Answer to Question #285977 in Molecular Physics | Thermodynamics for Sadeeq

Question #285977

Find the minimum kinetic energy (in MeV) in the lab system needed by an alpha particle to cause


the reaction:


7𝑁


14 + 𝛼 β†’ 8𝑂


17 + 1𝐻


1


Given that: m(


14N) = 14.00307 u; m(Ξ±) 4.003879 u; m(


1H) = 1.00783 u; m(


17O) = 17.0013 u;


1 𝑒 = 931.5 𝑀𝑒𝑉/𝑐


2

1
Expert's answer
2022-01-10T09:16:36-0500

Solution:

N14+Ξ±β†’O17+pN^{14}+ \alpha \to O^{17}+p


Energy=E=Ξ”mc2Energy=E= \Delta mc^2


Ξ”m=14.00307+4.003879βˆ’1.00783βˆ’17.0013=βˆ’0.002181\Delta m=14.00307+4.003879-1.00783-17.0013=-0.002181


E=Ξ”mc2=0.002182Γ—931.5𝑀𝑒𝑉=βˆ’2.0316015β€…β€ŠMeVE= \Delta mc^2=0.002182 \times931.5 𝑀𝑒𝑉=-2.0316015\;MeV


So, -2.0316015 MeV is needed by an alpha particle to cause the reaction.



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