When a uranium isotope \$^{235}_{92}U\$ is bombarded with a neutron, it generates \$^{89}_{36}Kr\$, three neutrons and :

\$_{92}U^{235} + \ _on^1 \to \ _QX^P + _{36}Kr^{89} + 3\ _on^1+ Q(energy)\$\$\sum\$ Atomic number on \$LHS = 89 + 0 = 89 + 3\$\$\sum\$ Atomic number on \$RHS = Q + 36 + 3 \times 0\$\$\therefore LHS = RHS\$\$Q+36 = 92\$\$Q = 56\$also, \$\sum\$ atomic mass number on \$LHS = 235 + 1 = 236\$\$\sum\$ Atomic mass number on \$RHS=P + 89 + 3 \times 1\$\$\therefore LHS = RHS\$\$P + 92= 236\$\$P = 144\$So, the element is \$_{56}Ba^{144}\$

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Pair Production and Pair Animation

Q: What is the wavelength of a photon emitted during a transition from \$n = 5\$ state to \$n = 2\$ state in the hydrogen atom?

\$\displaystyle \dfrac { 1}{ \lambda }=R [\dfrac { 1 }{n^2_1 } - \dfrac { 1}{ n^2_2 }] \$\$\displaystyle \dfrac { 1}{ \lambda }=109677.6 \ cm^{-1} \times [\dfrac { 1 }{2^2 } - \dfrac { 1}{ 5^2 }]\$\$\displaystyle \dfrac { 1}{ \lambda }=23032.3 \ cm^{-1} \$\$\displaystyle \lambda = 4.34 \times 10^{-5} \ cm\$Convert unit from cm to nm1 cm \$\displaystyle =10^7 \ nm\$\$\displaystyle \lambda = 4.34 \times 10^{-5} \ cm \times \dfrac {10^7 \ nm}{ 1 \ cm}\$\$\displaystyle \lambda = 434 \ nm\$

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.

Physics

definition

Pair production and annihilation

Pair production is the creation of an elementary particle and its antiparticle, for example creating an electron and positron, a muon and antimuon, or a proton and antiproton. Pair production often refers specifically to a photon creating an electron-positron pair near a nucleus but can more generally refer to any neutral boson creating a particle-antiparticle pair. In order for pair production to occur, the incoming energy of the interaction must be above a threshold in order to create the pair, at least the total rest mass energy of the two particles and that the situation allows, both energy and momentum to be conserved.There exists an inverse process to pair production called pair annihilation, in which a particle and its antiparticle collide and annihilate each other, the total energy of the two particles appearing as electromagnetic radiation.

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