Consider the following nuclear reaction :
$^{238}\text{U}\rightarrow{}^{234}\text{Th}+{}^4\text{He}$
Take masses of $^{238}\text{U}$ , $^{234}\text{Th}$ and $^4\text{He}$ as $238.050$ u, $234.043$ u and $4.003$ u, respectively. The Q value for the reaction, in keV, is :
[Given : $1$ u $=931.5$ MeV c $^{-2}$ ]

Question

Consider the following nuclear reaction :

^{238}\text{U}\rightarrow{}^{234}\text{Th}+{}^4\text{He}

Take masses of

^{238}\text{U}

,

^{234}\text{Th}

and

^4\text{He}

as

238.050

u,

234.043

u and

4.003

u, respectively. The Q value for the reaction, in keV, is :
[Given :

1

u

=931.5

MeV c

^{-2}

]

Quantrex Academy · Accepted Answer

Correct answer: B. 3726

The mass defect Δ m for the given nuclear reaction is: Δ m = m(^ 238 U) - [m(^ 234 Th) + m(^4He)] Δ m = 238.050 - (234.043 + 4.003) Δ m = 238.050 - 238.046 = 0.004 u The Q value of the reaction is given by: Q = Δ m × 931.5 MeV Q = 0.004 × 931.5 MeV Q = 3.726 MeV Converting to keV: Q = 3.726 × 10^3 keV = 3726 keV Answer: 3726

Consider the following nuclear reaction : ^ 238 Uarrow ^ 234 Th+...

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