For the reaction ^{14}N + alpha longrightarro | vedclass

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Question

(B) The reaction is $^{14}N + {}_{2}^{4}He \rightarrow {}^{17}O + {}_{1}^{1}H$. Since energy is absorbed, the mass of the products is greater than the

Vedclass · Accepted Answer

$16.9991$

Vedclass · Answer

(B) The reaction is $^{14}N + {}_{2}^{4}He \rightarrow {}^{17}O + {}_{1}^{1}H$. Since energy is absorbed, the mass of the products is greater than the mass of the reactants. Mass defect $(\Delta m) = 0.00124 \ amu$. $(m_{{}^{17}O} + m_{p}) - (m_{reactants}) = \Delta m$. $m_{{}^{17}O} + 1.00782 \ amu - 18.00567 \ amu = 0.00124 \ amu$. $m_{{}^{17}O} = 18.00567 + 0.00124 - 1.00782 = 16.99909 \ amu$. Rounding to four decimal places, we get $16.9991 \ amu$.

For the reaction $^{14}N + \alpha \longrightarrow {}^{17}O + p$, $1.16 \ MeV$ (Mass equivalent $= 0.00124 \ amu$) of energy is absorbed. Mass on the reactant side is $18.00567 \ amu$ and proton mass $= 1.00782 \ amu$. The atomic mass of ${}^{17}O$ will be (in $amu$)

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