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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+clear;
+clc;
+printf("\nEx3.13\n");
+//page no.-130
+//given
+x=10^-14;......//diameter of nucleus in m
+h=6.626*10^-34;......//planck's constant in J-s
+m=1.67*10^-27;......//proton mass in kg
+e=1.6*10^-19;.......//charge in C
+c=3*10^8;...........//speed in m/sec
+
+p=h/(2*%pi*x).......//momentum in kg*m/s from uncertainty principle
+
+printf("\n minimum value of momentum is 1.055*10^-20 kg*m/s\n");
+
+E=p^2/(2*m*e).........//minimum kinetic energy in MeV
+
+printf("\nminimum K.E. is 0.23 Mev\n");
+
+r=m*c^2/e...........//rest mass energy of proton in MeV
+
+printf("\nrest mass energy is 942 MeV\n");
+printf("\nAS REST MASS IS VERY LARGE THAN K.E. SO,PRESENCE OF SUCH PROTONS IN NUCLEUS IS PERMITTED\n");