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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+clc;
+clear;
+h=6.63*10^-34 //Plancks constant in J-s
+v=3*10^8 // frequency of light in Hz
+e=1.6*10^-19 //charge in C
+lambda=200*10^-9 //wavelength in m
+W=4.2 //work function in Joule
+c=3*10^8 //velocity of light in m/s
+
+//calculation
+E=(h*v)/(lambda)//energy in J
+E_v=E/e //energy in eV
+
+//case (1)
+E_k=E_v-W
+mprintf("\nThe kinetic energy of the fastest electrons is = %d eV\n",E_k)
+
+//case(2)
+mprintf("\nThe kinetic energy of slowest electrons is zero.As the emitted electrons have all possible energies from 0 to certain maximun value is E_k\n")
+
+//case(3)
+mprintf("If V_s is the stopping potential then E_k=e*V_s.Since the electrons have a maximum kinetoc energy of 2eV,the stopping potential is 2V.\n")
+
+//case(4)
+lambda_0=(h*c)/(W*e)
+mprintf("The cut off wavelength for aluminium is %1.2e m",lambda_0) //The answer provided in the textbook is wrong.