Determine the mass of `\ ^6Li` when it is bombarded by deuteron, disintegrating into two alpha particles by the release of 22.3 MeV (0.023940 u) energy. The mass of deuteron is 4.002603 u and that of an alpha particle is 4.002603 u. (Answer: 6.01504 u)  



Given Data:

Mass of energy released = Δm = 0.023940 u

Mass of deuteron `\ _{1}^{2}H` = `\m_H` = 2.014102 u

Mass of deuteron `\ _{2}^{4}He` = `\m_{He}` = 4.002603 u



To Find:

Mass of `\ _{3}^{6}Li` = `\m_{Li}` = ?



Solution:

According to the given conditions, the reaction is

 `\ _{3}^{6}Li`  +  `\ _{1}^{2}H`   →  2(`\ _{2}^{4}He`) + Energy

From this reaction we can calculate the mass `\ _{3}^{6}Li` as 

`\m_{Li}` + `\m_H` = 2 (`\m_{He}`) + Δm 

or


`\m_{Li}`  = 2 (`\m_{He}`) +  Î”m   - `\m_H`

by putting values

`\m_{Li}`  = 2 (4.002603 u) + 0.023940 u - 2.014102 u

`\m_{Li}`  = 8.005206 u + 0.023940 u - 2.014102 u

`\m_{Li}`  = 6.015044 u  ------------- Ans.



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