答:Emax = E×4m M/(m+M)2,即Emax/E = 4m M/(m+M)2 = 4×1×3/(1+3)2 = 0.75
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(補充)
碰撞後轉移給氚核能量為Emax=E×4m M/(m+M)2,即Emax/E = 4m M/(m+M)2
(證明)
※當14M eV快中子與一靜止的氚核/碳核發生正面碰撞時(一次碰撞)
(1)此快中子將轉移其入射能量之多少百分比給氚核/碳核?轉移多少MeV能量給氚核/碳核?
Emax=E×4m M/(m+M)2,即Emax/E = 4m M/(m+M)2
(2)此快中子剩下多少百分比能量?剩下多少MeV能量?
Emin=E×(m-M)2/(m+M)2,即Emax/E = (m-M)2/(m+M)2
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