Properties of Alpha, Beta and Gamma Rays and Decay Process Questions in English

(C) The intensity of radiation passing through a material is given by the formula $I = I_0 e^{-\mu t}$,where $I$ is the final intensity,$I_0$ is the initial intensity,$\mu$ is the absorption coefficient,and $t$ is the thickness of the material.
Given that the intensity is reduced to half $(I = I_0 / 2)$ for a thickness $t = 2.5 \, mm$.
Substituting these values into the equation: $I_0 / 2 = I_0 e^{-\mu (2.5)}$.
This simplifies to $1/2 = e^{-2.5 \mu}$.
Taking the natural logarithm on both sides: $\ln(1/2) = -2.5 \mu$.
$-\ln(2) = -2.5 \mu$.
$0.693 = 2.5 \mu$.
$\mu = 0.693 / 2.5 = 0.2772 \, mm^{-1}$.
Rounding to two decimal places,we get $\mu \approx 0.28 \, mm^{-1}$.

(A) In the given nuclear reaction,the atomic number increases by $1$ $(z \to z+1)$,while the mass number $A$ remains constant.
An electron $(_{-1}e^0)$ and an antineutrino $(\bar \nu)$ are emitted.
This process is characteristic of $\beta^-$ decay,where a neutron transforms into a proton,emitting an electron and an antineutrino.

(B) Let the mass number of the parent nucleus be $A = 220$. After emitting an $\alpha$-particle $(m_2 = 4)$,the daughter nucleus has a mass number $m_1 = 220 - 4 = 216$.
The total energy released in the reaction is $Q = K_1 + K_2 = 5.5 \, MeV$,where $K_1$ is the kinetic energy of the daughter nucleus and $K_2$ is the kinetic energy of the $\alpha$-particle.
By the law of conservation of linear momentum,the initial momentum is zero,so the magnitudes of the momenta of the daughter nucleus and the $\alpha$-particle must be equal:
$p_1 = p_2$
$\sqrt{2 m_1 K_1} = \sqrt{2 m_2 K_2}$
$m_1 K_1 = m_2 K_2$
$216 K_1 = 4 K_2$
$K_1 = \frac{4}{216} K_2 = \frac{1}{54} K_2$
Substituting this into the $Q$-value equation:
$K_1 + K_2 = 5.5$
$\frac{1}{54} K_2 + K_2 = 5.5$
$\frac{55}{54} K_2 = 5.5$
$K_2 = 5.5 \times \frac{54}{55} = 0.1 \times 54 = 5.4 \, MeV$.
Thus,the kinetic energy of the $\alpha$-particle is $5.4 \, MeV$.

(B) During $\beta^-$ decay,a neutron inside the nucleus transforms into a proton,an electron $(e^-)$,and an antineutrino $(\bar{\nu}_e)$.
The reaction is: $n \to p + e^- + \bar{\nu}_e$.
As a result of this process,the number of neutrons $(n)$ decreases by $1$ and the number of protons $(p)$ increases by $1$.
Since the numerator $(n)$ decreases and the denominator $(p)$ increases,the overall ratio $[n/p]$ will decrease.

(C) The initial nucleus is $_{72}X^{180}$.
$1$. After $\alpha$-decay ($_{2}He^{4}$ emission),the mass number decreases by $4$ and the atomic number decreases by $2$: $_{72}X^{180} \xrightarrow{\alpha} _{70}X_1^{176}$.
$2$. After $\beta$-decay (emission of an electron,$_{ -1}e^{0}$),the mass number remains unchanged and the atomic number increases by $1$: $_{70}X_1^{176} \xrightarrow{\beta} _{71}X_2^{176}$.
$3$. After another $\alpha$-decay,the mass number decreases by $4$ and the atomic number decreases by $2$: $_{71}X_2^{176} \xrightarrow{\alpha} _{69}X_3^{172}$.
$4$. After $\gamma$-decay,there is no change in mass number or atomic number: $_{69}X_3^{172} \xrightarrow{\gamma} _{69}X_4^{172}$.
Therefore,the mass number of $X_4$ is $172$ and the atomic number is $69$.

(D) The initial nucleus is $_Z{X^A}$.
When a nucleus emits an $\alpha$-particle $(_{2}^{4}He)$,the mass number decreases by $4$ and the atomic number decreases by $2$.
When it emits a $\beta$-particle $(_{-1}^{0}e)$,the mass number remains unchanged and the atomic number increases by $1$.
After emitting $3 \alpha$-particles and $5 \beta$-particles,the new mass number $A'$ and atomic number $Z'$ are:
$A' = A - 3(4) = A - 12$
$Z' = Z - 3(2) + 5(1) = Z - 6 + 5 = Z - 1$
Number of protons $P = Z' = Z - 1$
Number of neutrons $N = A' - Z' = (A - 12) - (Z - 1) = A - Z - 11$
Therefore,the ratio of neutrons to protons is $\frac{N}{P} = \frac{A - Z - 11}{Z - 1}$.

(B) The initial atomic number is $Z = 92$.
An $\alpha$-decay $(_{2}He^{4})$ decreases $Z$ by $2$.
$A$ $\beta^-$-decay $(_{-1}e^{0})$ increases $Z$ by $1$.
$A$ $\beta^+$-decay $(_{+1}e^{0})$ decreases $Z$ by $1$.
Counting the particles:
- Total $\alpha$ particles: $8$ (emitted at steps $1$,$2$,$5$,$6$,$7$,$8$,$11$,$14$).
- Total $\beta^-$ particles: $4$ (emitted at steps $3$,$4$,$9$,$10$).
- Total $\beta^+$ particles: $2$ (emitted at steps $12$,$13$).
Calculation of final $Z$:
$Z_{final} = Z_{initial} - (8 \times 2) + (4 \times 1) - (2 \times 1)$
$Z_{final} = 92 - 16 + 4 - 2$
$Z_{final} = 78$.

(C) According to the law of conservation of linear momentum,the initial momentum of the system is equal to the final momentum.
Initially,the $U^{238}$ nucleus is at rest,so the initial momentum is $0$.
The decay process is: $U^{238} \to Th^{234} + He^{4}$.
Let $M_D = 234$ be the mass of the daughter nucleus and $M_{\alpha} = 4$ be the mass of the $\alpha$-particle.
Let $v$ be the recoil speed of the daughter nucleus and $u$ be the speed of the $\alpha$-particle.
Applying conservation of momentum: $0 = M_D \cdot v + M_{\alpha} \cdot u$.
$0 = 234 \cdot v + 4 \cdot u$.
Solving for $v$: $v = -\frac{4u}{234}$.
The magnitude of the recoil speed is $\frac{4u}{234}$.

(A) In an alpha decay process,the mass number decreases by $4$ and the atomic number decreases by $2$.
Given the reaction: $_{92}U^{238} \to _{Z}X^{A} + _{2}He^{4}$.
According to the law of conservation of mass number: $238 = A + 4 \implies A = 238 - 4 = 234$.
According to the law of conservation of atomic number: $92 = Z + 2 \implies Z = 92 - 2 = 90$.
Therefore,$Z = 90$ and $A = 234$.

(D) The uranium series starts with an unstable radioactive parent nuclide and undergoes a series of alpha and beta decays until it reaches a stable isotope.
By definition,a stable nuclide does not undergo radioactive decay.
Since the decay constant $\lambda$ is defined as the probability of decay per unit time,for a stable nuclide,the probability of decay is zero.
Therefore,the decay constant of the stable end product is $0$.

(A) Let the atomic number of $A$ be $Z$ and its mass number be $M$.
In the first step: $A \to B + _{2}^{4}He$.
The atomic number of $B$ becomes $Z - 2$ and the mass number becomes $M - 4$.
In the second step: $B \to C + 2e^{-}$.
Since each beta decay $(e^{-})$ increases the atomic number by $1$ without changing the mass number,the emission of $2e^{-}$ increases the atomic number of $B$ by $2$.
Therefore,the atomic number of $C = (Z - 2) + 2 = Z$.
The mass number of $C$ remains $M - 4$.
Since $A$ and $C$ have the same atomic number $Z$ but different mass numbers,they are isotopes.

(A) $1$. In the first step,${}_{Z}^{A}X \to {}_{Z+1}^{A}Y$,the atomic number $Z$ increases by $1$ while the mass number $A$ remains constant. This corresponds to the emission of a $\beta^-$ particle.
$2$. In the second step,${}_{Z+1}^{A}Y \to {}_{Z-1}^{A-4}K$,the mass number $A$ decreases by $4$ and the atomic number $Z$ decreases by $2$. This corresponds to the emission of an $\alpha$ particle.
$3$. In the third step,${}_{Z-1}^{A-4}K \to {}_{Z-1}^{A-4}K$,there is no change in the mass number $A$ or the atomic number $Z$. This corresponds to the emission of a $\gamma$ ray (de-excitation of the nucleus).
Therefore,the order of emission is $\beta^-$,$\alpha$,$\gamma$.

(D) The number of $\alpha$-particles emitted is given by $n_{\alpha} = \frac{A - A'}{4} = \frac{200 - 168}{4} = \frac{32}{4} = 8$.
The change in atomic number due to $8$ $\alpha$-particles is $8 \times 2 = 16$. The new atomic number would be $90 - 16 = 74$.
Since the final atomic number is $80$,the number of $\beta$-particles emitted is $n_{\beta} = Z' - (Z - 2n_{\alpha}) = 80 - (90 - 16) = 80 - 74 = 6$.
Thus,the number of $\alpha$ and $\beta$ particles are $8$ and $6$ respectively.

(B) The ionizing power of radioactive radiations follows the order: $\alpha > \beta > \gamma$. Thus,the increasing order of ionizing power is $\gamma < \beta < \alpha$.
The penetrating power of radioactive radiations follows the order: $\gamma > \beta > \alpha$. Thus,the increasing order of penetrating power is $\alpha < \beta < \gamma$.

(A) The initial nucleus is $_a^b X$.
$1$. Positron emission $(_{+1}^0 \beta)$: $_a^b X \rightarrow _{a-1}^b X_1 + _{+1}^0 \beta$.
$2$. Emission of two $\alpha$-particles $(_{2}^4 \alpha)$: $_{a-1}^b X_1 \rightarrow _{a-1-4}^{b-8} X_2 + 2(_{2}^4 \alpha) = _{a-5}^{b-8} X_2$.
$3$. Emission of two $\beta$-particles $(_{-1}^0 \beta)$: $_{a-5}^{b-8} X_2 \rightarrow _{a-5+2}^{b-8} X_3 + 2(_{-1}^0 \beta) = _{a-3}^{b-8} X_3$.
$4$. Emission of one $\alpha$-particle $(_{2}^4 \alpha)$: $_{a-3}^{b-8} X_3 \rightarrow _{a-3-2}^{b-8-4} Y = _{a-5}^{b-12} Y$.
Comparing this with $_d^c Y$,we get $c = b - 12$ and $d = a - 5$.

(B) The intensity of radiation after passing through a material of thickness $x$ is given by the formula: $I = I_0 e^{-kx}$,where $k$ is the absorption coefficient.
From the given data,when $x_1 = 36 \, mm$,$I_1 = I_0/8$.
Substituting these values: $I_0/8 = I_0 e^{-k(36)} \Rightarrow 1/8 = e^{-36k}$.
Taking the natural logarithm on both sides: $\ln(1/8) = -36k \Rightarrow -\ln(8) = -36k \Rightarrow \ln(2^3) = 36k \Rightarrow 3 \ln(2) = 36k \Rightarrow k = \frac{3 \ln(2)}{36} = \frac{\ln(2)}{12}$.
Now,we need to find the thickness $x_2$ when $I_2 = I_0/2$.
Using the formula: $I_0/2 = I_0 e^{-kx_2} \Rightarrow 1/2 = e^{-kx_2}$.
Taking the natural logarithm: $\ln(1/2) = -kx_2 \Rightarrow -\ln(2) = -kx_2 \Rightarrow \ln(2) = kx_2$.
Substituting the value of $k$: $\ln(2) = (\frac{\ln(2)}{12}) x_2$.
Solving for $x_2$: $x_2 = 12 \, mm$.

(A) Let the number of $\alpha$-particles emitted be $n_{\alpha}$ and the number of $\beta$-particles emitted be $n_{\beta}$.
The decay equation is: $_{90}Th^{238} \rightarrow _{83}Bi^{222} + n_{\alpha}(_{2}He^{4}) + n_{\beta}(-_{1}e^{0})$.
Equating the mass numbers: $238 = 222 + 4n_{\alpha} \Rightarrow 4n_{\alpha} = 16 \Rightarrow n_{\alpha} = 4$.
Equating the atomic numbers: $90 = 83 + 2n_{\alpha} - n_{\beta}$.
Substituting $n_{\alpha} = 4$: $90 = 83 + 2(4) - n_{\beta} \Rightarrow 90 = 83 + 8 - n_{\beta} \Rightarrow 90 = 91 - n_{\beta} \Rightarrow n_{\beta} = 1$.
Thus,$4 \alpha$ particles and $1 \beta$ particle are emitted.

(B) The $rad$ (radiation absorbed dose) is a unit of absorbed dose of ionizing radiation.
It is defined as the absorption of $0.01 \ J$ of energy per kilogram of matter.
Therefore,it measures the energy absorbed by the target from radiation.

(B) Let $m_1 = 216$ be the mass number of the daughter nucleus and $m_2 = 4$ be the mass number of the $\alpha$-particle.
Let $k_1$ and $k_2$ be their respective kinetic energies.
According to the law of conservation of energy,$k_1 + k_2 = 5.5 \, MeV \dots (i)$
According to the law of conservation of linear momentum,the magnitudes of momenta must be equal: $p_1 = p_2$.
Since $p = \sqrt{2mk}$,we have $\sqrt{2(216)k_1} = \sqrt{2(4)k_2}$.
Squaring both sides,$216 k_1 = 4 k_2$,which gives $k_2 = 54 k_1$.
Substituting this into equation $(i)$,$k_1 + 54 k_1 = 5.5 \, MeV$.
$55 k_1 = 5.5 \, MeV$,so $k_1 = 0.1 \, MeV$.
Therefore,the kinetic energy of the $\alpha$-particle is $k_2 = 54 \times 0.1 = 5.4 \, MeV$.

(A) Let the number of $\alpha$-particles emitted be $n_{\alpha}$ and the number of $\beta$-particles emitted be $n_{\beta}$.
The change in mass number is given by $A - A' = 4n_{\alpha}$.
$228 - 212 = 4n_{\alpha} \implies 16 = 4n_{\alpha} \implies n_{\alpha} = 4$.
The change in atomic number is given by $Z - Z' = 2n_{\alpha} - n_{\beta}$.
$90 - 83 = 2(4) - n_{\beta}$.
$7 = 8 - n_{\beta} \implies n_{\beta} = 1$.
Thus,$4$ $\alpha$-particles and $1$ $\beta$-particle are emitted.

(B) The intensity of radiation after traveling a distance $x$ is given by $I' = I_0 e^{-\mu x}$,where $I_0$ is the initial intensity and $\mu$ is the absorption coefficient.
Given that after $x_1 = 36 \, mm$,$I_1 = I_0 / 8$.
Substituting this into the formula: $I_0 / 8 = I_0 e^{-\mu (36)} \implies 8 = e^{36\mu} \implies \ln(8) = 36\mu \implies 3 \ln(2) = 36\mu \implies \mu = \frac{\ln(2)}{12}$.
We need to find the distance $x_2$ where the intensity becomes half,i.e.,$I_2 = I_0 / 2$.
Using the formula: $I_0 / 2 = I_0 e^{-\mu x_2} \implies 2 = e^{\mu x_2} \implies \ln(2) = \mu x_2$.
Substituting the value of $\mu$: $\ln(2) = \left( \frac{\ln(2)}{12} \right) x_2$.
Solving for $x_2$: $x_2 = 12 \, mm$.

(C) In beta minus decay $(\beta^{-})$, a neutron is transformed into a proton, and an electron is emitted from the nucleus along with an antineutrino.
The process is represented by the equation: $n \rightarrow p + e^{-} + \bar{\nu}$
where $n$ is the neutron, $p$ is the proton, $e^{-}$ is the emitted electron ($\beta$-particle), and $\bar{\nu}$ is the antineutrino.

(B) The given decay sequence is: ${}_Z^AX \to {}_{Z + 1}^AY \to {}_{Z - 1}^{A - 4}K \to {}_{Z - 1}^{A - 4}K$.
$1$. In the first step,${}_Z^AX \to {}_{Z + 1}^AY$: The atomic number increases by $1$ while the mass number remains the same. This corresponds to the emission of a $\beta^-$ particle.
$2$. In the second step,${}_{Z + 1}^AY \to {}_{Z - 1}^{A - 4}K$: The atomic number decreases by $2$ and the mass number decreases by $4$. This corresponds to the emission of an $\alpha$ particle.
$3$. In the third step,${}_{Z - 1}^{A - 4}K \to {}_{Z - 1}^{A - 4}K$: There is no change in the atomic number or mass number,which indicates the emission of a $\gamma$ ray (de-excitation of the nucleus).
Therefore,the sequence of particles emitted is $\beta, \alpha, \gamma$.

(C) Let the parent nucleus be represented as ${}_Z^AX$.
When one $\alpha$ particle $({}_2^4He)$ is emitted,the atomic number $Z$ decreases by $2$ and the mass number $A$ decreases by $4$.
When one $\beta^-$ particle $({}_{-1}^0e)$ is emitted,the atomic number $Z$ increases by $1$ and the mass number $A$ remains unchanged.
Given that the number of $\beta$ particles emitted is twice the number of $\alpha$ particles,let the number of $\alpha$ particles be $n$. Then the number of $\beta$ particles is $2n$.
The change in atomic number $Z'$ of the daughter nucleus is given by: $Z' = Z - 2(n) + 1(2n) = Z - 2n + 2n = Z$.
The change in mass number $A'$ of the daughter nucleus is given by: $A' = A - 4(n) + 0(2n) = A - 4n$.
Since the atomic number $Z$ remains the same,the resulting daughter nucleus is an isotope of the parent nucleus.

(D) When an alpha particle $({}_2^4He)$ is emitted,the mass number decreases by $4$ and the atomic number decreases by $2$.
When a beta particle $(\beta^-)$ is emitted,the atomic number increases by $1$ and the mass number remains unchanged.
Starting with $_n^mX$:
$1$. After emitting one $\alpha$ particle: $_n^mX \rightarrow {}_{n-2}^{m-4}Y + {}_2^4He$.
$2$. After emitting two $\beta$ particles: ${}_{n-2}^{m-4}Y \rightarrow {}_{n-2+2}^{m-4}Z + 2_{-1}^0e = {}_n^{m-4}Z$.
Thus,the resulting nucleus is $_n^{m-4}Z$.

(A) The initial nucleus is $_Z{X^A}$.
When it emits $9 \alpha$-particles,the atomic number $Z$ decreases by $9 \times 2 = 18$ and the mass number $A$ decreases by $9 \times 4 = 36$. The nucleus becomes $_{Z-18}{X^{A-36}}$.
When this nucleus emits $5 \beta$-particles,the atomic number $Z$ increases by $5 \times 1 = 5$,while the mass number $A$ remains unchanged. The final nucleus is $_{Z-18+5}{X^{A-36}} = _{Z-13}{X^{A-36}}$.
In the final nucleus,the number of protons $P = Z - 13$.
The number of neutrons $N = (A - 36) - (Z - 13) = A - Z - 23$.
Therefore,the ratio of protons to neutrons is $\frac{P}{N} = \frac{Z - 13}{A - Z - 23}$.

(A) The radioactive decay chain starting from $U^{238}$ is known as the uranium series or the $(4n+2)$ series.
In this series,the mass number of the parent nucleus is $238$,which can be expressed as $4n+2$ where $n=59$.
Through a sequence of $8$ $\alpha$-decays and $6$ $\beta$-decays,the nucleus reaches a stable state.
The final stable end product of this decay chain is $Pb^{206}$.

(B) Let the number of $\alpha$-particles emitted be $x$ and the number of $\beta$-particles emitted be $y$.
The change in mass number is due only to $\alpha$-particles,as each $\alpha$-particle reduces the mass number by $4$.
$4x = 238 - 206 = 32$
$x = 8$
The change in atomic number (charge) is given by $2x - y = 92 - 82$,where $2x$ is the decrease due to $\alpha$-decay and $y$ is the increase due to $\beta$-decay.
$2(8) - y = 10$
$16 - y = 10$
$y = 6$
Thus,the number of $\alpha$-particles is $8$ and the number of $\beta$-particles is $6$.

(D) In radioactive decay involving the emission of an electron (beta decay,$\beta^-$),a neutron in the nucleus is converted into a proton and an electron (and an antineutrino).
The process is represented as: $n \rightarrow p + e^- + \bar{\nu}_e$.
The mass number $A$ is the total number of protons and neutrons.
Since one neutron is lost and one proton is gained,the total number of nucleons remains constant.
The emission of $\gamma$ radiation involves the release of energy from an excited nucleus and does not change the mass number or atomic number.
Therefore,the mass number of the daughter nucleus $(m_y)$ is equal to the mass number of the parent nucleus $(m_x)$.
Thus,$m_y = m_x$.

(C) Let the number of $\alpha$ particles emitted be $x$ and the number of $\beta^-$ particles emitted be $y$.
The decay reaction is: ${}_{88}^{226}Ra \rightarrow {}_{82}^{206}Pb + x({}_{2}^{4}\alpha) + y({}_{-1}^{0}\beta^-)$.
Balancing the mass number:
$226 = 206 + 4x + 0$
$4x = 226 - 206$
$4x = 20$
$x = 5$.
Balancing the atomic number:
$88 = 82 + 2x - y$
$88 = 82 + 2(5) - y$
$88 = 82 + 10 - y$
$88 = 92 - y$
$y = 92 - 88$
$y = 4$.
Therefore,$5$ $\alpha$ particles and $4$ $\beta^-$ particles are emitted.

(A) radioactive series consists of a sequence of radioactive decays where a parent nucleus decays into a daughter nucleus,which then decays further until a stable nucleus is reached.
By definition,the end product of a radioactive series is a stable nucleus.
$A$ stable nucleus does not undergo radioactive decay.
The decay constant $\lambda$ is defined as the probability of decay per unit time.
Since a stable nucleus does not decay,its decay constant is $0$.

(D) In $\alpha$ decay,the nucleus emits an $\alpha$ particle $(^4_2He)$,so the atomic number $Z$ decreases by $2$ and the mass number $A$ decreases by $4$.
In $\beta^+$ decay (positron emission),a proton converts into a neutron,so the atomic number $Z$ decreases by $1$ while the mass number $A$ remains unchanged.
In $\gamma$ decay,the nucleus transitions from an excited state to a lower energy state by emitting a photon,so neither $Z$ nor $A$ changes.
Since all the given statements are correct,the correct option is $D$.

(D) In $\beta$-decay,the energy released is shared between the $\beta$-particle and the antineutrino (or neutrino). Because the energy is shared in a continuous manner,the $\beta$-particles are emitted with a continuous range of kinetic energies,from zero up to a maximum value $E_0$. The distribution curve $N(E)$ versus $E$ starts from zero at $E=0$,reaches a peak at an intermediate energy,and drops to zero at the maximum energy $E_0$. This corresponds to the graph shown in option $D$.

(C) The '$rad$' (radiation absorbed dose) is a unit of absorbed radiation dose.
It is defined as the absorption of $100 \ erg$ of energy per gram of matter (usually tissue).
Therefore,it measures the energy delivered by radiation to a target material.
One '$rad$' is equivalent to $0.01 \ J/kg$ or $0.01 \ Gray$ $(Gy)$.

(A) Gamma $(\gamma)$ emission occurs when a nucleus in an excited state transitions to a lower energy state or the ground state. During this process,the nucleus emits a high-energy photon known as a $\gamma$-ray. Since $\gamma$-rays have zero charge and zero rest mass,the emission does not alter the atomic number $(Z)$ or the mass number $(A)$ of the nucleus. Therefore,the number of protons and neutrons remains unchanged.

(B) The initial nucleus is represented as $_Z^AX$.
An $\alpha$-particle is $_2^4He$ and a positron is $_1^0e^+$.
After emitting $3$ $\alpha$-particles and $2$ positrons, the final nucleus $_Z^AY$ is formed.
The new mass number $A' = A - (3 \times 4) = A - 12$.
The new atomic number $Z' = Z - (3 \times 2) + (2 \times 1) = Z - 6 + 2 = Z - 4$.
The number of protons in the final nucleus is $P = Z' = Z - 4$.
The number of neutrons in the final nucleus is $N = A' - Z' = (A - 12) - (Z - 4) = A - Z - 8$.
The ratio of neutrons to protons is $\frac{N}{P} = \frac{A - Z - 8}{Z - 4}$.

(B) According to the law of conservation of momentum,the momentum of the emitted photon must be equal to the momentum of the recoiling nucleus.
Momentum of photon $p = \frac{hv}{c}$.
Let $V$ be the recoil velocity of the nucleus. Then,the momentum of the nucleus is $p = MV$.
Equating the two: $\frac{hv}{c} = MV$,which gives $V = \frac{hv}{Mc}$.
The recoil kinetic energy of the nucleus is $K = \frac{1}{2}MV^2$.
Substituting the value of $V$: $K = \frac{1}{2}M \left( \frac{hv}{Mc} \right)^2 = \frac{1}{2}M \frac{h^2 v^2}{M^2 c^2} = \frac{h^2 v^2}{2Mc^2}$.

(A) In $\beta$-decay,the energy released is shared between the $\beta$-particle (electron or positron) and an antineutrino (or neutrino).
Because the energy is shared between two particles,the $\beta$-particles are emitted with a continuous range of energies from $0$ up to a maximum value $E_0$.
The number of $\beta$-particles $N(E)$ emitted with a particular energy $E$ follows a characteristic distribution curve that starts at $N(E)=0$ for $E=0$,rises to a peak,and then drops to zero at the maximum energy $E_0$.
This continuous spectrum is a fundamental feature of $\beta$-decay,as shown in option $A$.

(A) Let $N_1$ and $N_2$ be the number of nuclei of $A_1$ and $A_2$ at time $t$.
Given $\frac{dN_1}{dt} = -\lambda_1 N_1$,which integrates to $N_1(t) = N_{10} e^{-\lambda_1 t}$.
The rate of change of $N_2$ is $\frac{dN_2}{dt} = \lambda_1 N_1 - \lambda_2 N_2$.
Substituting $N_1$,we get $\frac{dN_2}{dt} + \lambda_2 N_2 = \lambda_1 N_{10} e^{-\lambda_1 t}$.
Solving this linear differential equation with initial condition $N_2(0) = 0$,we get $N_2(t) = \frac{\lambda_1 N_{10}}{\lambda_2 - \lambda_1} (e^{-\lambda_1 t} - e^{-\lambda_2 t})$.
The activity of $A_2$ is $R_2 = \lambda_2 N_2$. For $R_2$ to be maximum,$\frac{dR_2}{dt} = 0$,which implies $\frac{dN_2}{dt} = 0$.
Setting $\lambda_1 N_1 = \lambda_2 N_2$ gives $\lambda_1 N_{10} e^{-\lambda_1 t} = \lambda_2 \left[ \frac{\lambda_1 N_{10}}{\lambda_2 - \lambda_1} (e^{-\lambda_1 t} - e^{-\lambda_2 t}) \right]$.
Simplifying,we get $e^{-\lambda_1 t} = \frac{\lambda_2}{\lambda_2 - \lambda_1} (e^{-\lambda_1 t} - e^{-\lambda_2 t})$.
Rearranging leads to $e^{(\lambda_2 - \lambda_1)t} = \frac{\lambda_2}{\lambda_1}$.
Taking the natural logarithm on both sides,$t = \frac{\ln(\lambda_2/\lambda_1)}{\lambda_2 - \lambda_1}$.

(C) The $Q$-value of the $\beta$ decay reaction is given by the mass difference between the parent nucleus and the products.
Since the atomic mass of ${}_1^3H$ includes one electron and the atomic mass of ${}_2^3He$ includes two electrons,we must account for the electron emitted in the decay.
The $Q$-value is calculated as:
$Q = [M({}_1^3H) - M({}_2^3He)] \times c^2$
Note: The mass of the emitted electron is already accounted for in the atomic mass difference of the neutral atoms.
$Q = (3.016050 \, u - 3.016030 \, u) \times 931.5 \, MeV/u$
$Q = 0.000020 \, u \times 931.5 \, MeV/u$
$Q = 0.01863 \, MeV$
The maximum kinetic energy of the electron is approximately equal to the $Q$-value of the reaction.
Therefore,the maximum energy is $0.0186 \, MeV$.

(C) The rate of decay for each process is given by $\frac{dN_1}{dt} = -\lambda_1 N$,$\frac{dN_2}{dt} = -\lambda_2 N$,and $\frac{dN_3}{dt} = -\lambda_3 N$.
The total rate of decay is the sum of the individual rates: $\frac{dN}{dt} = \frac{dN_1}{dt} + \frac{dN_2}{dt} + \frac{dN_3}{dt}$.
Substituting the expressions,we get: $\frac{dN}{dt} = -\lambda_1 N - \lambda_2 N - \lambda_3 N = -(\lambda_1 + \lambda_2 + \lambda_3) N$.
Comparing this with the general decay equation $\frac{dN}{dt} = -\lambda_{eff} N$,we find that the effective decay constant is $\lambda_{eff} = \lambda_1 + \lambda_2 + \lambda_3$.

(C) The initial nucleus is ${}_{84}X^{202}$.
$1$. After $\alpha$-decay: The atomic number decreases by $2$ and the mass number decreases by $4$.
${}_{84}X^{202} \rightarrow {}_{82}Y^{198} + {}_{2}He^{4}$.
$2$. After $\beta$-decay: The atomic number increases by $1$ and the mass number remains unchanged.
${}_{82}Y^{198} \rightarrow {}_{83}Z^{198} + {}_{-1}e^{0}$.
$3$. After $\gamma$-emission: The atomic number and mass number remain unchanged.
${}_{83}Z^{198} \rightarrow {}_{83}Z^{198} + \gamma$.
The final nucleus formed is ${}_{83}Z^{198}$,which has an atomic number of $83$.

(A) The initial nucleus $A$ has mass number $M = 180$ and atomic number $Z = 72$.
$1$. Emission of an $\alpha$-particle $(_{2}^{4}He)$ reduces the mass number by $4$ and the atomic number by $2$.
$2$. Emission of a $\beta$-particle $(_{-1}^{0}e)$ does not change the mass number but increases the atomic number by $1$.
$3$. Emission of a $\gamma$-ray does not change the mass number or the atomic number.
Sequence of decays:
$A(180, 72) \xrightarrow{\alpha} A_1(176, 70) \xrightarrow{\beta} A_2(176, 71) \xrightarrow{\alpha} A_3(172, 69) \xrightarrow{\gamma} A_4(172, 69)$.
Therefore,the mass number and atomic number for $A_4$ are $172$ and $69$ respectively.

(C) The given nuclear reaction is: ${}_3^7Li + {}_1^1p \to {}_4^8Be + x$.
According to the law of conservation of mass number:
$7 + 1 = 8 + A_x \Rightarrow A_x = 0$.
According to the law of conservation of atomic number (charge):
$3 + 1 = 4 + Z_x \Rightarrow Z_x = 0$.
$A$ particle with mass number $0$ and atomic number $0$ is a gamma photon $(\gamma)$.
Therefore,the correct option is $C$.

(D) Let the number of $\alpha$-particles emitted be $n_{\alpha}$ and the number of $\beta^-$-particles emitted be $n_{\beta}$.
In an $\alpha$-decay,the mass number decreases by $4$ and the atomic number decreases by $2$. In a $\beta^-$-decay,the mass number remains unchanged and the atomic number increases by $1$.
The change in mass number is $238 - 206 = 32$. Since only $\alpha$-decay affects the mass number,$4n_{\alpha} = 32$,which gives $n_{\alpha} = 8$.
The change in atomic number is $92 - 82 = 10$. The net change in atomic number is given by $2n_{\alpha} - n_{\beta} = 10$.
Substituting $n_{\alpha} = 8$,we get $2(8) - n_{\beta} = 10$,which implies $16 - n_{\beta} = 10$,so $n_{\beta} = 6$.
Thus,$8$ $\alpha$-particles and $6$ $\beta^-$-particles are emitted.

(B) In an $\alpha$-decay,the energy released ($Q$-value) is shared between the $\alpha$-particle and the daughter nucleus to conserve momentum.
The kinetic energy of the $\alpha$-particle $(K_{\alpha})$ is given by the formula: $K_{\alpha} = \left( \frac{A - 4}{A} \right) Q$,where $A$ is the mass number of the mother nucleus.
Given: $K_{\alpha} = 48 \, MeV$ and $Q = 50 \, MeV$.
Substituting the values: $48 = \left( \frac{A - 4}{A} \right) \times 50$.
Divide both sides by $50$: $\frac{48}{50} = \frac{A - 4}{A}$.
Simplify the fraction: $0.96 = 1 - \frac{4}{A}$.
Rearrange to solve for $A$: $\frac{4}{A} = 1 - 0.96 = 0.04$.
$A = \frac{4}{0.04} = 100$.
Thus,the mass number of the mother nucleus is $100$.

(B) Let the number of $\alpha$-particles emitted be $n$ and the number of $\beta$-particles emitted be $m$.
The change in mass number is given by $232 - 208 = 24$.
Since each $\alpha$-particle reduces the mass number by $4$,the number of $\alpha$-particles emitted is $n = \frac{24}{4} = 6$.
The change in atomic number is given by $90 - 82 = 8$.
Each $\alpha$-particle reduces the atomic number by $2$,and each $\beta$-particle increases it by $1$.
Therefore,$90 - 2n + m = 82$.
Substituting $n = 6$ into the equation: $90 - 2(6) + m = 82$.
$90 - 12 + m = 82 \Rightarrow 78 + m = 82$.
$m = 82 - 78 = 4$.
Thus,the nucleus emitted $6 \,\alpha$ and $4 \,\beta$ particles.

(C) The penetrating power of radiation depends on its energy and interaction with matter.
$\alpha$-particles are heavy and charged,leading to high ionization but low penetration.
$\beta$-particles have higher penetrating power than $\alpha$-particles.
$\gamma$-rays are high-energy electromagnetic waves with no charge and very small mass,allowing them to pass through most materials.
The penetrating power of $\gamma$-rays is approximately $1000$ times that of $\beta$-particles,and $\beta$-particles have about $100$ times the penetrating power of $\alpha$-particles.
Therefore,$\gamma$-rays have the highest penetrating power among the given options.

(C) The initial nucleus is ${}_{90}^{232}Th$.
Each $\alpha$-particle emission reduces the mass number $A$ by $4$ and the atomic number $Z$ by $2$.
Each $\beta$-particle emission does not change the mass number $A$ and increases the atomic number $Z$ by $1$.
After $6$ $\alpha$-decays:
$A' = 232 - (6 \times 4) = 232 - 24 = 208$
$Z' = 90 - (6 \times 2) = 90 - 12 = 78$
After $4$ $\beta$-decays:
$A = A' = 208$
$Z = Z' + (4 \times 1) = 78 + 4 = 82$
Thus,the final nucleus is ${}_{82}^{208}X$,where $A = 208$ and $Z = 82$.