Question

The star ship Voyager, like many other vessels of the newly designed 24th-century fleet, uses antimatter as fuel.

(a) How much energy is released when 1.00kg each of antimatter and matter annihilate each other?

(b) When the antimatter is atomic antihydrogen, a small amount of it is mixed with excess atomic hydrogen (gathered from interstellar space during flight). The annihilation releases so much heat that the remaining hydrogen nuclei fuse to form ⁴He. If each hydrogen-antihydrogen collision releases enough heat to fuse$1.00\times {10}^{-5}$hydrogen atoms, how much energy (in kJ) is released per kilogram of antihydrogen?

(c) Which produces more energy per kilogram of antihydrogen, the procedure in part (a) or that in part (b)?

Short answer

a. Energy is released when 1.00 kg each of antimatter and matter annihilate each other is $-1.7975\times {10}^{17}J$.

b. Energy (in kJ) is released per kilogram of antihydrogen is $1.0297\times {10}^{-10}kJ$.

c. It is observed that procedure (b) produces more energy per kg of antihydrogen

Step-by-step solution

Definition of Nuclear energy

The energy released by “protons and neutrons” present in the nucleus is termed as nuclear energy.

This type of energy can be generated in one of two ways:

• Fission (when atom nuclei split into several pieces) or
• Fusion (when nuclei fuse together).

Calculation for the determination of energy in part a.

From Einstein's equation:

$\Delta E=m{c}^2$

wheremis the mass difference between reactants and products:

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

Binding energies are commonly expressed in millions of electron volts, that is, in mega-electron volts (MeV):

$1MeV={10}^{6}eV=1.602x{10}^{-13}J$

A particularly useful factor converts the atomic mass unit to its energy equivalent in electron volts:

$1amu=931.5x{10}^{6}eV=931.5MeV$

Calculation for the determination of energy in part a

The change in mass when 1 Kg of antimatter annihilates 1 Kg matter:

$$\begin{gathered} \Delta m=0-2kg=-2kg \\ \Delta E=\Delta m{c}^2.....\left(i\right) \end{gathered}$$

The known values are substituted.

$$\begin{gathered} \Delta E=(-2kg){\left(2.99792x{10}^8\frac{m}{s}\right)}^2\left(\frac{J}{\frac{kg·m^2}{s^2}}\right) \\ \Delta E=-1.7975\times {10}^{17}J \end{gathered}$$

(Note: it is negative since the energy is being released)

Calculation for the determination of energy in part b and explanation of part c

b. Energy released per kilogram of antihydrogen.

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

$$\begin{gathered} \Delta m=\left[4\right(1.007825amu\left)\right]-[4.0026amu+2(0.000549amu\left)\right] \\ =0.02760amu \end{gathered}$$

$$\begin{gathered} T_{\Delta m}=\left(1x{10}^5\right.Hatoms)\left(\frac{0.02760amu}{{}^{4}He}\right)\left(\frac{1{}^{4}He}{4Hatoms}\right) \\ {T}_{\Delta m}=6.9x{10}^{2}amu \end{gathered}$$

$$\begin{gathered} H_{anti}=1kg\left[\frac{1000g}{1kg}\right]\left[\frac{1mol}{1.008g}\right]\left[\frac{6.022x{10}^{23}antiH}{1molantiH}\right] \\ {H}_{anti}=5.9742\times {10}^{26}antiH \end{gathered}$$

$$\begin{gathered} E=6.9x{10}^{2}amu\left[\frac{931.5MeV}{1amu}\right]\left[\frac{1.602x{10}^{-16}kJ}{1MeV}\right] \\ =1.0297x{10}^{-10}kJ \end{gathered}$$

$$\begin{gathered} E_{released}=\left[\frac{1.0297\times {10}^{-10}kJ}{antiH}\right]\left[5.9742\times {10}^{26}\right.anti\left.H\right] \\ =6.151\times {10}^{16}kJ \end{gathered}$$

c. From the calculation with excess hydrogen, it is observed that procedure (b) produces more energy per kg of antihydrogen.