Question 305: Atoms & Nuclei

Question

The energy released when $\frac{7}{17.13} \mathrm{~kg}$ of ${ }_3^7 \mathrm{Li}$ is converted into ${ }_2^4 \mathrm{He}$ by proton bombardment is $\alpha 	imes 10^{32} \mathrm{eV}$. The value of $\alpha$ is $\_\_\_\_$ . (Nearest integer) (Mass of ${ }_3^7 \mathrm{Li}=7.0183 \mathrm{u}$, mass of ${ }_2^4 \mathrm{He}=4.004 \mathrm{u}$, mass of proton $=1.008 \mathrm{u}$ and $1 \mathrm{u}=931 \mathrm{MeV} / \mathrm{c}^2$ and Avogadro number $=6.0 	imes 10^{23}$ )

Accepted Answer

The given reaction is the proton bombardment of lithium:
$ {}_3^7Li + {}_1^1p ightarrow 2\, {}_2^4He $
Mass of reactants:
$ m_{	ext{reactants}} = m({}_3^7Li) + m({}_1^1p) = 7.0183 + 1.008 = 8.0263~u $
Mass of products:
$ m_{	ext{products}} = 2 	imes m({}_2^4He) = 2 	imes 4.004 = 8.008~u $
Mass defect:
$ \Delta m = m_{	ext{reactants}} - m_{	ext{products}} = 8.0263 - 8.008 = 0.0183~u $
$ E_{	ext{per reaction}} = \Delta m 	imes 931~	ext{MeV} $
$ E_{	ext{per reaction}} = 0.0183 	imes 931 = 17.04~	ext{MeV} $
Convert this to eV:
$ E_{	ext{per reaction}} = 17.04 	imes 10^6~	ext{eV} = 1.704 	imes 10^7~	ext{eV} $
Given mass of Li:
$ m_{	ext{given}} = \frac{7}{17.13}~	ext{kg} $
Convert to grams:
$ m_{	ext{given}} = \frac{7}{17.13} 	imes 1000 = 408.6~	ext{g (approximately)} $
Molar mass of Li-7 = 7 g/mol.
Number of moles:
$ n = \frac{408.6}{7} = 58.37~	ext{mol} $
Number of atoms:
$ N = n 	imes N_A = 58.37 	imes 6.0 	imes 10^{23} = 3.50 	imes 10^{25} $
Each Li atom undergoes one reaction, so
$ E_{	ext{total}} = N 	imes E_{	ext{per reaction}} $
$ E_{	ext{total}} = 3.50 	imes 10^{25} 	imes 1.704 	imes 10^7 $
$ E_{	ext{total}} = 5.96 	imes 10^{32}~	ext{eV} $
The question states:
$ E_{	ext{total}} = \alpha 	imes 10^{32}~	ext{eV} $
$ \Rightarrow \alpha = 5.96 \approx 6 $
Final Answer
$ \boxed{\alpha = 6} $

Atoms & Nuclei

Integer / Numeric Answer Type