Nuclear reactions - class-XII

According to Einstein, energy and mass are related by the relation.
$E = mc^2$
where c is the speed of light in vacuum.

Here, $m = 1\ mg = 1\times 10^{-3} g $

Since the density of nucleus is fixed.

$1u=931.5 MeV$
we know $1u=1.66\times 10^{-21}kv$
$931.5 MeV=931.5\times 10^6\times 1.6\times 10^{-19}J=1.49\times 10^{-10}$
$E=(1u)(c^2)=(1.66\times 10^{-27})(3\times 10^8)^2J=1.494\times 10^{-10}J$
so $(1u)(c)^2=931.5MeV$

By conservation of momentum, the momentum of the alpha particle (mass 4, speed v) must equal the momentum of the daughter nucleus (mass A-4, speed V). Thus, 4v = (A-4)V, so V = 4v / (A-4).

The binding energy per nucleon curve rises sharply for light nuclei, peaks around iron (mass number 56), and then gradually decreases for heavier nuclei.

$ Mc^{2} + (B.E) = (N _{mN} + Z _{mP})c^{2}$
where,
         $M _{c} =  $  total mass of nucleus.
         $N _{mN} =  $  total mass of neutrons
         $N _{mP}  =  $  total mass of protons

$\begin{array}{l} { E _{ b } }+{ E _{ c } }>{ E _{ a } } \ , \because some, , energy, , is, , goen, , in, , breaking, , nuclie, , of, , A \ Hence, \ option, , A, , is, , correct, , answer. \end{array}$

Nuclear density is constant, so mass m = density * volume. Therefore, m is directly proportional to volume (m proportional to v).

Pair production :
Gamma rays of sufficient energy when passing near a nucleus disappear and materialize into pair of an electron and a positron.
To have pair production, minimum energy of $\gamma$ - ray radiation is 1.02MeV.

In a pair annihilation, atleast two gamma rays must be produced. For example, when an electron encounters a positron, they annihilate to produce two gamma rays each having $0.511$ MeV energy.

Mass of electron $ = 9.1 \times 10^{-31}  kg$

Rest mass energy $ = mc^{2}$

Rest mass energy of a positron (as well as electron) is $0.51$ MeV.

When a particle encounters its antiparticle, they annihilate with the transformation of their mass energies into two gamma rays obeying law of conservation of mass.

Minimum energy to produce pair production should be 1.02 MeV.

As, $m=10^{-6}\ kg$

Commonly used moderators in nuclear reactors include water, heavy water (D2O), and graphite, which slow down fast neutrons effectively. Molten sodium is used as a coolant to transfer heat from the reactor core, not as a moderator.

Handling the source with long tongs will prevent him from the exposure of $\beta$- particles. Since the range of $\beta$-particle is only $30cm$ so, long tongs can easily be a safer and effective precaution.

One electron volt is the energy gained by an electron accelerated through a potential difference of 1 volt, which is 1.6 * 10^-19 Joules.

$\Delta m = 1  u$
$E = 1 \times 931.5  MeV = 931.5  MeV$

Sir Einstein's mass-energy equation states that mass and energy can be converted into each other by the following relation.

Before Sir Einstien, mass and energy were two completely different physical quantities, which were not related to each other anyway. Sir Einstein told that energy and mass are related to each other i.e. energy and mass can be converted into each other by the following relation:

The mass of object as observed by a moving observer is given by $m=\dfrac{m _0}{\sqrt{1-(\dfrac{v}{c})^2}}$.

931 MeV

A free proton is stable. A proton inside a nucleus can change into a neutron via beta-plus decay or electron capture to reach a more stable state.

Protons within a nucleus experience strong electrostatic repulsion due to their positive charges. The strong nuclear force, which is an attractive force acting between nucleons at very short ranges, is significantly stronger than this electrostatic repulsion, thereby holding the nucleus together.

According to Einstein's mass-energy equivalence principle, 1 atomic mass unit (u) is equivalent to approximately 931.5 MeV of energy. Since 1 MeV equals 10^6 eV, 931.5 MeV equals 931.5 * 10^6 eV.

Mass defect represents the difference between the mass of a nucleus and the sum of the masses of its individual nucleons. A larger mass defect corresponds to a higher binding energy per nucleon, which indicates greater nuclear stability.

$^{12}C+^{12}C\rightarrow ^{24}Mg$
mass of $^{12}C=12.0642 amu$
Total input mass $m _1=2\times 12.0642$
$=24.1284$ amu
Mass of $^{24}Mg$=output mass
$=m _0=23.955$
mass loss $=m _1-m _o$
$Am=24.1284-23.985$
$=0.1434 amu$
Energy released $=0.1434\times 931 Mev$
$=13.965 Mev$

Number of mole of radium, $n=1$

The number of nuclei of radium, $N={{N} _{a}}=avagdaro\,\,number=6.023\times {{10}^{23}}$

Activity, $A=\lambda N$

$ \because 1\,\,curie=3.7\times {{10}^{10}}\,decay/\sec  $

$ \dfrac{1}{3.7}\,kilocurie=0.27\,kilocurie=9.99\times {{10}^{12}}\,dacay/\sec  $

So, $\lambda =\dfrac{A}{N}=\dfrac{9.99\times {{10}^{12}}}{6.023\times {{10}^{23}}}=1.65\times {{10}^{-11}}\,$

Power required is P = 1 kW = 1000 J/s. Energy released per fission is 200 MeV = 200 * 10^6 * 1.6 * 10^-19 J = 3.2 * 10^-11 J. The number of fissions per second n is given by P = n * E, so n = 1000 / (3.2 * 10^-11) = 3.125 * 10^13 fissions per second, which matches option A.

Cadmium rods are used in nuclear reactors to absorb neutrons, which helps control the rate of the fission chain reaction.

In alpha decay, the mass number decreases by 4 and the atomic number decreases by 2. For U-238, A becomes 238 - 4 = 234 and Z becomes 92 - 2 = 90.

Free neutrons outside a nucleus are unstable and decay with a half-life of about 10.2 minutes. Protons, however, are stable particles. The statement 'Neutrons is unstable' (despite the grammatical error) correctly identifies that free neutrons undergo beta decay.

$\begin{array}{l} \left( { { M _{ n } }+{ M _{ H } }-{ M _{ 0 } } } \right) \times 431.5 \ =\left[ { 15.000108+1.007827-15.994915 } \right] \times 431 \ =12.13, Mev \end{array}$

Energy released = BE(products) - BE(reactants). BE(He) = 4 * 7 = 28 MeV. BE(2 deuterons) = 2 * (2 * 1.1) = 4.4 MeV. Energy released = 28 - 4.4 = 23.6 MeV.

Energy released = BE(He) - 2 * BE(deuteron) = (4 * 7) - 2 * (2 * 1.1) = 28 - 4.4 = 23.6 MeV.

Initial BE = 240 * 7.6 = 1824 MeV. Final BE = 2 * (120 * 8.5) = 2 * 1020 = 2040 MeV. Energy released = 2040 - 1824 = 216 MeV.

$ E = \Delta m c^{2}$
    $ = (2 \times 1.67 \times 10^{-27}  kg) \times (3 \times 10^{8} \frac{m}{s})^{2} $
    $ = 3 \times 10^{-10}  J$

Two photons are produced during the annihilation of an electron and a positron along with $1.02$ MeV released energy.

Mass defect: Δm = (2mp + 2mn) - mα = 2(1.00785) + 2(1.00866) - 4.00274 = 0.03028 u. BE = Δm × 931.5 MeV/u = 0.03028 × 931.5 ≈ 28.21 MeV. Note: Problem shows mp = 1000785 u which appears to be a typo for 1.00785 u.

For production to occur, the energy must be greater than the rest mass of proton + electron.
$E > ( 2 \times 0.511\ MeV = 1.022\ MeV)$
$h \nu > 1.022 \times  10^{6} \times 1.6 \times 10^{-19} J$
$\nu > \dfrac{1.6352 \times  10^{-13}}{h} = 2.5 \times 10^{20} Hz$

The pair production becomes possible with gamma energies exceeding $1.02\ MeV$, and becomes important as an absorption mechanism at energies over $5\ MeV$.

$ e^{-} + e^{+} \Rightarrow  \gamma +\gamma $
This annihilation is in accordance to three laws:-
1] conservation of linear momentum
2] conservation of charge
3] conservation of angular momentum
Two $\gamma $ ray photons are released on annihilation of electron and positron.

Binding energy is the amount of energy required to separate a particle from a system of particles or to disperse all the particles of the system. Binding energy is especially applicable to subatomic particles in atomic nuclei, to electrons bound to nuclei in atoms, and to atoms and ions bound together in crystals. Binding energy is more if the nucleus is relatively more stable.

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

$m=1\ a.m.u.=1.66\times10^{-27}kg$

Total binding energy of two deuterons = 2 × 2.2 = 4.4 MeV. Total binding energy of helium = 4 × 7 = 28 MeV. Energy released = 28 - 4.4 = 23.6 MeV. Note: This is another duplicate with 'meV' typo (should be MeV).

$E=mC^2$

A positive energy release (Q > 0) indicates a stable product, while a negative energy release (Q < 0) indicates an unstable product. Since the first reaction releases 20 MeV, He-4 is more stable than He-3. Since the second reaction requires 0.9 MeV, He-5 is less stable than He-4.

In the reaction, we need to identify what M represents. Based on nuclear reaction notation and the fact that it appears after the product nucleus, M is a neutron that is emitted in the reaction. The reaction shows helium-4 bombarding target X, producing an excited nucleus that emits gamma rays and a neutron.

Using E = mc^2, m = 1 mg = 10^-6 kg, E = 10^-6 * (3 * 10^8)^2 = 9 * 10^10 J.

Alpha decay involves the emission of a helium nucleus, which has a mass number of 4 and an atomic number of 2. Therefore, the daughter nucleus must have a mass number reduced by 4 and an atomic number reduced by 2.

The reaction shows a heavy nucleus being bombarded and breaking down into a lighter nucleus with the emission of multiple protons and neutrons. This process of splitting or breaking down is characteristic of nuclear separation or fission-like processes.

Annihilation is the conversion of matter into energy.
Using equation $E={ m } _{ e }{ c }^{ 2 }$
$\dfrac { hc }{ \lambda  } =2{ m } _{ e }{ c }^{ 2 }$
$\Rightarrow \lambda =\dfrac { h }{ { 2m } _{ e }{ c } } $

In option A, the mass number change is 243-242 = 1, but the reaction is (alpha, 2n). An alpha particle adds 4 to the mass and 2 to the atomic number, while 2 neutrons remove 2 from the mass. The net change should be +2 mass and +2 atomic number, making the product Cm-242 + He-4 - 2n = Bk-244, not Bk-243.

The radius of a charged particle in a magnetic field is r = p / (qB). Since both particles have the same momentum p and enter the same field B, the ratio r1/r2 = q2/q1. A proton has charge 1 and an alpha particle has charge 2, so r1/r2 = 2/1 = 2.

Photon wavelength: λ = 5 × 10⁻⁴ Å = 5 × 10⁻¹⁴ m. Energy: E = hc/λ = (1240 eV·nm)/(5 × 10⁻⁵ nm) ≈ 24.8 MeV. Total rest energy of e⁻e⁺ pair: 2 × 0.511 = 1.022 MeV. Available kinetic energy: 24.8 - 1.022 ≈ 23.78 MeV. Each particle gets half: ≈ 11.89 MeV, approximately 12 MeV. Note: '0.5 11 MeV' appears to be typo for 0.511 MeV.

$E={ mc }^{ 2 }={ 10 }^{ -6 }\times \left( 3\times { 10 }^{ 8 } \right) ^{ 2 }={ 10 }^{ -6 }\times 9\times { 10 }^{ 16 }=9\times { 10 }^{ 10 }J$

$E={ mc }^{ 2 }={ 10 }^{ -3 }\times{ 10 }^{ -3 } \left( 3\times { 10 }^{ 8 } \right) ^{ 2 }=9\times { 10 }^{ 10 }J$

1 amu
It is defined as one twelfth of the mass of an unbound neutral atom of carbon-12 in its nuclear and electronic ground state.

Energy of electron $={ m } _{ e }{ c }^{ 2 }$
Energy of positron $={ m } _{ p }{ c }^{ 2 }$
${ m } _{ e }$ = ${ m } _{ p }$, $ c=$ speed of light.

Rest energy of an electron $=m _{ e }{ c }^{ 2 }$
Here $m _{ e }=9.1\times { 10 }^{ -31 }kg$ and $C$ = velocity of light

$E=m{ c }^{ 2 }=10^{ -3 }\times { 10 }^{ -3 }\times { \left( 3\times {
10 }^{ 8 } \right)  }^{ 2 }= 9\times { 10 }^{ 10 }J$

Mass defect in a nuclear reaction          $\Delta M = 0.3    g  =  3 \times 10^{-4}    kg$

Total B.E. for $C^{13}$ is $13\times7.5=97.5\ MeV$

Let $E _{0}$ be the initial energy of neutron, the energy of neutron after 1 collision reduces to $E _{0}/2=E _{1}(let)$ i.e. $E _{1}/E _{0}=1/2$. 

The mass of the reactant nuclei $= 7.0160 + 1.0078 = 8.0238  u$
The mass of the product nuclei $= 4.0026 + 4.0026 = 8.0052  u$
Mass defect $= \Delta m = 8.0238 - 8.0052 = 0.0186  u$
Energy released $= 0.0186  u \times 931.5  MeV = 17.326  MeV$

The nucleons present in $ _{3}{Li}^{7}$ is $7$.
The binding energy per nucleon for lithium is ${39.2}/{7} = 5.6  MeV$
The binding per nucleon for helium is ${28.24}/{4} = 7.06  MeV$
The binding energy per nucleon is the measure of stability of the nuclei. Therefore, helium is more stable than lithium.

In a nuclear reaction, there may be conversion of some mass into energy. So,both mass and energy are not conserved. It is the momentum which is conserved.a

The chain reaction in nuclear reactor is controlled 

$1\ amu =1.66\times 10^{-27} kg$

BE per nucleon $^{16}O=7.97MeV$
BE per nucleon $^{17}O=7.75MeV$
$^{17}O\rightarrow { _0n^1}+{^{16}O}$
Energy required to remove neutron
$=17\times 7.75-16\times 7.97$
$=4.23MeV$.

Mass defect  $\Delta M = 0.03$ amu
Binding energy  $E _{B} = \Delta Mc^2 = \Delta M\times 931.5 $  MeV
$\therefore \ E _{B} = 0.03\times 931.5$ MeV $ =27.93 $ MeV

A

When two nuclei of masses ${m _1}$ and ${m _2}$ are fused to form a stable nucleus of mass ${m _3}$ and some of the mass is converted in energy.

Therefore,

${m _3} < {m _1} + {m _2}$

For electron and positron pair production, minimum energy is $1.02\ MeV$.
Energy of photon is given: $ 1.7 \times 10^{-3} J=\dfrac{1.7 \times 10^{-13}}{1.6 \times 10^{-19}}$$=1.06 , MeV$.
Since energy of photon is greater than 1.02 MeV, electron positron pair will be created.