Question 351: Atoms & Nuclei

Question

The average energy released per fission for the nucleus of ${ }_{92}^{235} \mathrm{U}$ is 190 MeV . When all the atoms of 47 g pure ${ }_{92}^{235} \mathrm{U}$ undergo fission process, the energy released is $\alpha 	imes 10^{23} \mathrm{MeV}$. The value of $\alpha$ is $\_\_\_\_$ .
(Avogadro Number $=6 	imes 10^{23}$ per mole)

Accepted Answer

The number of moles (n) is given by the formula :
$$ \mathrm{n}=\frac{	ext { Given Mass }}{	ext { Molar Mass }} $$
Given :
Mass of Uranium $=47 \mathrm{~g}$
Molar Mass of $\mathrm{U}_{92}^{235}=235 \mathrm{~g} / \mathrm{mol}$
$$ \mathrm{n}=\frac{47}{235}=\frac{1}{5}=0.2 \mathrm{moles} $$
Each mole contains a number of atoms equal to Avogadro's Number ( $N_A$ ).
$$ \mathrm{N}=\mathrm{n} 	imes \mathrm{N}_{\mathrm{A}} $$
$$\Rightarrow $$ $\mathrm{N}=0.2 	imes\left(6 	imes 10^{23}ight)$
Since each atom (nucleus) releases 190 MeV upon fission :
$$ \mathrm{E}=\mathrm{N} 	imes 	ext { Energy per fission } $$
$$\Rightarrow $$ $\mathrm{E}=\left(1.2 	imes 10^{23}ight) 	imes 190 \mathrm{MeV}$
$$\Rightarrow $$ $\mathrm{E}=228 	imes 10^{23} \mathrm{MeV}$
$\Rightarrow \alpha 	imes 10^{23} \mathrm{MeV}=228 	imes 10^{23} \mathrm{MeV}$
$$ \Rightarrow \alpha=228 $$
Therefore, the correct answer is $\mathbf{2 2 8}$.

Atoms & Nuclei

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