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# We are given the following atomic masses $^{238}\text{U}=238\cdot 05076{{\text{u}}^{234}}\text{T h}=234\cdot 04363{{\text{u}}^{4}}\text{He}=4\cdot 00260\text{u}$ the energy released during the alpha decay of $^{238}\text{U}$ is:(A) $6\cdot 00\text{ MeV}$ (B) $4\cdot 25\text{ MeV}$ (C) $3\cdot 75\text{ MeV}$ (D) $5\cdot 03\text{ MeV}$

Last updated date: 23rd Jul 2024
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Hint: Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle and thereby transforms or decays into a different atomic nucleus, with a mass number that is reduced by four and an atomic number that is reduced by them.
Example: $^{238}\text{U}{{\to }^{238}}\text{T h }{{\text{+}}^{4}}\text{H e}+\text{Q}$
Where Q is energy released. We can find out Q by $\text{Q}=\vartriangle \text{M}{{\text{C}}^{2}}$
Where $\vartriangle \text{M}$ = mass difference.

Complete step by step solution
Given: mass of $^{238}\text{U}=238\cdot 05079\text{u}$
Mass of $^{234}\text{T h}=234\cdot 04363\text{u}$
Mass of $^{4}\text{H e}=4\cdot 002604$
And we know that
$1\text{ u}=9315\cdot \text{Me V/}{{\text{c}}^{2}}$
The energy released $\left( \text{ }\!\!\alpha\!\!\text{ }-\text{particle} \right)$
\begin{align} & \text{Q}=\left( {{\text{M}}_{\text{u}}}-{{\text{M}}_{\text{TH}}}-{{\text{M}}_{\text{He}}} \right){{\text{C}}^{2}} \\ & =\left( 238\cdot 05079-234\cdot 04363-4\cdot 00260 \right)\text{u}\times {{\text{C}}^{2}} \\ & =\left( 0\cdot 00456 \right)\text{u}\times {{\text{C}}^{2}} \\ & =0\cdot 00456\times 931\cdot 5\dfrac{\text{MeV}}{{{\text{C}}^{2}}}\times {{\text{C}}^{2}} \\ & =4\cdot 24\text{ MeV} \\ \end{align}
So, when $^{238}\text{U}$ is decayed, $4\cdot 24\text{ MeV}$ energy is released.
Therefore the option (B) is correct.

Note
In general, alpha particles have a very limited ability to penetrate other materials. In other words, those particles of ionizing radiation can be blocked by a sheet of proper, skin, or even a few inches of air. When Q is positive, that means reaction is endothermic and when Q is negative, reaction is exothermic.