Question

What is the energy absorbed in this endothermic nuclear reaction 14 7 N + 4 2 H e → 17 8 O + 1 1 H 7 14 N + 2 4 H e → 8 17 O + 1 1 H ? (The atomic mass of 14 N 14 N is 14.003074 u and that of 17 O 17 O is 16.999132 u)

Answer 1

Answer: The energy absorbed in the given nuclear reaction is 1.2072 MeV.

Explanation:

For the given nuclear reaction:

$_{7}^{14}\textrm{N}+_2^4\textrm{He}\rightarrow _{8}^{17}\textrm{O}+_{1}^{1}\textrm{H}$

We are given:

Mass of $_{7}^{14}\textrm{N}$ = 14.003074 u

Mass of $_{8}^{17}\textrm{O}$ = 16.999132 u

Mass of $_{2}^{4}\textrm{He}$ = 4.002602 u

Mass of $_{1}^{1}\textrm{H}$ = 1.00784 u u

To calculate the mass defect, we use the equation:

$\Delta m=\text{Mass of reactants}-\text{Mass of products}$

Putting values in above equation, we get:

$\Delta m=(m_N+m_{He})-(m_{O}+m_{H})\\\\\Delta m=(14.003074+4.002602)-(16.999132+1.00784)=-0.001296u$

To calculate the energy released, we use the equation:

$E=\Delta mc^2\\E=(-0.001296u)\times c^2$

$E=(-0.001296u)\times (931.5MeV)$    (Conversion factor:  $1u=931.5MeV/c^2$  )

$E=-1.2072MeV$    (negative sign indicates that energy is getting absorbed)

Hence, the energy absorbed in the given nuclear reaction is 1.2072 MeV.