ # PROBLEM 21 #
+Standard formula used
+ 1/2*m*v^2 = (h * c)/ lambda
+
+ Maximum kinetic energy is 2.314286e-19 J.
+ Number of electrons emitted per sec from 1cm^2 area is 2.160494e+12 .
diff --git a/1271/CH14/EX14.21/example14_21.sce b/1271/CH14/EX14.21/example14_21.sce
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+clc
+// Given that
+lambda = 3.5e-7 // wavelength of light in meter
+i = 1 // intensity in W/m^2
+p = 0.5 // percent of incident photon produce electron
+a = 1 // surface area of potassium in cm^2
+w = 2.1 // work function of potassium in eV
+h = 6.62e-34 // Planck constant in J-sec
+c = 3e8 // speed of light in m/sec
+e = 1.6e-19 // charge on an electron in C
+m = 9.1e-31 // mass of an electron in kg
+// Sample Problem 21 on page no. 14.28
+printf("\n # PROBLEM 21 # \n")
+printf("Standard formula used \n ")
+printf(" 1/2*m*v^2 = (h * c)/ lambda\n")
+E = (((h * c) / lambda) * (1 / e) - w) * e
+E_ = (p * a * 1e-4) / 100 // in W/cm^2
+n = E_ / E
+printf("\n Maximum kinetic energy is %e J.\n Number of electrons emitted per sec from 1cm^2 area is %e .",E,n)
+