24.118 CP - Chemistry: Molecular Nature Of Matter And Change

Question

The rd-century starship Enterprise uses a substance called “dilithium crystals” as its fuel. 

(a) Assuming this material is the result of fusion, what is the product of the fusion of two ${}^{6}Li$nuclei ?

(b) How much energy is released per kilogram of dilithium formed? (Mass of one ${}^{6}Li$ atom is 6.01512 amu.) 

(c) When four ${}^{1}H$ atoms fuse to form ${}^{4}He$, how many positrons are released? 

(d) To determine the energy potential of the fusion processes in parts (b) and (c), compare the changes in mass per kilogram of dilithium and of . 

(e) Compare the change in mass per kilogram in part (b) to that for the formation of ${}^{4}He$by the method used in current fusion reactors 

(f) Using early st-century fusion technology, how much tritium can be produced per kilogram of ${}^{6}Li$in the following reaction: ${}_3{}^{6}Li{+}_0^{1}n{\to }_2^{4}He{+}_1^{3}H$? When this amount of tritium is fused with deuterium, what is the change in mass? How does this quantity compare with the use of dilithium in part (b)?

Step-by-Step Solution

Answer

1. Dilithium is the product of the fusion of two ${}^{6}Li$ nuclei.
2. $2.26\times {10}^{14}\frac{J}{kg}$energy is released per kilogram of dilithium.
3. 2 electrons are released.
4. He has a larger mass change.
5. The current procedure produces a smaller change in mass.
6. Mass $=3.125\times {10}^{-3}kg$;Dilithium has a smaller change in mass.

Step 1: Definition of Dilithium crystal

The rare crystal dilithium is employed in tactical warp drives. It's semi-permeable to both deuterium and anti-deuterium, and it acts as a natural chamber for a regulated matter-antimatter reaction, concentrating the energy and allowing it to be harvested and used for power.

Step 2: Determining product of fusion

The balanced reaction of two ${}^{6}Li$ nuclei fusing is:

${}_3{}^{6}Li{+}_3^6{\to }_6^{12}C$

Dilithium is the result of the fusion of two ${}^{6}Li$nuclei.

Step 3: Determining energy released per Kg of dilithium

Using Einstein's equation, we calculate the amount of energy released per kilogram of dilithium formed:

$E = 4 m{c}^2$

When the mass difference between the reactants and the products is 4 m:

$4m = m_{products }-m_{reac\tan ts }$

One ${}^{6}Li$ atom has a mass of 6.015121 amu.

One ${}^{12}C$  atom has a mass of 12 amu.

$4m = 2(6.015121amu)-12amu = 0.0302\frac{amu}{ atom }\left(\frac{1.66054x{10}^{-27}}{1amu}\right)$

$4m = 5.0148x{10}^{-29}\frac{kg}{ atom }$

$E = 4m{c}^2$

$E = 5.0148x{10}^{-29}\frac{kg}{ atom }{\left(\frac{2.99792x{10}^{8}m}{s}\right)}^2\left(\frac{1J}{\frac{kg\times m^2}{s^2}}\right)$

$E = 4.5071x{10}^{-12}\frac{J}{atom}\left(\frac{atom}{12amu}\right)\left(\frac{1amu}{1.66054x{10}^{-27}}\right)$

$E = 2.2619\times {10}^{14}\frac{J}{kg}$

Step 4: Determining how many positrons are release

The equation for four ${}_1{}^{1}H$ atoms fusing to generate ${}_2{}^{4}He$ is:

$4_1^{1}H{\to }_2^{4}He+2_1^0\beta$

We can see that two positrons are released in this equation.

Step 5: Comparing change in mass in Kg of dilithium and of H 4 e

Compare the changes in mass per kilogram of dilithium and ${}^{4}He$ to estimate the energy potential of the fusion reactions in parts (b) and (c).

$Mass =\left(5.02080507x{10}^{-29}\frac{kg}{atom}\right)\left(\frac{1atom}{12amu}\right)\left(\frac{1amu}{1.66054x{10}^{-27}kg}\right)$

$Mass = 2.5202\times {10}^{-3}\frac{kg}{k{g}^{12}C}$

$4m = 4\left(mas{s}_1^{1}H\right)-\left(mas{s}_2^{4}He+2mas{s}_{-1}^{0}e\right)$

$4m = 4(1.007825amu)-\left[4.0026amu+2\left(5.4858x{10}^{-4}amu\right)\right.$

$4m = 0.0276\frac{amu}{atom}\left(\frac{1.66054x{10}^{-27}kg}{amu}\right)= 4.5831\times {10}^{-29}\frac{kg}{ atom }$

We must convert the computed value for 4m to obtain the mass of ${}^{4}He$.

$Mass = 4.5831x{10}^{-29}\frac{kg}{ atom }\left(\frac{1atom}{4.0026amu}\right)\left(\frac{1amu}{1.66054x{10}^{-27}k{g}^{4}He}\right.$

 

$Mass = 6.8955\times {10}^{-3}\frac{kg}{k{g}^{4}He}$

Step 6: Comparing change in mass in Kg in part (b)

The change in mass per kilogram in section (b) compared to the approach employed in existing fusion reactors to generate ${}^{4}He$.

${}_1{}^{2}H{+}_1^{3}H{\to }_2^{4}He{+}_0^{1}n$

$4m = (2.0140 amu+3.01605 amu)-(4.0026 amu+1.008665 amu)$

$4m = 0.0188\frac{amu}{atom}\left(\frac{1.66054x{10}^{-27}kg}{amu}\right) = 3.1218x{10}^{-29}\frac{kg}{ atom }$

$Mass = 3.1218x{10}^{-29}\frac{kg}{ atom }\left(\frac{1atom}{4.0026amu}\right)\left(\frac{1amu}{1.66054x{10}^{-27}}k{g}^{4}He\right)$

$Mass = 4.6969\times {10}^{-3}\frac{kg}{k{g}^{4}He}$

 

 

Step 7: Determining Change in mass

Reaction given:

${}_3{}^{6}Li{+}_0^{1}n{\to }_2^{4}He{+}_1^{3}H$

Mass

$Mass\frac{{}_1{}^{3}H}{k{g}^{6}Li}=\frac{1mo{l}^{3}H}{1mo{l}^{6}Li}\left(\frac{3.01605g^{3}H}{1mo{l}^{3}H}\right)\left(\frac{1mo{l}^{6}Li}{6.015121g^{6}Li}\right)\left(\frac{1000g^{6}Li}{1k{g}^{6}Li}\right)\left(\frac{1k{g}^{3}H}{1000g^{3}H}\right)$

$Mass\left(\frac{{}_1{}^{3}H}{k{g}^{6}Li}\right) = 0.5014\frac{k{g}^{3}H}{k{g}^{6}Li}$

$Mass = 0.5014k{g}^{3}H\left(\frac{1000g}{1kg}\right)\left(\frac{1mo{l}^{3}H}{3.01605g^{6}H}\right)\left(\frac{6.022x{10}^{23}a om{ }^{3}H}{mo{l}^{3}H}\right)\left(\frac{3.121852x{10}^{-29}kg}{a tom }\right)$

$Mass = 3.1254\times 10-3kg$

Reaction with dilithium:

$Mass = 5.02180507x{10}^{-29}\frac{kg}{ato{m}^{12}C}\left(\frac{1ato{m}^{12}C}{2ato{m}^{6}Li}\right)\left(\frac{6.022x{10}^{23}a om{ }^{6}Li}{mol}\right)\left(\frac{1mo{l}^{6}Li}{6.0515121g^{6}Li}\right)\left(\frac{1000g}{1kg}\right)$

$Mass = 2.5138x{10}^{-3}kg$

Dilithium fusion produces less mass change than tritium fusion.