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The energy released in the following process is:
$A + B \to C + D + Q$
(mass of $A$ is $1.002{\text{ amu}}$; mass of $B$ is $1.004{\text{ amu}}$; mass of $C$ is $1.001{\text{ amu}}$; mass of $D$ is $1.003{\text{ amu}}$)
(A) $1.254MeV$
(B) $0.931MeV$
(C) $0.465MeV$
(D) $1.862MeV$

Last updated date: 25th Jul 2024
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Hint The energy released in any chemical process is in the form of the difference in mass of the reactants and the products. This disappeared mass from the reaction gets converted to energy from the mass-energy equivalence theorem. So by calculating the difference in mass from the given mass of reactants and products, we can then convert that mass to energy by multiplying it by $931.478MeV$.

Formula Used To solve this problem we use the formula,
$E = \Delta m \times {c^2}$
where $E$ is the energy released in the process, $\Delta m$ is the difference in mass between the reactants and products and $c$ is the speed of light given by $3 \times {10^8}m/s$

Complete step-by-step answer:
The reaction given in the question is
$A + B \to C + D + Q$
where the variables in the L.H.S of the equation, $A$ and $B$ are the reactants and the variables on the R.H.S of the chemical equation $C$ and $D$ are the products of the reaction, where the $Q$ denotes the energy released in the reaction.
In the question, we can see that the masses are given in the unit of amu or the atomic mass unit.
The amu is defined as the mass of one-twelfth of a ${C^{12}}$ carbon atom. In nuclear science, the unit amu is more conveniently used than the S.I. or the C.G.S unit of mass. The conversion factor from amu to kg is given by
$1amu = 1.6606 \times {10^{ - 27}}kg$
So in the question, we are given,
$A = 1.002amu$, $B = 1.004amu$, $C = 1.001amu$ and $D = 1.003amu$.
So the sum of the masses of the reactants is the sum of masses of $A$ and $B$,
$\left( {1.002 + 1.004} \right)amu = 2.006amu$
And the sum of the masses of the products is the sum of masses of $C$ and $D$,
$\left( {1.001 + 1.003} \right)amu = 2.004amu$
So, from here, we can see that the mass of the reactants is much less than that of the products.
The difference in mass is denoted by $\Delta m$ and is given by,
$\Delta m = \left( {2.006 - 2.004} \right)amu = 0.002amu$
Now we can convert this mass unit to S.I. unit by multiplying it with a factor of $1.6606 \times {10^{ - 27}}$
$\Delta m = 0.002 \times 1.6606 \times {10^{ - 27}}kg$
$ \Rightarrow \Delta m = 3.3212 \times {10^{ - 30}}kg$
This mass is converted to energy by the formula, $E = \Delta m \times {c^2}$
$\therefore E = 3.3212 \times {10^{ - 30}} \times {\left( {3 \times {{10}^8}} \right)^2}J$
$ \Rightarrow E = 2.98908 \times {10^{ - 13}}J$
But the answer is given in the units of MeV, so we can convert the energy in MeV by,
\[1MeV = 1.6 \times {10^{ - 19}}J\]
So, we can get the answer by dividing the energy by a factor of \[1.6 \times {10^{ - 19}}\].
\[\therefore E = \dfrac{{2.98098 \times {{10}^{13}}}}{{1.6 \times {{10}^{ - 19}}}}MeV\]
\[ \Rightarrow E = 1.862MeV\]

So, the correct answer is option D.

Note We can solve this problem in an alternative way by directly converting the mass difference to the energy by the conversion factor,
\[1amu = 931.5MeV\]
So the energy will be given by,
\[E = \Delta m \times 931.5Mev\]
Now putting the value of $\Delta m$ we get,
$E = 0.002 \times 931.5MeV$
$ \Rightarrow E = 1.862MeV$