Beta Decay and Radioactive Processes

The radioactive element undergoes four alpha decays and one beta decay as follows:

Radioactive series are classified by their mass number modulo 4. For a mass number 223, 223 / 4 = 55 with a remainder of 3. Thus, it belongs to the 4n+3 series (Actinium series).

Time taken in falling a height $h=3000-1000=2000m$ 

$ _{A}^{Z}\textrm{X}$ $\rightarrow  _{A-1}^{Z}\textrm{Y} $  $+  \beta^{-1}$

The given nuclear reaction is-

The nuclear reaction :   $ _{15}P^{30}\rightarrow$   ${14}Si^{30} + $  $ _{+1}e^0$

When a nucleus of mass number 220 at rest decays by emitting an alpha particle, momentum is conserved. The alpha particle (mass 4) and daughter nucleus (mass 216) recoil in opposite directions. The Q-value is the sum of kinetic energies of the alpha particle and the daughter nucleus. Using momentum conservation, K_alpha / K_daughter = M_daughter / M_alpha = 216 / 4 = 54. Given K_daughter = 0.2 MeV, K_alpha = 54 * 0.2 = 10.8 MeV. Thus, Q = K_alpha + K_daughter = 10.8 + 0.2 = 11 MeV.

Antiparticle of electron $ _{ -1 }{ e }^{ 0 }$ is positron $ _{ +1 }{ e }^{ 0 }$

$\beta^-$ decay : $^A _ZX \rightarrow ^A _{Z+1}Y $ $+ $  $ _{-1}^0e $

In an endothermic nuclear reaction, heat energy is given to the reaction to proceed and thus according to Einstein's mass-energy equivalence law, heat energy given gets converted into additional mass which results in the formation of products with more mass than the reactants.

$\alpha $ - decay 

$\beta^-$ emission is due to decay of neutron in the nucleus $n\rightarrow p+e^-$.

The rate equation is dN/dt = alpha - lambdaN. Solving this with N(0)=N0 and alpha=2N0lambda leads to N(t) = 2N0 - N0exp(-lambdat). At t = half-life (ln2/lambda), N = 2N0 - N0/2 = 1.5N0. As t approaches infinity, N approaches 2*N0.

After time t, 90% remains (N/N0 = 0.9). After time 2t, the fraction remaining is (0.9)^2 = 0.81. The amount decayed is 1 - 0.81 = 0.19, or 19%.

$84P _o^{210}\xrightarrow 82Pb^{206}+\alpha \Rightarrow $ mass of remaining nucleus $(Pb)=206$

When the neutron to proton ratio in the nucleus is too great a beta particle is emitted. In basic beta decay a neutron is transformed into a proton and an electron. The electron is then emitted as a beta particle which increases the atomic number by one and the molar mass is unchanged.

Hence, the proton- neutron ratio increase.

A free neutron undergoes beta decay into a proton, an electron, and an antineutrino, represented by the equation n -> p + e^- + anti-v. Thus, option A correctly describes the decay process.

$Remaining$ nuclei after $5$ years will be $N=5\times 10^{16} \times (\dfrac{1}{2})^{5/10}=\dfrac{5\times 10^{16}}{\sqrt[2]{2}}=\dfrac{5\times 10^{16}}{1.414}=3.54\times 10^{16}$

N1(t) = 40 * (1/2)^(t/20). N2(t) = 160 * (1/2)^(t/10). Set N1(t) = N2(t): 40 * (1/2)^(t/20) = 160 * (1/2)^(t/10). Dividing by 40: (1/2)^(t/20) = 4 * (1/2)^(t/10). This simplifies to (1/2)^(t/20) = 2^2 * (1/2)^(t/10). Solving for t gives 40s.

$\begin{array}{l} N={ N _{ 0 } }{ e^{ -\lambda t } } \ \therefore if, initial, rate, is, same, , and, { \lambda _{ A } }={ \lambda _{ B } } \end{array}$

Nuclear radius R = R0 * A^(1/3), where R0 is approx 1.2 fm. 36 = 1.2 * A^(1/3). A^(1/3) = 30. A = 30^3 = 27000. This calculation suggests the value 36 fm is very large. If R0 = 1.2, A = 27 is R = 1.2 * 3 = 3.6 fm. The question likely meant 3.6 fm.

The initial activity of 200 g of carbon-14 for a living organism is 200 g * 15 decay/min/g = 3000 decay/min. The current activity is given as 375 decay/min. Using the radioactive decay law, A = A_0 * (1/2)^(t / T_1/2), we have 375 = 3000 * (1/2)^(t / 5730), which simplifies to (1/2)^3 = (1/2)^(t / 5730). Thus, t / 5730 = 3, giving t = 3 * 5730 = 17190 years.

A free neutron will decay with a half life of about $10.3$ minutes but it is stable if combined into a nucleus. The stability of a nucleus is determined by number of neutrons as well as protons, Only fast moving neutrons are capable of penetrating the nucleus.A few neutron decays into a proton, an electron and anti-neutrino, and other options are known facts.

$ _{3}Li^{7} + ^{1} _{1}P \Rightarrow  _{4}Be^{8} + _{0}^{0}\gamma (energy)$
As, the emitted particles is only energy and gamma photons radiation is nothing but energy so, option  C is correct.

Electron capture :   $^{57} _{27}$  $ + $ $ ^0 _{-1}e$ $\rightarrow$  $^{57} _{26}Fe$

$\beta^{+}$ decay cannot occur in an isolated proton because it requires energy due to the mass of the neutron being greater than the mass of the proton. $\beta^{+}$ decay can only happen inside nuclei when the daughter nucleus has a greater binding energy.

Energy released Q = delta_m * c^2. This energy is shared as kinetic energy between the two daughter nuclei. K1 + K2 = Q. Since masses are equal, velocities are equal. 2 * (1/2 * (M/2) * v^2) = delta_m * c^2. (M/2) * v^2 = delta_m * c^2. v^2 = (2 * delta_m / M) * c^2. v = c * sqrt(2 * delta_m / M).

The given reaction :      $^{13} _7N\rightarrow ^{13} _6C + ^0 _1e + ^A _Z X$

The emission rate of a radioactive material is proportional to the number of radioactive nuclei present (activity A = lambda * N). Since the material decreased from 10g to 3g, the amount of radioactive material is now 0.3 times the initial amount. Therefore, the emission rate also becomes 0.3 times the original rate.

 Positive beta decay $(β+\ \ \text{decay})$ also called the  positron emission. In this decay  a proton in the parent nucleus decays into a neutron that remains in the daughter nucleus, and the nucleus emits a neutrino and a positron, which is a positive particle like an ordinary electron in mass but of opposite charge. Thus, positive beta decay produces a daughter nucleus, the atomic number of which is one less than its parent and the mass number of which is the same.

We know that in any $\beta -$ emission the mass number of the nucleus does not changes .

The reaction of beta decay is shown as :

A (d, 2n) reaction on U-238 (Z=92, A=238) involves adding a deuteron (Z=1, A=2) and removing two neutrons (Z=0, A=1 each), resulting in Z=93, A=238 (Np-238). A subsequent beta decay (Z increases by 1, A stays same) results in Z=94, A=238, which is Pu-238.

${ \beta  }^{ - }\quad decay$ :
Example: $ _{ 6 }^{ 14 }{ C }\longrightarrow _{ 7 }^{ 14 }{ N }+{ e }^{ - }+\bar { { v } _{ e } }$
Atomic number increases by 1 unit, which implies increase in proton.

$Zn$ with higher no.of nucleons has a lower mass than $Cu$, which means that the binding energy/nucleon is higher in $Zn$.
Which means that $Zn$ is more stable than $Cu$.
Hence, $Cu$ will have a tendency to convert to $Zn$ by radioactive decay
$\beta$ decay to change the atomic number.
$\gamma$ decay can't help in changing the no. of protons in the nucleus.

The electron emitted in beta radiation may originates from neutron and it increases the atomic number $1$.

Same no. of nucleons for $Cu^{64}$ and $Zn^{64}$
However, $M _{Cu}> M _{Zn}$
which indicates that the mass defect/ nuclear binding energy per nucleon is lesser of $Cu$, hence it will have a tendency to get to more stable form by $\beta $ decay.

Since the charge is already conserved in the decay process, proton cannot be ejected.

The positron or antielectron is the antiparticle or the antimatter counterpart of the electron. The positron has an electric charge of $+1e$, a spin of $\cfrac{1}{2}$ and has the same mass as an electron.

Beta decay can involve the emission of either electrons or positrons. The electrons or positrons emitted in a $\beta$ - decay do not exist inside the nucleus. They are only created at the time of emission, just as photons are created when an atom makes a transition from higher to a lower energy state.
In negative $\beta$- decay a neutron in the nucleus is transformed into a proton, an electron and an antineutrino. Hence, in radioactive decay process, the negatively charged emitted $\beta$- particles are the electrons produced as a result of the decay of neutrons present inside the nucleus.

Stable nucleus has mass deficit: binding energy means M_nucleus < Σ(m_nucleons). This mass defect equals binding energy/c². Option B is wrong. Option C mentions fusion, not fission. Option D has 'very low nucleus' which is garbled; should say 'very heavy nucleus'.

The given reaction :    $^2 _1 H$  $+$  $^{63} _{29} Cu  \rightarrow      ^{64} _{30} Zn  $  $+$  $^A _Z X$

We know that for $\beta^-$ decay, one neutron is converted into proton and electron and the electron is ejected in the reaction. For this, the number of neutrons will decrease by one unit. 

When beta decay occurs, the atomic number of nucleus will increase by one unit and mass number will remain same. Thus, here the after beta decay the daughter nucleus has same mass number as product nuclei and one unit atomic number will increase than parent nuclei.    

$\beta^+$ decay: $^{60} _{28}Co \rightarrow$ $^{A} _{Z}X$ $+$ $^0 _{+1}e$ 

Beta decay of $ _Z^A X$ :    $ _Z^A X \rightarrow   _{Z+1}^A Y$ $+$ $  ^o _{-1}e$

$^{57}Co$ is undergoing beta decay i.e electron is being produced.But an electron has very less mass ($9.11\times10^{-31}kg$) as compared to the $Co$ atom.therefore,after $570$ days,even the atom go under large beta decay,the weight of the material in the container will be nearly $10 g$.

$^1 _0n\rightarrow$ $^1 _1p+$ $^0 _{-1}e$
$Q=(m _n=m _p-m _e)c^2$

The following principles hold true in any situation in the universe
1. Conservation of linear momentum (if no external force)
2. Conservation of angular momentum (no external moment)
2. Conservation of energy
3. Conservation of charge

When a positron is emitted from nucleus, proton will convert into a neutron. So atomic number will be decrease by one  but atomic mass will be constant. 

$ _{6}^{11}\textrm{C}\rightarrow _{5}^{11}\textrm{B}+ _{+1}^{0}\textrm{e}$

Since mass of the isobar has decreased, an electron has been ejected.

Since the no. of nucleons will conserved .