Question

An isotopic species of lithium hydride $LiH$ is a potential nuclear fuel, on the basis of the reaction calculate the expected power production in MW, associated with 1.00 g of $LiH$ per day(process is 100% efficient).

${}_3^{6}Li{+}_1^{2}H⟶2_2^{4}He$
${[}_3^{6}Li=6.01512u,{}_1^{2}H=2.01410u,{}_2^{4}He=4.00260u]$, u is amu and $1u=931MeV$]

• A. $3.125MW{g}^{-1}$
• B. $4.135MW{g}^{-1}$
• C. $5.144MW{g}^{-1}$
• D. None of these

Answer 1

The correct option is A $3.125MW{g}^{-1}$

$△m=massdecay=m{(}_3^{6}Li)+m{(}_1^{2}H)-2m{(}_2^{4}He)$
$=(6.01512+2.01410)-2\times 4.0026$
$=0.02402$ u
thus, energy of one atomic event(one mass decay)
$=0.0240\times 931\times {10}^6$ eV
$=0.0240\times 931\times {10}^6\times 1.6\times {10}^{-19}$ J
$=3.59\times {10}^{-12}$ J
energy per mol of $LiH=3.59\times {10}^{-12}\times 6.02\times {10}^{23}$
$2.16\times {10}^{12}$ J $mo{l}^{-1}$
energy per g of $LiH=\frac{2.16\times {10}^{12}}{8}$ J $g^{-1}$
energy per g of LiH per second $=\frac{2.16\times {10}^{12}}{8\times 24\times 3600}$ J $g^{-1}{s}^{-1}$
$=3.125\times {10}^6$ J $s^{-1}{g}^{-1}$
$=3.125\times {10}^6$ W $g^{-1}$ $(J{s}^{-1}=1W)$
$=3.125MWg-1$ $(MW={10}^{6}W)$