Lamarsh Solution Chapter 2

Lamarsh Solutions Chapter-2 2.5 This is a question of fortune, For molecules which have an approximate weight of 2, on that point are two 1H and we kitty find the probability or the percentage over 1 as, 0.99985*0.99985=0.99970 The same computer science digest be made for the mol. weights of 3 and 4 For 3 on that point are one 1H and one 2 H and so, 0.99985*0.00015=1.49e-4 For 4 there are two 2 H and so, 0.00015*0.00015=2.25e-8 2.7 From table of nuclides we can find the atomic weights of O and H using the abundances 1 x[99.759 x15.99492 ? 0.097 x16.99913 ? 0.204 x17.99916] 100 m(O) ? 15.99938 1 m( H ) ? x[99.985 x1.007825 ? 0.015 x 2.0141] 100 m( H ) ? 1.007975 m( H 2O) ? 18.01533 m(O) ? (a) # of moles of water= 50 =2.76 moles 18.01533 (b) # of 1H atoms=2.76 moles x 0.6022e24 x 2 atoms of H x 0.99985 abund. =3.32e24 atoms of 1H (c) # of 2 H atoms=2.76 moles x 0.6022e24 x 2 atoms of H x 0.015e-2 abund. =4.98e20 atoms of 2 H 2.16 The nuclear fission of the nucleus of 235 U releases approximately 200MeV. How much energy (in kilowatt- hours and megawatt-days) is released when 1 g of 235 U undergoes fission? Solution: 1g ?235 U ? 1g ? 0.6022 ?1024 fissioned ?235 Uatoms ? 2.563 ?1021 atoms 235 g 2.563 ?1021 atoms ? Released brawniness: ? 512.6 ?1021 MeV ? ? 22810kWhr ? 200MeV ? 512.6 ?1021 MeV 1atoms 4.450 ?10?26 kWhr ? 22810kWhr 1MeV 1day ? 0.9505MWd 24hr 2.

20 Erest ? m0 c 2 Etot ? mc 2 m? m0 1? ,in question we see the square of E_rest and E_tot ,so we do ?2 c2 Erest 2 m0c 2 2 ( ) ?( 2 ) Etot mc m0 2 Erest 2 ( ) ! ?( ) 2 and from here we find m Etot comparison between m_0 and m as m0 2 ?2 ( ) ? 1? 2 m c we find, ?2 Erest 2 1? 2 ? c Etot 2 and finally Erest 2 ? ? c (1 ? ) Etot 2 2 2 and Erest 2 ? ? c 1? Etot 2 2.22 (a) wavelength of a 1 MeV photon can be demonstrate as, as wanted. and knowing the ?? 1.24e ? 6 1.24e ? 6 ? ?...If you want to liquidate a full essay, order it on our website: BestEssayCheap.com

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