PC Chapter 14

PC Chapter 14 - 14 Nuclear Physics Applications 14-1 18 O =...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
225 14 Nuclear Physics Applications 14-1 = 18 O 17.999 160 = 18 F 18.000 938 = 1.008 664 9 n m = 1 H 1.007 825 all in u. (a) [ ] [ ] [ ] = + + - = - =- 2 OHF n 0.002 617 9 u 931.494 3 MeV u 2.438 6 MeV Q M M M mc compared to 2.453 0.000 2 MeV . (b) ( 29 χ   = - + = +=     1.007 825 1 2.438 6 MeV 1 2.575 1 MeV 17.999 160 a th M KQ M 14-2 (a) At 6 600 10 K T , each carbon nuclei has thermal energy of approximately (29 ( ( - = × × 5 64 B 3 1.5 8.62 10 eV K 600 10 K 7.7 10 eV 2 kT . (b) ( 29 ( 29 ( 29 ( 29 ( 29 = -- = = = -= 1 2 42 12 24 Energy released 2 C Ne He 24.000 000 19.992 440 4.002 603 931.49 MeV 4.617 MeV Energy released 2 C Mg 931.49 MeV 24.000 000 23.985 042 931.49 MeV 13.93 MeV mm mc (c) Energy released = the # of carbon nuclei in a 1 kg sample times ( 29 ( × = × ×× × × = × 19 3 23 19 26 6 19 4.62 MeV nucleus times 1 kW h 2.25 10 MeV 1 KWh 10 g 6.02 10 atoms mole 12 g mole 4.62 MeV 2.25 10 MeV 0.5 10 4.62 kWh 10.3 10 kWh 2.25 10
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
226 CHAPTER 14 NUCLEAR PHYSICS APPLICATIONS 14-3 ( 29 ( 29 ( 29 ( ( 29 ( α = +-- =+ -- = 9 12 n B eC 931.5 MeV u 4.002 603 u 9.012 182 u 12.000 000 u 1.008 665 u 931.5 MeV u 5.70 MeV Q M MMM Q 14-4 ( 29 ( 29 ( ( ( 29 ( α = + =+-- =- 27 30 n A lP 931.5 MeV u 4.002 603 u 26.981 538 u 29.978 310 u 1.008 665 u 931.5 MeV u 2.64 MeV Q M M MM Q 14-5 ( 29 [ ] = 931.5 MeV u aXYb Qmmmm ( 29 ( 29 ( 29 [ ] [ ] [ ] α   = = 174 H Li He u 931.5 MeV u 1.007 825 u 7.016 004 u 4.002 603 u 4.002 603 u 931.5 MeV u 17.35 MeV Q m m mm Q Q 14-6 ( 29 ( 29 ( 29 ( 29 = + 4 1 0 1 1 32 1 H eB HC Q M M MMc ( 29 ( 29 ( 29 ( 29 = = -= 13 1 10 4 2 2 1 2 12 C H B He ; Q Q Q QQ 14-7 (a) ( 29 ( 29 ( 29 ( 29 ( = + 14 4 17 1 N He O H 931.5 MeV u Q Using Table 13.6 for the masses. ( 29 ( ( 29 ( 29 ( 29 ( 29 = + +  =- - + =   4 14 14 14.003 074 u 4.002 603 u 16.999 132 u 1.007 825 u 931.5 MeV u 1.19 MeV H eN 4.002 603 1.19 MeV 1 1.53 MeV 14.003 074 N th Q Q Q K m (b) ( 29 ( 29 ( 29 ( = +- 7 11 Li H 2 He 931.5 MeV u Q m ( 29 (29 ( [ ] ( = = 7.016 004 u 1.007 825 u 2 4.002 603 u 931.5 MeV u 17.35 MeV Q Q 14-8 (a) ( 29 ( 29 ( 29 ( 29 ( = + 9 4 1 21 Be He C n 931.5 MeV u Q [ ] ( = 9.012 182 u 4.002 603 u 12.000 000 u 1.008 665 u 931.5 MeV u 5.70 MeV Q Q (b) ( 29 ( 29 ( 29 ( = 2 31 2 H He n 931.5 MeV u Q m (29 ( [ ] ( = = 2 2.014 102 u 3.016 029 u 1.008 665 u 931.5 MeV u 3.27 MeV; exothermic Q Q
Background image of page 2
MODERN PHYSICS 227 14-9 (a) CM SYSTEM M a M X v V χ χ χχ ν == +  =+=   2 22 CM 2 a a aa p M MV MM p pp K M M LAB SYSTEM M a M X v + V (29 ( 29 ( χ =+ + = + lab 2 2 2 lab lab Eq. 1 for substituting and in Eq. 1. a a a a P MvV p vV M p M MM p K Comparing to CM K , χ χ + = lab CM a KK M or χ  = -+   1 a th M KQ M (b) First calculate the Q -value ( 29 ( 29 ( 29 ( 29 ( [ ] ( = + -- = + =- 14 4 17 1 N He O H 931.5 MeV u 14.003 074 u 4.002 603 u 16.999 132 u 1.007 825 u 931.5 MeV u 1.19 MeV Q mm Q Q Then ( 29 ( 29 = = - - += 4 14 He 1 N 4.002 603 1.19 MeV 1 1.53 MeV 14.003 074 th th m m K 14-10 If 1 N is the number of particles emerging from the first layer, then σ - = 11 10 nx N Ne . Also, σ - = 1 N Ne and therefore, ( σ = 0 nx nx . For a material consisting of k layers: σ =    0 1 exp k ii i N N .
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
228 CHAPTER 14 NUCLEAR PHYSICS APPLICATIONS 14-11 σ - = 0 nx R Re , = 2 m x , = 0 0.8 RR , ρ - - = = × 3 2 83 27 atom 70 kg m 4.19 10 m 1.67 10 kg n m , σ - = 00 0.8 R Re , σ - = 0.8 e , σ =- ln 0.8 , (29 σ - - - = = = ×= ×× 3 02 2 1 0.223 ln 0.8 2.66 10 m 0.026 6 b 4.19 10 m 2 m nx 14-12 (a) σ - = 0 I Ie or ( 29 σ -  = ==   1 0 1 11 ln ln 1.25 cm 1 cm 0.286 5 I n xI To reduce to - 4 10 , ( 29 ( 29 σ = 44 1 cm ln 10 ln 10 7.37 cm 1.25 x n . (b) σ - = 1 1.25 cm n where ( 29 ( 29 ( σ σ - - - = × = 3 23 2 23 1 2 2 32 11.35 g cm 6.02 10 atoms mole 207 g mole 3.30 10 atoms cm 1.25 cm 3.30 10 cm 3.70 10 cm 38 b n 14-13 Equation 14.4 gives ( σ = 0 R R nx . Using values of E and σ , we have (a) σ σ 10 10 0.037 3 R R , (b) = 1 0.1 0.066 3 R R , and (c) 0.1 0.01 1 R R (d) Therefore we can use cadmium as an energy selector in the range 0.1 eV to 10 eV to detect order of magnitude changes in energy.
Background image of page 4
MODERN PHYSICS 229 14-14 σ= 2 000 barns (See Figure 14.3) the attenuation length is L
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 20

PC Chapter 14 - 14 Nuclear Physics Applications 14-1 18 O =...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online