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authorpriyanka2015-06-24 15:03:17 +0530
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+ # PROBLEM 20 #
+Standard formula used
+ E = (h * c) * (1 / lambda1 - 1 / lambda2)
+ E = (h * c)/ lambda
+
+ Threshold wavelength is 5642 Angstrom.
+ Incident electromagnetic wavelength is 2002.016129 Angstrom
diff --git a/1271/CH14/EX14.20/example14_20.sce b/1271/CH14/EX14.20/example14_20.sce
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+clc
+// Given that
+k = 4 // maximum kinetic energy of electron in eV
+w = 2.2 // work function of sodium 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
+// Sample Problem 20 on page no. 14.27
+printf("\n # PROBLEM 20 # \n")
+printf("Standard formula used \n ")
+printf(" E = (h * c) * (1 / lambda1 - 1 / lambda2) \n E = (h * c)/ lambda \n")
+lambda_ = ((h * c) / (w * e))
+lambda = (1 / ((((k * e) / (h * c))) + (1 / lambda_)))
+printf("\n Threshold wavelength is %d Angstrom.\n Incident electromagnetic wavelength is %f Angstrom",lambda_ * 1e10,lambda * 1e10)
+