Question 10. How Were the Elements from Iron to Uranium Made?
“Scientists’ understanding of the production of elements up to iron in stars and supernovae is fairly complete. The precise origin of the heavier elements from iron to uranium remains a mystery."
Scientific Significance: While we have a relatively an atomic number higher than iron—such as silver and good understanding of the origin of elements lighter gold—presents a mystery. than iron, important details in the production of structures:elements from iron to uranium remain a puzzle.existence Picture A sequence of rapid neutron captures stars by nuclei, the periodic table. The existence of every element with known as the r-process (where r is for rapid), is
Researchers using an Oak Ridge National Laboratory supercomputer have found that the organized flow beneath the outgoing shock wave in a supernova explosion persists in three dimensions. This fluid flow develops beneath the shock and, when coupled to nuclear and neutrino ineractions, is believed to lead to the explosion. The enormous energies that this generates creates an environment where elements of life are synthesized.
45 Kelvin’s paradox
“Some unknown source of energy laid down in the store house of creation” –1–
- a.k.a. Astronomy 111 Order of MagnitudeKelvin’s Astrophysics –1–paradox
Order of Magnitude Astrophysics - a.k.a. Astronomy 111
The gravitationalPhysical binding Conditions energy of of–1– the Astrophysical mass of the Matter Sun after it falls from infinity to the radius of the Sun: - a.k.a. Astronomy 111 OrderPhysical of Magnitude Conditions Astrophysics2 of Astrophysical Matter 3 GM 48 U = " = 3 10 erg (1) −5 R × " L = 3.9 1033 erg/s (2) " 2 Since the Sun produces: 3 GM× 48 PhysicalU Conditions= of" = Astrophysical3 10 erg Matter (1) " =−25 R1014 s = ×107 yrs. (3) L × " " L = 3.9 1033 erg/s (2) " × 2 U 3 GM 14 748 U =" = 2 10" =s3= 1010 yrserg. (1)(3) the gravitational energy releasedL −5 ×R could have× provided the energy " " radiated for L = 3.9 1033 erg/s (2) " × U " = 2 1014 s = 107 yrs. (3) L × "
10 That is a lot of energy but our Sun has been glowing for almost 10 yrs. M 1/3 T 1.5 107 (36) i ≈ × M ! ! "
T 1.8 107 K (37) i ∼ × M 1/3 T 1.5 107 (36) i ≈ × M ! ! "
√L 8 R = ! =7.1 10 m (38) free fall timescale! 2 7 √4πσ T × Ti 1.8 10 K× ! (37) ∼ × Let!s take away the pressure of the sun!
t =(R/v ) (39) √L 8 ff ff R = ! =7.1 10 m (38) ! √4πσ T 2 × × ! free fall velocity:
GM v (40) tff =(R/vff )ff ∼ R (39)
GM 3 v R (40) ff ∼ R tff =(R/vff ) (41) ∼ #GM
R3 tff =(R/vff ) 1 hour (41) ∼ #GM ∼ nuclear energy
“what is possible in the Cavendish Laboratory cannot be too complicated for a star” Arthur Eddington 1/3 10 1/3 10 Rplanet NG a0 10Rplanetcm N a0 10 cm R ≈N 1/3a ≈1010 cm ≈ G ≈ planet ≈ G 0 ≈ 1/3 10 Rplanet NG a0 10 cm ≈! ! ≈ a F !a !F !a ∼!F ∼ ∼1/3 10 1/3 10 R N a R10planetcmNG a0 10 cm planet ≈ !G 0!≈ ≈ ≈ a ∼ F kBT !a !F k T ! ! kBT ≤!a ≈!F B a F structures: existence≤ ≈ !a stars!≤F ≈ !a !F ∼ k T ∼ ! ! B ≤ a ≈ F ! = !f (n) + kBT ! = !f (n) + kBT kB!T = !af (n!)F + kBT kBT !a !F ≤ ≈ When the mass of the system is Pincreased,nk≤BT the+≈ n gravitational!F pressure increases and - P ≈! =nk!Bf (Tn)++nk!FBT P nk T + n! to balance it - both the Fermi pressure≈ and the thermal pressure≈ B will increase:F ! = !f (n) + kBT P nk T + n! ! = !f (nB) + kBT F ≈ P nk T + n! (kBT + !F ) !g ≈ B F (kPBT +nk!FT)+≈n!!g (k T + ! ) ! ≈ B ≈ F B F ≈ g (kBT + !F ) !g This pressure can balance gravitational pressure 2if 2( /k 3 B T 2 + ! F ) ! g . For a classical 2 2/3 1/3 (3≈π2 )2/3 !2 2/3≈ system we get kBT Gm2pN2/3 n1/3 (3π ) ! n2/3 . 2 2/3 2 kBT ≈ GmpN n − 2 2 m2en/3 1./3 (3π ) ! 2/3 ≈ kB−T 2Gm2 2p/m3Ne 2 n n . 2 2/3 1/3 ≈(3π ) ! 2/3 − 2 me kBT GmpN n n . ≈ − 2 me n = nc n = nc The maximum temperature of the system is reached when n = n c , with
2/3 2 αG N2/3 α2G 4/3 2 n αG N ; k T αG (N4/3 m c2 ), nc 2 2/3 ; k BT 2/23 2/3(N m ec ), 2 c ≈ (3π2 )2/3 λe αB ≈ 2(N3π2 )2/3 e α ≈ (3π ) ! λen " G ≈ 2(3π ) ; k T G (N 4/3m c2), ! c"≈ (3π2)2/3 λ B ≈ 2(3π2)2/3 e ! e " λe = (!/mec) 1/3 10 Rplanet1/3 NG a100 10 cm Rplanet NG ≈a0 10 ≈cm 1/3 10 ≈ ≈ Rplanet NG a0 10 cm R N 1/3a 1010 cm≈ ≈ planet ≈ G 0 ≈ 1/3 10 Rplanet N a0 10 cm ≈ G !≈a !F !a !F ∼ !a !F ∼ 1/3 10 R N !a a !F 10 cm ∼ planet ≈ G ∼0 ≈ !a !F 1/3 10 R N a 10 cm ∼kBT !a !F planet G 0 kBT !a ≤!F ≈ ≈ ≈ ≤ ≈ kBT !a !F kB!aT !!Fa !F ≤ ≈ R N 1/3a 10k10Tcm ! ∼≤! ≈ planet ≈ G 0 ≈ B ≤ a ≈ F ! ! ! = ! (n) + k T a ∼ F ! = !f (n) +fkBT B ! = !f (n) + kBT ! = ! (n) + k T kPBT nkf!a T +!FnB! !a !F P! = nk!f (BnT) +≤+knBB!TF≈ F ∼ ≈ ≈ P nkBT + n!F kBT !a !F P nkP T +nknB! T + n!F ≈ ≤ ≈ ≈ B ≈ F ! = !f (n) + kBT kBT !a !(FkBT +(k!BFT) +!!gF ) !g ≤ ≈ ≈ ≈ (kBT + !F ) !g ! = !f (n) + kBT (k T +(k!BT) +!!F ) !g ≈ BP Fnk≈BTg+ n≈!F ≈ 2 2/3 22 2/3 2 P nk T + n! 2 2/3 21/32/3 (31/π3 ) (3!π ) 2/3 ! 2/3 B F ! k=BT!f (nk)Gm+TkBNTGmn N n n . n . 2 2/3 2 ≈ B p p 2 2/3 2 2 2/3 2 (3π ) ! ≈ ≈ − (3π 2)− !me2 2 2m/3 1/3 2/3 k T Gm2N 2/3n21/3 2/3 k1B/3T (3Gmπn)2p/N3. ! en 2/3 n . B kBT p Gm N n n . P nkB≈T + n!F (kpBT−+ !F2) ≈!mg − 2 me ≈ ≈ −≈ e2 me (kBT + !F ) !g n = n ≈ cn = nc n = nc 2 2/3 2 n = n 2 2/3 1/3 (3π ) ! 2/3 c (kBTk+ !TF ) Gm!g N n n = nc n . B ≈ p 2 2/3 2 ≈2/3 − 2 2 me 2 2/3 1/3 (3π ) ! 2/3 2/3 2 αG α N 2/3 N α2G α 4/3 2 kBT GmpN n nn . α GN ; k T α 2(/N3 G m c )4,/3 2 2 ≈ − 2 m c nc 2 2G/3 B ; αkBT G2 2N/3 4/3 (e2N mec ),α e n≈c (3π ) 2 2λ/e3 2 2;/32/3k2BT ≈ G2(3π ) (N 22 m2/e3c ), G 4/3 2 2 2≈/32 (321//3π3α)G (3π )Nλn ! 2/3 2≈2/2(3 3απG); k 4T/3 2 (N m c ), k T Gm≈Nn(3π n) ! λe " ec n; ≈.k22(23T/π3 ) (NB m c ), 2 2/3 e B structures:p c !2 2/3 !" existence≈"n(3=πn)B λe 2 2 /stars3 ≈e 2(3π ) ≈ ≈ (3π−) 2 λ me c ≈ 2(3π ) ! e " ! " n = n c λe = ( /mec) λe = (!λ/m=ec() /m c) 2/3 ! e ! e 2 α N α λe = (!/mec) n =G nc λe = (!/mec) G 4/3 2 nc 2 2/3 ; kBT 2 2/3 (N mec ), 2/3 2 ≈ (3π ) λe ≈ 2(3π ) αG N An interestingαG phenomena4/3 2 arises when the maximum temperature is sufficiently high nc ; kBT (N mec ), ! 22 " 22 2 2/3 2 2/3 !n!uclnucl==ηηαα mmppcc 2 ((η2η 00.1).1) ≈ (3π ) λe to≈ trigger2(3π nuclear) 2/ 3fusion 2 , with . The energy corresponding to the ! " αG N !nuclα=G ηα m4p/c3 ≈≈2 (η 0.21) 2 nmaximumc temperature; k B T will be larger(2N than2me c! n ) ucl , =when≈ηα m pc (η 0.1) ≈ (3π2)2/3 λ ≈!2(nucl3π=2)2η/α3 mpc (η 0.1) ≈ ! e " λe = (!/mec) ≈ λe = (!/mec) m 3/4 α 3/2 N > (2η)3/4(3π2)1/2 P 3/4 3/2 4 1056 λe = (!/mec) mP α N > (2η)3/4(3π2)1/22 m2e αG ≈4 ×1056 !nucl = ηα m!pc " (!η 0".1) me αG ≈ × 2 2 ! " ! ≈ " !nucl = ηα mpc for( η 0 . 1) . The corresponding condition on mass is ≈ 2 2 ! = ηα m c (η 0.1) 3/4 3/2 nucl p 3/4 3/2 3/4 2≈1/2 mP α 56 N > (2η) (33/π4 ) 2 1/2 mP α 4 10 32 M > M (2η) (3π ) mp 4 10 g ∗ ≈ mem αGα ≈ ×≈ × ! ! "e " !! "G "
which is comparable with the mass of the smallest3/4 stars3/2 observed in our Universe. 3/4 2 1/2 mP α 32 M > M (2η) (3π ) mp 4 10 g ∗ ≈ me αG ≈ × The mass of the Sun, for example, is! " ! "
M = 2 1033 g " × structures: existence stars
The lowest-mass star that can fuse H into He in its core has a mass of about 0.07 solar masses structures: existence stars
The lowest-mass star that can fuse H into He in its core has a mass of about 0.07 solar masses the sun’s source of power solar neutrinos main-sequence lifetime
Take the amount of H in the Sun!s core (7% of its total mass) and divide it by the amount of fuel consumed each second (360 million tons per second) to obtain an estimate of the sun!s main-sequence lifetime:
0.07 2 1030 kg t = × × = 3.9 1017 s = 1.2 1010 yr (12) ! 360 109 kg/s × × ×
For the values given, we obtain a lifetime for the Sun of about 12 billion years on the main sequence. Other, more detailedUth stellarkT evolution models indicate that (13) slightly less hydrogen will be fused in the∼ core before the Sun leaves the main sequence phase, so the Sun!s main-sequence lifetime is rounded off to about 10 billion years. 23 k = 1.38 10− J/K (14) ×
GM mp kT ! (15) i ∼ R !
T 1.8 107 K (16) i ∼ ×
√L 8 R = ! = 7.1 10 m (17) ! √4πσ T 2 × × ! stellar evolution massive stars eddington limit eddington6 3 limit n 10 pc− ∗ ∼ If the luminosity of the source is L, we have by spherical symmetry that the inward force on an electron-positron pair is Lacc = GM˙ M/R ∗
Lσ 1 GMm T . p − 4πc r2 ! "
There is a limiting luminosity for which this expression vanishes, the Eddingron limit. Dead Stars and Black Holes Dead Stars and Black Holes a.k.a.DeAstrad onomyStars and15 WinterBlack HQuarolester 2008 a.k.a. Astronomy 15 Winter Quarter 2008 a.k.a.GraviAstrty onomyTrium15phanWintert: KeyQuarCteronce2008pts GrGravivittyyTTrriumiumphanphant:tKey: KeyCoCnceonceptspts
GMm v2 F = = m (1) R2 R GMGMmm v2 v2 FF== = =m2 m (1) (1) R2 2 V RR M
5 M LEdd = 105 L M (10) LEdd = 10 L ! M (10) ! "M ! # " ! # Since we know experimentally, 5 M LEdd = 10 L (10) ! 3M 3 M " ! # L ML (11) L L ≈ ! forMM 10M (11) ≈ ! M " ! #≤ ! " ! # 3 M 2 M L L L forM > 10M (12)(11) ≈ ! M≈ ! M ! " M!!#180"M ! # (12) a star will be unstable if !
M 180M (12) ! ! most massive stars
List of the heaviest star
Star name Solar Mass Eta Carinae 150 Pistol Star 150 LBV 1806-20 130 A1 in NGC 3603 114 + 84 S Doradus 100 Pismis 24-1 100 Cyg OB2-12 92 WR 20a+b 80 + 80 R 66 70 eta carinae