Physics. X m (cm)

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1 Entranc xa Physics Duration: hours I- [ pts] An oscillator A chanical oscillator (C) is ford of a solid (S), of ass, attachd to th xtrity A of a horizontal spring of stiffnss (constant) = 80 N/ whos othr xtrity B is fixd. Th solid (S) can ov on a horizontal rail. Th position of its cntr of gravity G is locatd, at an instant t, by its abscissa x = OG, O bing its quilibriu position. x(c) 3 A) Thortical study W nglct all frictional forcs.. Driv th diffrntial quation that dscribs th oscillations of (C). 0. Dtrin th xprssion of th natural priod T0 of ths oscillations B) Exprintal study - - Fr oscillations An appropriat dvic, connctd to a coputr, givs us th curv of th abov figur rprsnting th variation of x in trs of ti. a. Dtrin, fro th graph, th duration T of on oscillation of th solid. b. Dtrin th avrag powr dissipatd btwn th instants 0 and 3T. - Forcd oscillations Now, th xtrity B is connctd to a vibrator of adjustabl frquncy f and for ach frquncy f a rcording is prford. Th tabl blow shows th aplitud X rlativ to ach f. a. Using th tabl, dtrin, with justification, an approxiat valu of T0. b. Dtrinr thn th valu of. f (Hz) c. What do w obtain: i. in th absnc of all frictional forcs? X (c) ii. in th cas whr th agnitud of th frictional forc is incrasd? C ) Th hydrogn chlorid olcul A hydrogn chlorid olcul (HCl) can b rprsntd by a haronic oscillator of ass H = g, and of stiffnss. Th potntial nrgy du to th intraction btwn th two atos can b rducd to: 0.7 EP(x) = x ; whr: = 90 9 SI; = C; and 3 r0 =.30-0 ; r0 bing th 40r0 40 distanc btwn th two atos at quilibriu and x is th displacnt of th hydrogn ato with rspct to its quilibriu position with x << r0. This olcul, whn xcitd with an lctroagntic wav of frquncy, oscillats with a axiu aplitud X whr X << r0. Dtrin, with justification, th valu of. t(s)

$II- [8 pts] Hydrogn ato and diffraction A- Hydrogn ato Th adjacnt figur shows so nrgtic lvls of a hydrogn ato.$ . Th downward transition of a hydrogn ato fro an xcitd lvl (n>) to th fundantal stat givs th Lyan sris. Calculat th shortst and longst wavlngths of th xtr radiations corrsponding to this sris. 3.

. Th downward transition of a hydrogn ato fro an xcitd lvl (n>) to th fundantal stat givs th Lyan sris. Calculat th shortst and longst wavlngths of th xtr radiations corrsponding to this sris. 3.

$60 n = B- Diffraction a A diffraction xprint is prford with a light ittd by a hydrogn lap L providd with four filtrs. Each filtr allows th passag of a onochroatic radiation.$

2 II- [8 pts] Hydrogn ato and diffraction A- Hydrogn ato Th adjacnt figur shows so nrgtic lvls of a hydrogn ato.. Dtrin, with justification, th bhavior of a hydrogn ato tan at th fundantal stat whn it rcivs a photon of nrgy: a).75 V; b) 0.99 V and c) 5.6 V.. Th downward transition of a hydrogn ato fro an xcitd lvl (n>) to th fundantal stat givs th Lyan sris. Calculat th shortst and longst wavlngths of th xtr radiations corrsponding to this sris. 3. Th downward transition of a hydrogn ato fro an xcitd lvl (n > ) to th first xcitd lvl givs th Balr sris. Calculat th longst wavlngth of th radiation corrsponding to this sris. 4. Spcify, with justification, th radiation that blongs to th visibl spctru. Givn: h = J.s; c = /s; V = J En (V) n = n = 6 n = 5 n = 4 n = 3 n = n = B- Diffraction a A diffraction xprint is prford with a light ittd by a hydrogn lap L providd with four filtrs. Each filtr allows th passag of a onochroatic radiation. Th hydrogn lap is placd in front of a slit of wih a = 0.5. For ach of ths four radiations, a diffraction pattrn is obsrvd on a scrn placd at a distanc D =.600 fro th slit. D Th asurnt of th wih L of th cntral fring givs, for th usd radiations, th rspctiv valus: L = 4.0 ; L = 3. ; L3 =.78 and L4 =.63. Lt b th angular wih of th cntral fring (s adjacnt figur).. Dtrin th wavlngths of th usd four onochroatic radiations.. Dtrin, with justification, th transition corrsponding to th ission of ach of ths radiations. Écran III- [ pts] An analogy First part: Exponntial dcay of th charg of a capacitor A capacitor (C), of capacitanc C = 0.0 F and of initial charg Q0, is put in sris in a circuit ford of a rsistor (R) of rsistanc R, a switch (K) and conncting wirs. W clos (K) at th instant to = 0. At th instant t, (C) has th charg q (q>0) and th circuit carris th currnt i. An appropriat dvic allows us to obtain th variation of q in trs of ti (graph of th adjacnt figur) q (C) t (s)

. Ma a diagra of th circuit showing th ral dirction of th currnt and spcifying th aratur that has th charg q.. Driv th diffrntial quation dscribing th volution of q in trs of ti. 3.

3 . Ma a diagra of th circuit showing th ral dirction of th currnt and spcifying th aratur that has th charg q.. Driv th diffrntial quation dscribing th volution of q in trs of ti. 3. Th solution of this quation is of th for q = A + B -t whr A, B and ar constants. a. Dtrin A, B and / and spcify th aning of /. b. Writ down, in trs of R, C and t, th xprssion giving th nubr N of th lctrons on th aratur that has an xcss of lctrons. 4. Us th adjacnt figur to find th valu of /. Dduc th valu of R. 5. Dtrin th rlation btwn i and N. 6. Dtrin th nrgy dlivrd by th capacitor btwn th instants t0 = 0 and t = /. Scond part: Exponntial dcay of th radon 0 In an xprintal sssion, w study th radioactiv dcay of 0 th activity A of a sapl of radon 0 ( Rn 86 ). A (Bq) Th adjacnt figur shows th shap of th curv giving th variation of A in trs of ti.. A radon nuclus, an ittr, disintgrats into a poloniu nuclus (Po). Writ down th quation of this disintgration.. Th instantanous xprssion of th activity A is givn by: A = A0 -t, whr is th radioactiv constant of th sapl. a. i. Dfin th activity of a radioactiv sapl of a substanc and giv th rlation btwn A and dn/, whr N rprsnts th avrag nubr of th radon nucli that ar prsnt at th instant t ii. Dtrin, with justification, th physical quantity in th first part (A) which is analogous to th activity in th scond part. b. How can w dtrin, fro th graph, th valu of /? Dtrin its valu. 3. Aftr what ti can w considr that th sapl is practically copltly disintgratd? 4. By coparing th two figurs, show that th radioactivity has a hazardous charactr. 5. Dtrin th nrgy libratd by th radon sapl btwn th instants t0 = 0 and t = /. Givn: (Rn) = u; (Po) = u; () = u; u = 93.5 MV/c. t (s)

4 Entranc xa Solution of Physics Duration: hours I- An oscillator A) Thortical study. No friction, consrvation of ME(C): ME = KE + EPE = ½ V + ½ x = constant. Lt us driv with rspct to ti: V V. This quation is of th for x + + x x = 0 ; x 0 x = 0, 0 = B) Exprintal study - Fr oscillations a. W hav 3T = 0.94 s T = 0.94/3 = 0.33 s. ME b. Pav = ; ME = ; PE(axiu) = ½ ( t final + X - X x = 0 t T0 = initial And t = 3T = 0.94 s. Thus : Pav = 0 - /0.94 =.30 - W. 0, thn T0 = ) = ½ 80(9 4)0-4 = 0 - J. Forcd oscillations a. According to th tabl, th aplitud of th oscillations tas a axiu valu (rsonanc of aplitud) whn th frquncy f of th xcitations is qual to f = 3. Hz. According to th graph, (fr oscillations) th oscillations ar slightly sall, thn f f0 T0 = T = /f = /3. = s T0 b. T0 = ; = 4 = /4 = 0.99 g c. i. In th absnc of any forc of friction, th aplitud X passs by a vry larg valu () for T = T0 and thr is ris of ruptur of th spring. ii. Whn w incras th agnitud of th forcs of friction, th aplitud X dcras and th psudopriod of rsonanc of aplitud is largr than T0. Th rsonanc, which was acut, bcos lss and lss acut to bco fuzzy. (As long as th critical od yt was not rachd). C ) Th hydrogn chlorid olcul Dtrination of : PE(x) = ½ x 0.7 = x = r ' Th propr frquncy of oscillations of th olcul is: 0 = 0 = =,930 3 Hz. Thus, th rsonanc of aplitud tas plac for = 0 = Hz 9 = N/

5 II- Hydrogn ato and diffraction A- Hydrogn ato. a) An nrgy gain of.75 V would lad th hydrogn ato to an nrgy of: = V. This nrgy is that of th lvl n = 4. Th photon is wll absorbd, th ato passs to th lvl n = 4. b) An nrgy gain of 0.99 V would lad th hydrogn ato to an nrgy of: = -.6 V. This valu dos not corrspond to any nrgy lvl of th hydrogn ato. This ato thus rains at th fundantal lvl, th photon in qustion is not absorbd. c) This contribution of nrgy (5.6 V) xcd th nrgy of ionization (3,60 V). Th ato is thus ionizd and th fr lctron lavs with a intic nrgy: =.0 V.. a- Th sallst nrgy ittd by th hydrogn ato corrsponds in th passag of th xcitd lvl n = (E = V) to th fundantal lvl (E = - 3,60 V). Th ittd nrgy is thus: E = E E = (- 3.60) = 0. V = J = J. Th wav associatd with th ittd photon has a frquncy and a wavlngth satisfying : E = h = h.c / = h.c/e = / ( ). =.40-7 =.4 n. b- Th gratst nrgy ittd by th hydrogn ato corrsponds in th passag fro th axiu nrgy lvl (Eax = 0 V) at th fundantal lvl (E = - 3,60 V). Th ittd nrgy is thus: Eax = 3.60 V = J = J. th wav associatd with th ittd photon has a wavlngth ax satisfying: ax = h.c/eax = /( ) ; ax = 9,0-8 = 9, n. 3. Th largst wavlngth thus corrsponds to th ission of a photon having th sallst nrgy which corrsponds in th passag fro th lvl n = 3 to th lvl n = : E3 =.89 V = J = J. Th wav associatd with th ittd photon has a wavlngth: 3 = h.c/e3 = /( ) ; 3 = = 656 n. 4- Th radiation of wavlngth 3 is visibl, bcaus its wavlngth in vacuu is btwn 400 n and 800 n. Whil th othrs ar invisibl (ultraviolt) B- Diffraction.. W now that = with = /a. According to th figur tan = L/D =, bcaus L/D is vry wa. But = ; thus: L/D = /a = La/D. By applying this rlation, w obtain: = /.6 = ; = ; 3 = and 4 = ;

6 . Thy blong to th Balr sris bcaus thy ar visibl (400 n < < 800 n). Th 4 transitions ust corrspond to th passag of th xcitd lvls n = 3, 4, 5 and 6 to th xcitd lvl n = : fro n = 3 to n = ; fro n = 4 to n = ; 3 fro n = 5 to n = and 4 fro n = 6 to n =. (Or by aing calculation) III- An analogy A- Exponntial dcay of th charg of a capacitor. S adjacnt figur.. W hav uc = Ri. But uc = Finally: + q = 0. q C and i = -, thus: q C = - R. q i C K R 3. a. For t0 = 0 : Q0 = A + B and for t =, q = 0 = A A = 0 and B = Q0; thus q = B -t t = - B -t - B -t + B -t = 0 = and / =. b. Th aratur which carris an xcss of lctrons at th dat t carris th charg q ; thus: N = -q/(-) = q/ ; (with Q0 =.00-4 C) N = Q 0 t N = t lctrons 4. On th graph / = = 80 s (point of ting of th tangnt at th origin with th asyptot). = = 80 s R = 80/0.0-3 = , 5. Th instantanous xprssion of i : i = - = - dn i = - dn. Q 6. Th nrgy providd by th capacitor W = W0 - W = C W = [(00-6 ) (800-6 ) ] = J q with q = 0.37Q0 = C C

7 B- Exponntial dcay of radon Rn 6 84 Po 4 H. a i. Avrag nubr of disintgrations pr unit of ti: A = dn/ ii. A = - dn is quivalnt to th currnt i dividd by : Thus A is quivalnt to i. i = - dn ; b. plays th sa rol as in th first part. Donc / is th intrsction of th tangnt to th curv at t0 = 0 with th axis of ti. / = 80 s and = 0.05 s By quivalnc with th first part, w can say that 5(/) = 400 s 4. For disintgration, th points ar distributd around th curv; thus its charactr is a rando charactr. 5. At th nd of t, th nubr of disintgratd nucli = N0 N = /[ A0(- -t )] = 80[0(- - ) N0 N =.0 4 nucli. Th nrgy rlasd by th disintgration of a nuclus: E = c = ( ) = u E = = MV. Th total nrgy rlasd btwn t0 = 0 and t = / = (N0 N) E = = 7033 MV.

fiziks Institute for NET/JRF, GATE, IIT JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics

fiziks Institute for NET/JRF, GATE, IIT JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Atoic and olcular Physics JEST Q. Th binding nrgy of th hydrogn ato (lctron bound to proton) is.6 V. Th binding nrgy of positroniu (lctron bound to positron) is (a).6 / V (b).6 / 8 V (c).6 8 V (d).6 V.6