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authorpriyanka2015-06-24 15:03:17 +0530
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+ # PROBLEM 15 #
+Standard formula used
+ E = (h * c) * (1 / lambda1 - 1 / lambda2)
+
+ Stopping potential is 0.411689 V.
+ Maximum kinetic energy is 6.587019e-20 J.
diff --git a/1271/CH14/EX14.15/example14_15.sce b/1271/CH14/EX14.15/example14_15.sce
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+clc
+// Given that
+lambda = 5.89e-7 // wavelength of light in meter
+lambda_ = 7.32e-7 // threshold wavelength of photoelectron in meter
+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 15 on page no. 14.25
+printf("\n # PROBLEM 15 # \n")
+printf("Standard formula used \n ")
+printf(" E = (h * c) * (1 / lambda1 - 1 / lambda2) \n")
+E = (h * c) * (1 / lambda - 1 / lambda_)
+V = E / e
+printf("\n Stopping potential is %f V.\n Maximum kinetic energy is %e J.",V,E)
+