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Mass - Energy Relation of Nuclei

Q: The energy equivalent of \$0.5 g\$ of a substance is :

The energy equivalent of the particular mass is given as:\$E=mc^2\$mass is given as \$=0.5g=0.5\times 10^{-3}\ kg\$and the speed of light \$(c)=3\times 10^8\ m/s\$;.So, \$E=0.5\times10^{-3}\times(3\times 10^{8})^2\$\$E=4.5\times 10^{!3}\ J\$

Q: The energy equivalent to a substance of mass \$1\$g is?

\$E=mC^2\$\$=1\times 10^{-3}\times 9\times 10^{16}\$\$=9\times 10^{13}\$J

Q: The energy equivalent of \$1\ amu\$ is

\$1\ amu =1.66\times 10^{-27} kg\$According to Einstein's mass energy equivalence, \$E=mc^2\$ where \$c=\$ velocity of light. So, \$E=1.66\times 10^{-27}\times (3\times 10^8)^2=14.94\times 10^{-11} J\$\$E=\dfrac{14.94\times 10^{-11}}{1.6\times 10^{-19}} eV\$ where \$1eV=1.6\times 10^{-19} J\$\$E=931\times 10^{6} eV=931\ MeV\$

Q: If both the number of protons and the number of neutrons are conserved in each nuclear reaction in what way is mass converted into energy (or vice-versa) in a nuclear reaction?

The sum of masses of nuclei of product element is less than sum of masses of reactants and hence, loss of mass takes place during the reaction. This difference of mass of product element and reactant converts into energy and liberated in the form of heat.\$H^2_1\$ +\$H^2_1\$→ \$H^3_2 \$+ \$n^2_1\$ + \$energy\$

Q: In nuclear fission, \$0.1\%\$ mass is converted into energy. The energy released in the fission of \$1\$Kg mass in KWH is

\$0.\$1% Mass is converted.\$E = (\dfrac{1}{1000}\times 1 kg)(3\times 10^8)^2 J\$ \$= 9\times 10^{13} J\$Let this energy is librated per second then,\$E\$ per hour \$= \dfrac{9\times 10^{13}}{60\times 60}\$\$= 0.25\times 10^{11} WH\$\$= 2.5\times 10^7 KWH\$

Q: One milligram of matter is converted into energy. The energy released will be

Here, \$m = 1\ mg = 1\times 10^{-3} g \$ \$= 1\times 10^{-6}kg\$According to Einstein mass-energy equivalence \$E = mc^{2}\$where \$c\$ is the speed of light in vacuum\$\therefore E = (1\times 10^{-6} kg)(3\times 10^{8}ms^{-1})^{2}\$\$ = 9\times 10^{10}J\$.

Q: The mass defect in a particular nuclear reaction in 0.3 grams.The amount of energy liberated in kilowatt hour is \$\left( Velocity\ of \ light=3\times { 10 }^{ 8 }m/s \right) \$

Mass defect in a nuclear reaction \$\Delta M = 0.3 g = 3 \times 10^{-4} kg\$Thus amount of energy released \$E = \Delta M c^2 = (3 \times 10^{-4}) \times (3 \times 10^8)^2 J\$\$\implies E = 27 \times 10^{12} J (1 kWh = 3.6 \times 10^6 J)\$\$\therefore E = 7.5 \times 10^{6} kWh \$

Q: What does mean by mass defect? Establish relation between mass defect and nuclear binding energy. And hence write the expression for binding energy per nucleon.

Mass defect: It is found that mass of a stable nucleus is alway less than the sum of masses of its constituent protons and neutrons in their free state. Consider the Nucleus \$^A_ZX\$. It has Z protons (A-Z) neutrons, therefore its mass defect will be \$\Delta m=Zm_p+(A-Z)m_n-m\$Expression for binding energy: The nucleus \$^A_ZX\$ contains Z protons and \$(A-Z)\$ neutrons.Its mass defect is \$\Delta m =Zm_p+(A-Z)m_n-m_N\$Where \$m_N \rightarrow\$ mass of nucleus \$\Delta E_b=\Delta m \times c^2=[Zm_p+(A-Z)m_n-m_N]c^2\$Mass defect always improve binding energy of nucleus. For mass defect is more stable nucleus.

Q: The energy equivalent to \$1kg\$ of matter in (in Joule)

Einstein mass -energy relation : \$E=mc^2=1\times (3\times 10^8)^2=9\times 10^{16} \sim 10^{17}\$ J

Q: A star initially has \$10^{40}\$ deuterons. It produces energy via the processes \$_1H^2+_1H^2\rightarrow _1H^3+p\$ and \$_1H^2+_1H^3\rightarrow _2He^4+n\$. If the average power radiated by the star is \$10^{16}\$ W, the deuteron supply of the star is exhausted in a time of the order of [The mass of nuclei are as follows: \$M(H^2)=2.014 amu, M(n)=1.008 amu,\$\$M(p)=1.007 amu, M(He^4)=4.001 amu\$ }

Adding two equations gives:\$3 _{ 1 }^{ 2 }{ H }\quad \longrightarrow \quad _{ 2 }^{ 4 }{ He }\quad +\quad n\quad +\quad p\$Mass-Energy conversion:\$\left( 3\times 2.014- 4.001 -2\times 1.008 \right) { 3\times { 10 }^{ 8 } }^{ 2 }\$Therefore, energy released by \$3\$ deuterons is \$3.73\times { 10 }^{ -12 }J\$\$1u\ = 1.66\times { 10 }^{ -27 }kg\$Given, average power of a star is \${10}^{16}W\$.If number of deuterons lost per second is \$N\$ then:\$N\times \dfrac { 3.73\times { 10 }^{ -12 } }{ 3 } ={ 10 }^{ 16 }\\ \Rightarrow \quad N= \dfrac { 1 }{ 1.24 } \times { 10 }^{ 28 }\$Deuteron supply will be exhausted in \$\dfrac { { 10 }^{ 40 } }{ \dfrac { 1 }{ 1.24 } \times { 10 }^{ 28 } } = 1.24\times { 10 }^{ 12 }s\$

Physics

definition

Relationship between mass and energy

By the theory of special relativity Einstein showed that mass is another form of energy and one can convert mass-energy into other forms of energy, say kinetic energy and vice-versa.Einstein gave the famous mass-energy equivalence relation

E = m c^{2}

where

c = 3 \times 1 0^{8} m / s

is the speed of light.

example

Energy equivalent of mass

Energy released if mass of 2 amu is converted into energy is

E = Δ m c^{2}

= (2 \times 1.67 \times 1 0^{- 27} k g) \times (3 \times 1 0^{8} \frac{m}{s})^{2}

= 3 \times 1 0^{- 10} J

definition

Mass Energy Equivalence

The "mass-energy equivalence" principle states that the mass of a system and its energy are the same property in any physical system.
This means that anything having mass has an equivalent amount of energy and vice versa.

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