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authorpriyanka2015-06-24 15:03:17 +0530
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+ # PROBLEM 13 #
+Standard formula used
+ E_k = h*c/lambda
+
+ From the above it is clear that the energy corresponding to wavelength 5000 A is i.e. 2.482500 found to be less than the work function i.e. 4.8 eV . So it will not be able to liberate an electron.
+ As the energy to wavelength 2000 A i.e. 6.206250 is greater than the work function. So it is sufficient to liberate electrons.
diff --git a/1271/CH14/EX14.13/example14_13.sce b/1271/CH14/EX14.13/example14_13.sce
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+clc
+// Given that
+w = 4.8 // work function in eV
+lambda1 = 5e-7 // wavelength of incident radiation in first case in meter
+lambda2 = 2e-7 // wavelength of incident radiation in second case 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
+// Sample Problem 13 on page no. 14.24
+printf("\n # PROBLEM 13 # \n")
+printf("Standard formula used \n ")
+printf(" E_k = h*c/lambda \n")
+E_k1 = h*c/lambda1
+E_k2 = h*c / lambda2
+printf("\n From the above it is clear that the energy corresponding to wavelength 5000 A is i.e. %f found to be less than the work function i.e. 4.8 eV . So it will not be able to liberate an electron.\n As the energy to wavelength 2000 A i.e. %f is greater than the work function. So it is sufficient to liberate electrons. ",E_k1/e,E_k2/e)
+