Dear Student,(a) The total momentum before the reaction is zero and afterward also it is zero Therefore, MLinLi=MHevHenHe=9 30×106 ms-1 is not close to the speed of light c and nLi will be even less since MLi>MHe Thus, we can write KLi=12MLiv2Li=12MLiMHevHe MLi2=MHe2vHe22MLiKLi=(4 0026)2(1 66×10−27)(9 30×106)22(7 0160)(1 66×10−27) =1 02MeV(B) We know that Kα=KX=0, so Q=KLi+KHe(As given in question)KHe =12MHev2HeKHe =12(4 0026)(1 66×10−27)(9 30×106)2KHe =1 80MeVso, Q=1 02MeV+1 80MeV=2 82MeVRegards

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D e a r S t u d e n t, (a) T h e t o t a l m o m e n t u m b e f o r e t h e r e a c t i o n i s z e r o a n d a f t e r w a r d a l s o i t i s z e r o . T h e r e f o r e, M_{L i} n_{L i} = M_{H e} v_{_{H e}} n_{_{H e}} = 9.30 \times 10^{6} m s^{- 1} i s n o t c l o s e t o t h e s p e e d o f l i g h t c a n d n_{L i} w i l l b e e v e n l e s s \sin c e M_{L i} > M_{H e} . T h u s, w e c a n w r i t e K_{L i} = \frac{1}{2} M_{L i} v^{2} L i = \frac{1}{2} M_{L i} {(\frac{M_{H e} v_{_{H e}}}{M_{L i}})}^{2} = \frac{{M_{H e}}^{2} {v_{H e}}^{2}}{2 M_{L i}} K_{L i} = {(4.0026)}^{2} (1.66 \times 10^{- 27}) (9.30 \times 10

⁶)22(7.0160)(1.66×10−27) =1.02MeV(B) We know that Kα=KX=0, so Q=KLi+KHe(As given in question)KHe =12MHev2HeKHe =12(4.0026)(1.66×10−27)(9.30×106)2KHe =1.80MeVso, Q=1.02MeV+1.80MeV=2.82MeVRegards

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