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Question: We are given the following atomic masses \( ^{238}\text{U}=238\cdot 05076{{\text{u}}^{234}}\text{T h...

We are given the following atomic masses 238U=238⋅05076u234T h=234⋅04363u4He=4⋅00260u^{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 238U^{238}\text{U} is:
(A) 6⋅00 MeV6\cdot 00\text{ MeV}
(B) 4⋅25 MeV4\cdot 25\text{ MeV}
(C) 3⋅75 MeV3\cdot 75\text{ MeV}
(D) 5⋅03 MeV5\cdot 03\text{ MeV}

Explanation

Solution

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: 238U→238T h +4H e+Q^{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 Q=△MC2\text{Q}=\vartriangle \text{M}{{\text{C}}^{2}}
Where △M\vartriangle \text{M} = mass difference.

Complete step by step solution
Given: mass of 238U=238⋅05079u^{238}\text{U}=238\cdot 05079\text{u}
Mass of 234T h=234⋅04363u^{234}\text{T h}=234\cdot 04363\text{u}
Mass of 4H e=4⋅002604^{4}\text{H e}=4\cdot 002604
And we know that
1 u=9315⋅Me V/c21\text{ u}=9315\cdot \text{Me V/}{{\text{c}}^{2}}
The energy released ( !!α!! −particle)\left( \text{ }\\!\\!\alpha\\!\\!\text{ }-\text{particle} \right)
Q=(Mu−MTH−MHe)C2 =(238⋅05079−234⋅04363−4⋅00260)u×C2 =(0⋅00456)u×C2 =0⋅00456×931⋅5MeVC2×C2 =4⋅24 MeV \begin{aligned} & \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{aligned}
So, when 238U^{238}\text{U} is decayed, 4⋅24 MeV4\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.