How much energy is released when the correct number of protons and neutrons come together to form deuterium? a) 0.52 MeV b) 1.7 MeV c) 2.5 MeV d) 6.1 MeV e) 9.3 MeV

Question

asked Apr 28, 2020 166k views

1 Answer

Buddhi Madarasinghe · Answer 1 · 2020-05-04T20:57:42+0000

Answer: The correct answer is 2.24 MeV.

Step-by-step explanation:

The chemical reaction for the formation of deuterium from proton and neutron follows:

$_(1)^1\textrm{H}+_0^1\textrm{n}\rightarrow _1^2\textrm{H}$

We are given:

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

Mass of
$_(0)^(1)\textrm{n}$ = 1.008665 u

Mass of
$_(1)^(2)\textrm{H}$ = 2.014102 u

To calculate the mass defect, we use the equation:

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

$\Delta m=(m_(_1^2H)+m_(n))-(m_(_1^2H))\\\\\Delta m=(1.00784+1.008665)-(2.014102)=0.002403u$

To calculate the energy released, we use the equation:

$E=\Delta mc^2\\E=(0.002403u)* c^2$

E=(0.002403u)* (931.5MeV) (Conversion factor:
)

E=2.24MeV

Hence, the energy released in the given nuclear reaction is 2.24 MeV.