Question

The atomic mass of uranium ${}_{92}{\text{U}}^{238}$ is $238.0508u$, that of thorium ${}_{90}{\text{Th}}^{234}$ is $234.0436u$ and that of an alpha particle ${}_2{\text{He}}^4$ is $4.0026u$. Determine the energy released when $\alpha -\text{decay}$ converts ${}_{92}{\text{U}}^{238}$ into ${}_{90}{\text{Th}}^{234}$. $[1u\times c^2=931.5\text{MeV}]$

• A. $1.1\text{MeV}$
• B. $4.3\text{MeV}$
• C. $8.2\text{MeV}$
• D. $15.9\text{MeV}$

Answer 1

The correct option is B $4.3\text{MeV}$

The given decay is represented as,

${}_{92}{\text{U}}^{238}\to {}_{90}{\text{Th}}^{234}{+}_2{\text{He}}^4$

The energy is released due to the difference in mass between the reactants and the products.

$\underset{238.0508u}{{}_{92}{\text{U}}^{238}}\to \underset{238.0462u}{\underset{}{\underset{234.0436u}{{}_{90}{\text{Th}}^{234}}+\underset{4.0026u}{{}_2{\text{He}}^4}}}$

The mass defect is given by,

$\Delta m=238.0508u-238.0462u$

$\Rightarrow \Delta m=0.0046u$.

Energy released, $E=\Delta m\times c^2$

$\Rightarrow E=0.0046u\times c^2$

We know that, $1u\times c^2=931.5\text{MeV}$

$\Rightarrow E=0.0046\times 931.5\text{MeV}$

$\Rightarrow E=4.28\text{MeV}\approx 4.3\text{MeV}$

Hence, option $\left(B\right)$ is correct.

Why this Question? ${}_Z{X}^A\to {}_{Z-2}{Y}^{A-4}+{}_{2}H{e}^4+Q$ $Q$ value: It is defined as the energy released during the decay process. $Q$ value = (rest mass energy of reactants - rest mass energy of products)