Question

An Isotopic species of $LiH$ is a potential nuclear fuel. On the basis of the reaction calculate the expected power production in MW, associated with $1.00gm$ of $LiH$ per day. ( process is $100\% $ efficient)
${}_3^6Li + {}_1^2H\xrightarrow{{}}2{}_2^4He$
$[{}_3^6Li = 6.01512u,\,{}_1^2H = 2.01410u,\,{}_2^4He = 4.00260u]$ $u$ is in $amu$ and $1\,u = 931MeV$

Answer 1

Hint: If there is a difference in the predicted mass value and the observed mass value, the excess mass is said to be associated with a certain amount of energy. This energy is responsible for the power production.

Complete step by step answer:
We need to first calculate if there is a mass defect and if there is what is the value of the mass defect.
$\Delta M = {M_{final}} - {M_{initial}}$
Where $\Delta M = $ mass defect
Hence, $\Delta M = 2 \times {M_{He}} - [{M_{Li}} + {M_H}]$
On substituting the values of masses of hydrogen , helium and lithium we get
$ \Rightarrow \Delta M = 2 \times 4.00260 - [6.01512 + 2.01410]$
$ \Rightarrow \Delta M = \; - 0.0240u$
The negative sign tells us that the initial mass is greater than final mass,
$\therefore $ There is a mass defect.
$E = m{c^2}$
$ \Rightarrow E = 0.0240 \times 931 \times {10^6}eV$
$ \Rightarrow E = 22.344eV$
To convert $eV$to $Joules$ multiply by $1.6 \times {10^{ - 19}}$
$ \Rightarrow E = 22.344 \times 1.6 \times {10^{ - 12}}J$
$ \Rightarrow E = 3.59 \times {10^{ - 12}}J$
For Molal energy of $LiH$ will be$ = 3.58 \times {10^{ - 12}} \times 6.022 \times {10^{23}}\;J{\kern 1pt} {\kern 1pt} mo{l^{ - 1}}$
$ = 2.16 \times {10^{12}}\;J{\kern 1pt} mo{l^{ - 1}}$
Mass of $LiH$= Mass of $Li$+ Mass of $H$
$[{M_{Li}} = 7g\quad {M_H} = 1g]$
$\therefore {M_{LiH}} = 7 + 1$
$ \Rightarrow 8g$
Energy Per gram of $LiH$=$\dfrac{{Molal{\kern 1pt} energy{\kern 1pt} of{\kern 1pt} LiH}}{{Mass{\kern 1pt} of{\kern 1pt} LiH}} \times Given{\kern 1pt} {\kern 1pt} mass{\kern 1pt} of{\kern 1pt} LiH$
On substituting the values of energy and mass , we get
$ \Rightarrow E = \dfrac{{2.16 \times {{10}^{ - 12}}}}{8} \times 1.00J{g^{ - 1}}$
$ \Rightarrow E = 0.27 \times {10^{12}}J{g^{ - 1}}$
Energy of $LiH$ produced per second = $\dfrac{{0.27 \times {{10}^{12}}}}{{24 \times 60 \times 60}}J{g^{ - 1}}{s^{ - 1}}$
$ \Rightarrow 3.125 \times {10^6}J{s^{ - 1}}{g^{ - 1}}$
$[1W = 1J{s^{ - 1}}]$
$\therefore Energy\,of\,LiH\, = 3.125 \times {10^6}W{g^{ - 1}}$
or $E = 3.125MW$


Note:
Since the efficiency is $100\% $ there is no need to multiply this value, but if the question had specified any other percentage of efficiency, it would have been multiplied by the same. For example; $96\% $ efficiency would mean only $96\% $ of the produced energy is being harnessed and hence the power production value would be multiplied by a factor of $0.96$.
Be careful with the conversion as there are multiple conversions taking place. In order to minimise error, write down the units at the end of every step and tally the units with the previous equation at every step and also mention the answer in the Units specified in the question only.