Suppose n electrons can be placed in either of two...

Suppose $N$ electrons can be placed in either of two configurations. In configuration $1,$ they are all placed on the cir- cumference of a narrow ring of radius $R$ and are uniformly distrributed so that the distance between adjacent electrons is the same everywhere. In configuration $2, N-1$ electrons are uniformly distributed on the ring and one electron is placed in the center of the ring. (a) What is the smallest value of $N$ for which the second configuration is less energetic than the first? (b) For that value of $N,$ consider any one circumference electron $-$ call it $e_{0} .$ How many other circumference electrons are closer to $e_{0}$ than the central electron is?

Transcript

00:01 Hello friends here it is given.

00:05 Suppose an electrons can be placed either of two configuration.

00:38 Configuration 1 electrons are placed on the circumference of a ring of radius r so that distance between two adjacent electrons to be same.

01:42 Configuration second n minus 1 electrons are uniformly distributed over the ring and one electron at its center you have to find the value of n for which energy of second configuration is less than that of first second for the given value of n consider any value of an circumference electron call it e not how many other circumference electrons are close to e not then that of central electron let us hear the solution let we have a system of e not e1 and e n electrons then total potential energy of the system can be defined as summation e square upon 4 pi a epsilon not r i rj here you may write e square upon 4 pi epsilon not summation of r i rj this can be defined as just a moment i have to correct n times summation of i is cut to 1 to n minus 1 1 upon r oi now configuration 1 r i j having the value 2 r sign of j minus i upon n into pi so n by 2 summation i is equal to 1 to n minus 1 1 upon r o i having the value n minus 1 upon i is called to 1 into 1 upon 2 r sign of j minus i upon n into pi so on solving it you may write n upon 4r submission of i skull to 1 to n minus 1 kossack of i upon n into pi so energy of one configuration you will get e square upon 4 pi epsilon n upon 4 r submission of i scut to 1 2 n minus 1 into cossack of i into pi upon n.

09:07 Now for configuration 2, from the center of the ring, we divide the circuit into n -minus -1 angles having value 2 -n -minus 1 upon, sorry, 2 -py upon n -minus 1.

10:14 So energy of the system in second configuration we can measure e square upon 4 pi epsilon not n minus 1 upon 4 r summation of i's equal to 1 to n minus 2 cossack i into pi upon n minus 1 plus n minus 1 into r now we have to form the table for initial energy of one configuration to other from electron n is called to 2 to 16.

11:28 See here, table for u1 and u2.

11:47 For n from 2 to 16 and u1 u2.

11:58 For n is called to 2, u1 is 0 .5, u1 is 1.

12:04 For 3 electron it is 1 .73 and 2 .5.

12:11 For 4 electron it is 3 .83, 4 .73.

12:19 For 5 electron, it is 6 .88, 7 .83.

12:28 For 7, 16 .13, 16 .96.

12:37 For 8.

12:38 22 .44, 23 .13 .13.

12:45 For 9.

12:48 29 .92.

12:51 30 .44.