1 APPLICATIONS OF SCHRÖDINGER S EQUATION AND BAND THEORY

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1 APPLICATIONS OF SCHRÖDINGER S EQUATION AND BAND THEORY 1.1 INTRODUCTION We hve lredy noted tht Schrödinger ws influenced by the mtter wve postulte of de Broglie.

1 1 APPLICATIONS OF SCHRÖDINGER S EQUATION AND BAND THEORY 1.1 INTRODUCTION We hve lredy noted tht Schrödinger ws influenced by the mtter wve postulte of de Broglie. In order to describe the behviour of prticle s wve, Schrödinger proposed n eqution in 196 to be known fter his nme s Schrödinger s Wve Eqution. The wve function ψ, which is solution of Schrödinger s eqution, is expected to describe the dynmic behviour or dynmicl stte of prticle. Schrödinger s eqution hs been found to be fundmentl eqution in Quntum Mechnics just s Newton s eqution F in Clssicl dp dt Mechnics or Mxwell s equtions in electromgnetism. Schrödinger gve two types of wve equtions nmely time dependent nd time independent. The time dependent Schrödinger s eqution is given s ψ ( rt, ) ψ t m Erwin Rudolf Josef Alexnder Schrödinger ( 1 August Jnury 1961) ws n Austrin physicist who chieved fme for his contributions to quntum mechnics. In Jnury 196, Schrödinger published in the Annlen der Physik the pper Quntistion s n Eigenvlue Problem on wve mechnics nd wht is now known s the Schrödinger eqution, for which he received the Nobel Prie in 19. i r t V r t r t (1.1) (, ) (, ) ψ(, ) nd time independent Schrödinger s eqution is given s m ψ () r E V () r ψ () r (1.) where the symbols crry their usul mening. Schrödinger s eqution is extremely useful for investigting vrious quntum mechnicl problems; vi prticle in one nd three dimensionl box, one dimensionl hrmonic oscilltor, step potentil nd potentil brrier. In the following subsections, we discuss two quntum mechnicl problems; prticle in three dimensionl box nd potentil brrier. 1. THE PARTICLE IN A THREE DIMENSIONAL BOX Let us consider prticle which is enclosed inside rectngulr box hving edges, b nd c in length s shown in Fig The potentil function V (x, y, ) is hving constnt vlue of ero in the regions given below: V (x, y, ) for < x <

2 Physics II V (x, y, ) for < y < b V (x, y, ) for < < c (1.) The potentil outside the box is infinite. The Schrodinger time independent wve eqution inside the box for such prticle my be written s ψ ψ ψ m E ψ y (1.4) To seprte out the bove eqution let us put Fig. 1.1 Three dimensionl box ψ ( xy,, ) Xx Yy Z XYZ (1.5) Substituting the vlue of ψ from Eq. (1.5), we get X Y Z m YZ ZX XY EXYZ y Dividing this eqution by XYZ, we get 1 X 1 Y 1 Z m E X Y y Z Z Y O c b X 1 X 1 Y 1 Z m E X Y y Z (1.6) me For the given energy of the prticle, the term is constnt nd ech term on the left side is function of one vrible only. If we llow only one of these (x or y or ) to vry t time, the sum of the three terms is still equl to the constnt on the right hnd side. This mens tht ech of the three terms on the left is itself constnt nd independent of the other vribles present in it. Let us represent ech term on left side of eqution, in succession, equl to the constnts α 1, α nd α. These hve minus sign becuse the term on right side of eqution hs minus sign. We, therefore, write 1 X α 1 X (1.7) X α 1 X (1.8) 1 Y α Y y (1.9) Y α Y y (1.1) 1 Z α Z (1.11)

3 Applictions of Schrödinger s Eqution nd Bnd Theory Z α Z On substituting equtions (1.7), (1.9) nd (1.11) in eqution (1.6), we obtin 1 (1.1) me α α α (1.1) The solution of equtions (1.8), (1.1) nd (1.1) re given by X A cosα x B sinα x (1.14) Y A cosα y B sinα y (1.15) Z A cosα B sinα (1.16) In the bove equtions A1, A, A nd B1, B, B re constnts. It is possible to obtin the vlues of these constnts by pplying the boundry conditions. As ψ vnishes t the surfces (boundries) of infinite potentil, it mens tht ψ when x, y, nd when x, y b, c If these boundry conditions re pplied, then we hve obtin A1 A A (1.17) n x π Also B1 sinα 1,i.e., α1 nxπ or α 1 (1.18) B n y π sinα b,i.e., αb nyπ or α (1.19) b n π B sinα c,i.e., αc nπ or α (1.) c On substituting equtions ( ) in equtions (1.14), (1.15) nd (1.16), we sin n π X B x 1 x (1.1) where n x 1,,... sin n π Y B y y (1.) b where n y 1,,... nd sin n π Z B c (1.) where n 1,,... Thus the complete wve function is given by ( xy,, ) nxπ nyπ nπ ψ n,, 1 sin sin sin x ny n BBB x y b c

4 4 Physics II ( xy,, ) nxπ nyπ nπ ψ n,, sin sin sin x ny n K x y (1.4) b c In the bove eqution, K is termed s normlition constnt. It is possible to obtin the vlue of K by using the normlition condition, i.e., bc ψψ * dxdyd 1 (1.5) But bc nxπx n yπ y nπ K sin sin sin dxdyd 1 (1.6) b c nxπx 1 1 nxπx sin dx cos dx b n yπ y b Similrly sin dy b nd c sin nπ c d c Thus eqution (1.6) becomes b c.. 1 K K (1.7) bc Therefore, the complete wve function becomes ( xy,, ) n y x, y, sin xπ x n π y n sin sin π ψ n n n (1.8) bc b c From eqution (1.1), we hve me α α α (1.9) 1 On substituting the vlue of α1, α nd α from equtions ( ) into eqution (1.9), we obtin nx π ny π n π me b c E nnn x y π n n x y n m b c (1.) If we consider box tht is cubicl in shpe such tht b c, energy cn be expressed s

5 Applictions of Schrödinger s Eqution nd Bnd Theory 5 π h E n x ny n n x ny n m 8m where E E 1 nx ny n (1.1) π h E1 m 8m (1.) 1..1 Degenercy of Energy Levels In the lowest quntum stte (1,1,1), in which n x, n y nd n respectively re equl to unity, it h is seen tht E E. There is only one set of quntum numbers tht gives this energy 1 8m stte, nd this level is sid to be non-degenerte. If we now consider the second energy stte s shown in Fig. 1., it is seen tht there re three sets (1,1,), (1,,1) nd (,1,1) of the quntum number n, n nd n tht will give x y 6h the sme energy level E such level is sid to be degenerte, nd in this prticulr cse 8m the level is triply degenerte. For cubicl box, it cn be concluded from the Fig. 1., tht virtully ll the energy levels re degenerte to some degree. x y Fig. 1. Energy levels, degree of degenercy nd quntum numbers of prticle in cubicl box Exmple 1.1 A free prticle is confined in cubicl box of side. Write the eigen vlues nd eigen functions for n energy stte represented by nx ny n 4. Wht is the order of degenercy in this cse? [RTU B.Tech. I sem. 7]

6 6 Physics II Solution: Given nx ny n 4 for this cse, there re three possible combintion of nx, n y nd n s (, 1, 1); (1,, 1); (1, 1, ) The eigen vlues nd eigen functions for -dimensionl cubicl box of side, re given by nd ( xy,, ) π E n n n m ψ n x y 8 n ny sin xπ π n sin sin π x y There re three eigen functions corresponding to single energy vlue Hence the order of degenercy is. The corresponding eigen functions re ψ (,1,1) 8 π π π sin xsin ysin 6π E. m ψ (1,,1) 8 π π π sin xsin ysin nd ψ 8 (1,1,) π π π sin xsin ysin Ans. Exmple 1. Answer the following questions with respect to prticulr cubicl box of side. (i) Is n n n 1 stte degenerte? x y (ii) Wht shll hppen to the degenercies for n n n 4 if the box is not x y cubicl but rectngulr prllelopiped with sides, b nd c such tht b c? Solution: The eigen vlue of prticle in three dimensionl box of sides, b nd c is π n n x y n En m b c nd the eigen function for tht prticle is 8 nxπ nyπ nπ ψ ( xy,, ) sin xsin ysin bc b c (i) For cubicl box b c nd for n n n 1, there is only one eigen function corresponding to eigen vlue x y π, so it is non-degenerte. m (ii) Given tht n n n 4 nd b c. Therefore, there re three sttes x y (, 1, 1), (1,, 1) nd (1, 1, ).

7 Applictions of Schrödinger s Eqution nd Bnd Theory 7 E E E π π 5 1 m c m c,1,1 π π 5 1 m c m c 1,, 1 π π 4 m c m c 1, 1, Thus the energy π 5 1 m c π 4 is doubly degenerte corresponding to eigen functions ψ,1,1 nd ψ 1,, 1 while energy is non-degenerte. m c 1. THE POTENTIAL BARRIER (TUNNEL EFFECT) Let us consider the motion of prticle in the x-direction, under the ction of potentil which is ero every where except over certin region AB where it hs constnt vlue V. Such region AB is clled squre potentil brrier (Fig. 1.). According to clssicl mechnics prticle trvelling in region I (on the left of the brrier) will overcome the brrier if its initil kinetic energy E is V V(x) greter thn the brrier height V. If its energy is smller thn V, it will be totlly reflected nd the I II III prticle cn never be found is region III (on the right of the brrier). Quntum mechnicl clcultions predict tht the prticle cn penetrte through the brrier even when its energy is less thn V. This A B O phenomenon is known s tunnelling nd the effect x is known s Tunnel Effect. Fig. 1. Potentil brrier Consider prticle of mss m nd energy E in the region I. The potentil field cn be represented s V( x ) x < Region I V( x) V x Region II V( x ) x> Region III (1.) The sitution represents the encounter between free prticle nd potentil brrier. The prticle cn be turned bck by the brrier or it cn penetrte through. Thus in the region I the desired wve function hs to represent prticle moving to the right (incident prticle) nd prticle moving to the left (reflected prticle). In the region III the wve function hs to represent only prticle moving to the right (trnsmitted prticle). In the force free, region prticle moving in definite direction (x-direction) with definite energy E hs definite momentum P x. However its position is completely uncertin nd it cn be represented by trvelling wve. In the region I nd III where V( x ), the wve eqution is given by

8 8 Physics II where ψ m E ψ ψ k ψ me k (1.4) The solutions of this eqution re ψ 1 nd ( x) for x < (Region I) (1.5) ikx ikx x Ae Be ikx ψ Ce for x > (Region III) (1.6) For x>, ikx e term hs been omitted s there is no wve trvelling in - x direction. The physicl mening of the coefficients A, B nd C cn be understood in terms of probbility current density. If v is the speed of prticle with propgtion constnt k, then the flux of prticles in the incident bem is given by f 1 probbility density of prticles in the incident bem prticle velocity. written s ( ikx ikx 1 * ) f Ae A e v ( *) AA v A v Similrly the flux of prticles in the reflected bem nd trnsmitted bem my be f B v nd trnsmission coefficient (T) re defined s f C v, respectively. Therefore reflection coefficient (R) nd B v B R A v A nd C v C T A v respec- A tively. Thus the bsolute vlues of A, B nd C re not importnt only their rtios re of more significnce. In the region II x nd V( x) V, the solution of Schrödinger wve eqution will depend on the energy of prticle, i.e., E < V or E > V. Therefore both cses re discussed seprtely. Cse I When E < V (Tunnel Effect) In this cse the wve eqution is ψ m ( V E ) ψ ψ αψ (1.7) where ( ) 1 E m V α

9 Applictions of Schrödinger s Eqution nd Bnd Theory 9 The solution of this wve eqution is ψ (1.8) αx αx x Fe Ge Now we will use the boundry condition to clculte the reflection coefficient nd trnsmission coefficient. ψ1 ψ At x, ψ1 ψ nd (1.9) Applying this boundry condition on equtions (1.5) nd (1.8) A B F G (1.4) i k (A B) α (F G) (1.41) ψ Similrly x, ψ ψ ψ, nd using this boundry condition on equtions (1.6) nd (1.8) (1.4) ik α α Ce Fe Ge (1.4) α α ik ikce α Fe Ge (1.44) From equtions (1.4) nd (1.41) α α i α iα A F G F G 1 ik ik k k (1.45) α α nd i α iα B F G F G 1 ik ik k k ( k iα) F G( k iα) ( α) ( α) B A k i F G k i G F G F ( k iα) ( k iα) ( k iα) ( k iα) Now dividing eqution (1.44) by eqution (1.4) α ik Fe α α ( Fe Ge ) α Ge α α α α α ikfe ikge αfe αge α ( α) ( α) ik Fe ik Ge α (1.46) (1.47) G ik α e F ik α α k iα e k iα α

10 1 Physics II G k iα α e F k iα Put this vlue in eqution (1.47), we get (1.48) k iα α ( k iα) ( k iα) e B k iα A k iα α ( k iα) ( k iα) e k iα α ( k iα) 1 e ( k iα) α ( k iα) ( k iα) e α ( k α ) 1 e ( α α ) ( α α ) k i k k i k e α ( k α ) ( α ) α α 1 e α k 1 e iαk 1 e ( k α ) ( α ) α α B e α A k e 1 iαk e 1 Since x x e e e sinh x nd cosh x ( α ) ( α ) x e x α α B k e e α α α α A k e e iαk e e ( k α ) ( ) B sinhα A k α sinhα iαkcoshα Now eqution (1.4) dividing by eqution (1.45), we get ik α α Ce Fe Ge A iα iα F 1 G 1 k k (1.49) C A k e α G e F G F α ( k iα) ( k iα)

11 Applictions of Schrödinger s Eqution nd Bnd Theory 11 Put the vlue of G F using eqution (1.48) k iα k e e e e C k iα A k iα α ( k iα) ( k iα) e k iα α α α ik α ke ( k iα) ( k iα) α k iα k iα e 4iα ke α ik ( α α) ( α α) k i k i e α 4iα ke α ( )( α 1 ) ( α k α e iαk e 1) 4iα ke α e e ik ik ( )( α 1) ( α k α e iαk e 1) ( k α ) e ik iα ke α α α α e e e e iαk ik C iα ke A k α sinhα iαkcosh α (1.5) Now Reflection coefficient * B B B R A A A From eqution (1.49) ( k α ) sinh α R k α sinhα iαkcoshα k α sinhα iαkcoshα ( k α ) sinh α α α α α α 4 4 k k sinh 4k cosh

12 1 Physics II ( k α ) sinh α 4 4 ( k α k α ) sinh α 4k α ( 1 sinh α) R R ( k α ) sinh α k α sinh α 4k α 1 1 4k α ( k α ) sinh α (1.51) (1.5) As we know tht me k nd α ( E ) m V R 1 4EV ( E) ( V ) sinh α C C C nd Trnsmission coefficient T A A A From eqution (1.5) T 4α k k α sinh α 4α k * (1.5) (1.54) 1 T 1 ( k α ) 4α k sinh ( V ) sinh α 4EV ( E) α 1 (1.55) Thus T, this implies tht the prticle hs finite probbility of penetrting the potentil brrier. This penetrtion is due to the wve nture of mtter nd is known s tunnel effect or quntum tunneling. This effect provides explntion for the following phenomen: (i) the electricl brekdown of insultors (ii) the switching ction of tunnel diode (iii) the emission of α-prticles from rdioctive nuclei (iv) the emission of electrons from cold metllic surfce, i.e., electron emission from metl under the ction of n electric field. (v) the reverse brekdown of semiconductor diodes From equtions (1.51) nd (1.54) it cn be proved tht R T 1. (1.56)

13 Applictions of Schrödinger s Eqution nd Bnd Theory 1 The trnsmission coefficient is mximum when E pproches V. So E V, α sinhα α Using eqution (1.55) C ( V ) α T A EV E 1 4 ( ) E V ( ) ( ) 4EV ( E) V m V E 1 mv 1 E mv T 1 E V E V α is very smll, i.e., (1.57) From eqution (1.57) it is cler tht the width or the height of potentil brrier ( V ) or both increses, the trnsmission probbility of prticle decreses. 1 If α >> 1, then sinhα e α Put this vlue in eqution (1.55), the trnsmission coefficient becomes α ( V ) e T 1 16 EV ( E) ( E) ( V ) α 16EV T e (1.58) As α increses, T decreses exponentilly. In the cse of E < V, the wve function will be of the form s depicted in Fig. (1.4).

14 14 Physics II ψ I II III Trnsmitted Wve E V x x Fig. 1.4 Brrier penetrtion Cse II When E > V the Schrödinger equtions for ll three regions ψ1 m E ψ1 or ψ 1 k ψ 1 ψ k ' ψ x < (1.59) < x < (1.6) nd where ψ k ψ me m k nd k' ( E V ) m V As we know tht α ( E ) m E Therefore α ( V ) x > (1.61) in previous cse E < V. (1.6) α ik α ik ' sinhα sin k' (1.6) Thus the vlue of reflection coefficient (R) nd trnsmission coefficient (T) cn be clculted by putting α ik ' in equtions (1.5) nd (1.54). R 1 4 kk' ( ) k k' sin k' nd Put the vlues of k nd k then R nd T become T 1 ( ) k k k ' sin ' 4 kk' (1.64)

15 Applictions of Schrödinger s Eqution nd Bnd Theory 15 R 1 ( V ) ( V ) 4E E sin K' (1.65) nd T 1 ( V ) sin ' 4E( E V ) K (1.66) From eqution (1.64) it cn be proved tht R T 1. Keeping E > V, s E pproches its lower limit V, In this limiting condition K ' becomes smll so tht ( ) 4 ( ) C V m E V T 1 A E E V ( V ) me sin K ' K'. E V mv T 1 (1.67) i.e., the trnsmission probbility of wve pcket is less thn one, even when the energy of the prticle is greter thn the height of potentil brrier. When k ' π, π, π,... nπ then sin k ' nd T 1, i.e., there is perfect trnsmission. Since λ π k ' π nπ λ nλ Hence when the brrier thickness is equl to n integrl multiple of hlf wvelength, the brrier does not obstruct the motion of prticle, i.e., T 1. O 1 4 (EN ) Fig. 1.5 Oscilltions of trnsmission coefficient As E increses (with E > V ) the trnsmission coefficient oscilltes nd for lrge vlues of E it pproches unity (Fig. 1.5). Exmple 1. The potentil brrier problem is good pproximtion to the problem of n electron trpped inside but ner the surfce of metl. Clculte the probbility of trnsmission tht 1. ev electron will penetrte potentil brrier of 4. ev when the brrier width is. Å. T

16 16 Physics II Solution: The trnsmission coefficient is given by ( ) 16EV E T exp m( V E) V Here E 1 ev, V 4 ev, Å 1 m, J-s, m kg 9 ( V E) ( ) 4 1 ev J 1 1 T 16 1 exp ~ % Ans. Thus, only bout 8 electrons of energy 1. ev out of every hundred, penetrte the brrier. Exmple 1.4 Consider n electron whose totl energy is 5 ev pproching brrier whose height is 6 ev nd width is 7Å. Find out the de Broglie wvelength of incident electron nd 1 probbility of trnsmission through the brrier. (Mss of electron m kg., Plnck s 4 constnt J-s). [Rj. B.E.I.] Solution: The de Broglie wvelength of n electron is given by 1.7 λ Å V Here V 5 Volt (s electron energy is 5 ev) 1.7 λ 5.48 Å 5 Ans. The trnsmission coefficient is given by ( ) 16EV E T exp m( V E) V Here V 6 ev, 7 1 m, E 5 ev J-s T exp % Ans. Exmple 1.5 Electrons of energy ev re incident on potentil brrier of height 5 ev nd width 5 Å. Find tunnel probbility for these electrons. [RTU B.Tech. I sem. 8]

17 Applictions of Schrödinger s Eqution nd Bnd Theory Solution: Given E ev J 19 V 5 ev J nd 5 Å m The tunnel probbility is given by EV T ( ) 16 E e V m V α where α ( E ) ( 5 ) ( ) 1 Now T exp exp [ ].84 exp( 8.867) T % Ans. Exmple 1.6 A bem of electrons is incident on potentil brrier of height 5 ev nd width. 1 nm. Wht should be the energy of the electrons so tht of them re ble to penetrte through the brrier? Solution: Given V 6 ev nd. nm 1 1 m The probbility of trnsmission is given by ( E) 16 EV T e V α Te α

18 18 Physics II where T V ( E) 16E V mximum trnsmission probbility m V nd α ( E ) Given T T T Te α e α α ln ( α ) (ln) mv ( E) 4. (ln ) (ln ) ( V E) 8m 4 ( ) (1.98) ( V E) ( 1 ) 1 V 9 E.1 1 J V E ev.15ev E ( V.15)eV 6.15 E 5.875eV Ans. 1.4 THEORY OF α DECAY The theory of α-decy ws explined by George Gmow in 198 on the bsis of quntum mechnicl brrier penetrtion. There re certin nuclei which emit α prticles nd get converted into new nuclei with tomic number less by two nd mss number less by four. This process is known s α -decy nd is depicted s follows: A A 4 4 ZX Z Y He When nα -prticle is present inside the nucleus, the energy of α -prticle cn not be greter thn the height of the potentil brrier which is existing t the surfce of the nucleus. It implies tht ccording to clssicl mechnics it is never possible for the α -prticle to come out of the nucleus. The first difficulty involves the energy of α -prticle in the field of the nucleus. It ws concluded from Rutherford s scttering experiments tht the force experienced by α -prticules

19 Applictions of Schrödinger s Eqution nd Bnd Theory 19 4 is Coulomb force which cts t very smll distnce ( 1 m) from the centre of the nucleus. After this direct distnce, Coulomb s lw breks down nd the boundry of the nucleus strts. Thus, there is only Coulomb potentil outside the nucleus which is depicted in Fig As soon s the α -prticle enters the nucleus ( < r < ), it will be under the influence of very strong ttrctive nucler forces. These forces re represented by potentil well s depicted in Fig If the rdius of the rdium nucleus is V m, the Coulomb potentil outside the nucleus is 7.8 MeV. However, the energy of α-prticle emitted by rdium nucleus is 4.88 MeV only. Now the question rises : How cn α -prticle of 4.88 MeV energy go through potentil brrier of 7.8 MeV? The second difficulty is concerned with the lw of cuslity. According to the rdioctive decy lw, the mount of rdium left fter 16 yers should be hlf. After severl hlf lives, smll frction of rdium should be left. Now the question rises. Why re only some rdium toms decying in the first few yers nd why re some others surviving for thousnds of yers? When quntum mechnics ws pplied to the problem of α-decy, the difficulties of clssicl mechnics s described bove dispper. According to quntum mechnics, every prticle is ssigned wve spect. If this fct is tken into considertion, then the electrosttic potentil brrier, lthough very high, cn not completely rule out the pssge of wve through it. There lwys exists certin probbility of the prticle to penetrte through the brrier, how smll the energy might be. It is possible to obtin some quntittive fetures of the theory from the expression of the trnsmission coefficient. ( ) 16EV E T exp m( V E) V (1.68) Now n ttempt is being mde to clculte the hlf life time in cse of urnium nucleus 4 which is hving rdius of bout 1 m. There is n evidence tht the α -prticle moves bck 7 nd forth freely with n verge speed of 1 m/s. Thus the α -prticle will strike the brrier /1 1 times per second. The probbility tht it penetrtes the brrier is equl to the trnsmittnce (T). Hence the probbility (P) tht α -prticle leks out from nucleus in one second, which is clled decy constnt ( λ ), i.e., λ 1 1 T per sec ( ) 16EV E mv V 1 1 exp V O ~ Fig. 1.6 Potentil for the α-prticle in the nucleus ( E) 7.8 MeV r V ( )e 4 r πε E α 4.88 MeV r (1.69)

20 Physics II For urnium, E 4. MeV nd V MeV. 7 Mss of α -prticle kg J-s, 1 m Thus, ( MeV) 1 4.MeV 4 MeV λ 1 16 exp mv ( E) λ 1 exp ( 89) 1 9 e per sec per yer Therefore hlf-life time of nucleus is.69 T1 λ yers (1.7) But experimentlly, T 19 1 hd been found to be 1 yers. The discrepncy hs been scribed to the fct tht the brrier is not rectngulr. This result explins why most energetic prticles re decying in first few yers nd why lest energetic prticles re surviving for thousnds of yers. Exmple 1.7 Find the height of the potentil brrier for α-prticles emitted from rdon ( 86Rn ) ssuming tht the effective nucler rdius is given by Solution: ( Z ) e V 4πε r 9 Here Z 86, e c r πε, 1 r A m V 5 1/ ev 6.6 MeV Ans.

Name Solutions to Test 3 November 8, 2017

Name Solutions to Test 3 November 8, 2017 Nme Solutions to Test 3 November 8, 07 This test consists of three prts. Plese note tht in prts II nd III, you cn skip one question of those offered. Some possibly useful formuls cn be found below. Brrier