Radioactivity and a,b and g rays Questions in English

(A) The radioactive decay reaction is given by: $_{92}X^{232} \to _{89}Y^{220} + x(_{2}\alpha^{4}) + y(_{-1}\beta^{0})$.
First,calculate the number of $\alpha$-particles $(x)$ by comparing the mass numbers: $232 = 220 + 4x \implies 4x = 12 \implies x = 3$.
Next,calculate the number of $\beta$-particles $(y)$ by comparing the atomic numbers: $92 = 89 + 2x - y \implies 92 = 89 + 2(3) - y \implies 92 = 89 + 6 - y \implies 92 = 95 - y \implies y = 3$.
Therefore,$3$ $\alpha$-particles and $3$ $\beta$-particles are ejected.

(D) In $\alpha$-decay,the atomic number decreases by $2$ and the mass number decreases by $4$.
In option $A$,$B$,and $C$,the atomic number decreases by $2$ and mass number by $4$,which represents $\alpha$-decay.
In option $D$,$_{83}Bi^{213} \rightarrow _{84}Po^{213} + _{-1}e^{0}$,the atomic number increases by $1$ and the mass number remains constant. This is a characteristic of $\beta$-decay.
Therefore,the correct option is $D$.

(C) The speed of particles emitted during radioactive decay depends on their mass and energy.
$\alpha$-particles are helium nuclei $(^4_2He^{2+})$ and are relatively heavy.
$\beta$-particles are high-speed electrons.
$\gamma$-rays are electromagnetic radiation (photons) that travel at the speed of light $(c \approx 3 \times 10^8 \ m/s)$.
Therefore,$\gamma$-rays move the fastest among the given options.

(A) $^{14}_{6}C$ undergoes $\beta$-decay to form $^{14}_{7}N$ and an electron ($\beta$-particle).
The nuclear reaction is: $^{14}_{6}C \to ^{14}_{7}N + ^{0}_{-1}e$.
Therefore,it is $\beta$-active and the product is $^{14}_{7}N$.

(C) In negative $\beta$-decay,a neutron $(n)$ inside the nucleus transforms into a proton $(p)$,an electron $(_{-1}e^0)$,and an antineutrino $(\bar{\nu})$.
The process is represented as: $n \to p + _{-1}e^0 + \bar{\nu}$.
Therefore,the correct statement is that a neutron in the nucleus decays,emitting an electron.

(A) An isobar is a nuclide that has the same mass number $(A)$ but a different atomic number $(Z)$.
In $\beta$-decay,positron emission,and electron capture,the mass number $(A)$ remains constant,resulting in the formation of an isobar.
In $\alpha$-decay,the mass number decreases by $4$ units $(A \rightarrow A-4)$,so the daughter nucleus is not an isobar of the parent nucleus.

(C) In a $\beta$-decay process,the mass number $(m)$ remains constant while the atomic number $(Z)$ increases by $1$ for each $\beta$-particle emitted.
The initial nucleus is $_{6}{X^{14}}$.
After the emission of $3\beta$-particles:
$1$. The mass number remains $14$,so $m = 14$.
$2$. The atomic number changes as $Z = 6 + 3 = 9$.
Therefore,the resulting nucleus is $_{9}{Y^{14}}$.

(B) Tritium $_1^3H \to _2^3He + _{-1}^0e$ is a $\beta$-emitter.
Radioactive decay of tritium involves the emission of a $\beta$-particle (electron) to form helium-$3$.

(A) The method of radioactive carbon dating was developed by $W. F. Libby$ in the late $1940s$.
For his pioneering work in this field,he was awarded the Nobel Prize in Chemistry in $1960$.
Therefore,the correct option is $(A)$.

(B) Carbon dating is a method used to determine the age of organic materials by measuring the amount of the radioactive isotope $_6^{14}C$ present in the sample.
This isotope is continuously produced in the upper atmosphere by cosmic rays and is incorporated into living organisms.
After the organism dies,the uptake of $_6^{14}C$ stops,and it begins to decay at a known rate,allowing scientists to estimate the time elapsed since death.

(B) It is a fundamental principle of nuclear chemistry that the emission of an $\alpha$-particle or a $\beta$-particle leaves the daughter nucleus in an excited state.
To return to the ground state,the nucleus releases the excess energy in the form of $\gamma$-rays.
Therefore,$\gamma$-rays are always emitted following the emission of $\alpha$ or $\beta$ particles.
It is impossible for a radioisotope to emit $\gamma$-rays alone without a preceding $\alpha$ or $\beta$ decay process.

(B) The radioactive isotope $_{6}C^{14}$ is used in radiocarbon dating to determine the age of organic archaeological findings and fossils.

(D) $G.M.$ counter (Geiger-$M$üller counter) is a device used to detect and measure ionizing radiation,specifically the rate of radioactive decay.

(B) The correct answer is $B$.
Radioactive isotope $^{14}C$ (Carbon-$14$) is used for radiocarbon dating to determine the age of ancient organic remains.

(A) An $\alpha$-particle is a helium nucleus,represented as $_{2}He^{4}$.
When a radioactive isotope $_{A}X^{M}$ emits an $\alpha$-particle,the atomic number decreases by $2$ and the mass number decreases by $4$.
The nuclear reaction is: $_{A}X^{M} \rightarrow _{A-2}Y^{M-4} + _{2}He^{4}$.
Therefore,the new atomic number is $A - 2$ and the new mass number is $M - 4$.

(A) In $\beta$-decay,a neutron is converted into a proton,increasing the atomic number by $1$ while the mass number remains constant.
The reaction is: $_{11}Na^{24} \to {}_{12}Mg^{24} + {}_{-1}e^{0}$.
Since the product is $_{12}Mg^{24}$,it is an isotope of $Mg$.

(A) For $_{11}^{23}Na$,the neutron-to-proton ratio is $\frac{n}{p} = \frac{12}{11} \approx 1.09$.
For $_{11}^{24}Na$,the neutron-to-proton ratio is $\frac{n}{p} = \frac{13}{11} \approx 1.18$.
Since $_{11}^{24}Na$ has a higher $\frac{n}{p}$ ratio than the stable isotope,it undergoes radioactive decay to decrease this ratio.
This is achieved by the emission of a $\beta^-$-particle,where a neutron converts into a proton: $n \to p + e^- + \bar{\nu}$ ($\beta^-$ emission).

(C) The activity of $1 \ g$ of Radium is defined as $1 \ Curie (Ci) = 3.70 \times 10^{10} \ disintegration \ s^{-1}$.
Given,the activity of nuclide $X$ is $A_X = 1.00 \times 10^5 \ disintegration \ s^{-1} \ g^{-1}$.
To find the activity in $Ci \ g^{-1}$,we divide the activity of $X$ by the activity of $1 \ g$ of Radium:
$Activity (in \ Ci \ g^{-1}) = \frac{1.00 \times 10^5}{3.70 \times 10^{10}} = 0.270 \times 10^{-5} \ Ci \ g^{-1}$.
Since $1 \ Ci = 1000 \ mCi$,the activity in $mCi \ g^{-1}$ is $0.270 \times 10^{-5} \times 10^3 = 0.00270 \ mCi \ g^{-1}$.
Therefore,the correct option is $C$.

(D) The correct option is $(D)$.
Cobalt-$60$ $(^{60}Co)$ is a radioactive isotope that emits gamma rays,which are used in radiotherapy to treat cancer.

(B) Radiation shielding is used to block harmful nuclear radiation,which consists of $\alpha$,$\beta$,and $\gamma$ rays. $Pb$ (Lead) is a dense metal that effectively absorbs and blocks these harmful radiations,making it the standard material for radiation shielding.

(A) $\gamma$ rays are high-energy electromagnetic radiation emitted from the nucleus of a radioactive atom. They have no mass and no charge. They are represented by the symbol $\gamma$ and their energy is given by $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency. Therefore,option $A$ is correct.

(C) During $\beta$-particle emission,a neutron in the nucleus decays into a proton and an electron (the $\beta$-particle). The electron is emitted,while the proton remains in the nucleus. Therefore,the atomic number increases by $1$,meaning one proton increases.

(C) The unit of radioactivity is the curie $(Ci)$.
$1 \ Ci = 3.7 \times 10^{10} \ \text{disintegrations per second (dps)}$.
One microcurie is defined as $1 \ \mu Ci = 10^{-6} \ Ci$.
Therefore,$1 \ \mu Ci = 10^{-6} \times 3.7 \times 10^{10} \ dps = 3.7 \times 10^4 \ dps$.

(B) The stable isotope of Aluminum is $^{27}_{13}A$,which has $13$ protons and $14$ neutrons. The ratio of neutrons to protons $(n/p)$ is $14/13 \approx 1.077$.
For the isotope $^{29}_{13}A$,the number of neutrons is $16$. The $n/p$ ratio is $16/13 \approx 1.23$.
Since this isotope has an excess of neutrons compared to the stable isotope,it will undergo $\beta$-decay to convert a neutron into a proton,thereby increasing the atomic number and moving towards a more stable $n/p$ ratio.
The reaction is: $^{29}_{13}A \rightarrow ^{29}_{14}Si + ^{0}_{-1}e$.

(D) $Co-60$ is used as an anticancerous agent among the given radioactive isotopes.
It emits $\beta$-particles and high-energy gamma rays,which is why it is widely used in radiation therapy for cancer treatment.

(N/A) Radioactive elements emit three types of rays: $\alpha$,$\beta$,and $\gamma$. Their characteristics are summarized below:

Property	$\alpha$-rays	$\beta$-rays	$\gamma$-rays
Nature	Helium nuclei $(He^{2+})$	High-speed electrons $(e^-)$	Electromagnetic radiation
Charge	$+2$	$-1$	$0$ (Neutral)
Mass	$4 \ amu$	Negligible $(1/1837 \ amu)$	$0$
Penetrating Power	Least	$100$ times more than $\alpha$	$1000$ times more than $\alpha$

(N/A)

Property	$\alpha$-rays	$\beta$-rays	$\gamma$-rays
Nature	Helium nuclei $(He^{2+})$	Fast-moving electrons $(e^-)$	Electromagnetic radiation (photons)
Charge	$+2$ units	$-1$ unit	Neutral $(0)$
Mass	$4 \ amu$	Negligible $(1/1837 \ amu)$	No mass
Penetrating Power	Least	$100$ times more than $\alpha$	$1000$ times more than $\alpha$
Formation	$He^{2+} + 2e^- \rightarrow He_{(g)}$	-	-

(N/A) The study of radioactivity was developed by $Marie \ Curie$,$Pierre \ Curie$,$Ernest \ Rutherford$,and $Frederick \ Soddy$.

(A) Tritium $(^{3}_{1}H)$ is a radioactive isotope of hydrogen. It undergoes radioactive decay by emitting a beta particle $(\beta^{-})$ to form Helium-$3$ $(^{3}_{2}He)$. The nuclear reaction is: $^{3}_{1}H \rightarrow ^{3}_{2}He + ^{0}_{-1}e + \bar{\nu}_{e}$.

(A) Analyze the decay sequence:
$1$. $^{238}_{92}U \xrightarrow{- x_1} ^{234}_{90}Th$: Mass number decreases by $4$ and atomic number by $2$,so $x_1$ is an $\alpha$-particle $(^{4}_{2}He^{2+})$. $\alpha$-particles are positively charged and deflect towards the negatively charged plate. Statement $(3)$ is correct.
$2$. $^{234}_{90}Th \xrightarrow{- x_2} ^{234}_{91}Pa$: Atomic number increases by $1$,so $x_2$ is a $\beta^{-}$-particle $(^{0}_{-1}e)$. Statement $(2)$ is correct.
$3$. $^{234}_{91}Pa \xrightarrow{- x_3} ^{234}_{92}Z$: Atomic number increases by $1$,so $x_3$ is a $\beta^{-}$-particle. Statement $(4)$ is incorrect.
$4$. $^{234}_{92}Z$ has atomic number $92$,same as $U$,so $Z$ is an isotope of uranium. Statement $(1)$ is correct.
Thus,statements $(1)$,$(2)$,and $(3)$ are correct.

(C) Radioactive decay is a first-order nuclear process.
The decay constant $(\lambda)$ is a characteristic property of the radioactive nucleus and is independent of external physical conditions such as temperature,pressure,or chemical environment.
Therefore,the statement that the decay constant increases with an increase in temperature is false.

(C) $\gamma$-radiations are electromagnetic waves with no mass and no charge.
Due to their high energy and neutral nature,they have the highest penetrating power compared to charged particles like $\alpha$-particles,protons,or positrons.

(B) The initial nucleus is ${ }_{83}^{214} Bi$.
Emission of a $\beta$-particle $({ }_{-1}^{0} e)$ increases the atomic number by $1$ while the mass number remains unchanged:
${ }_{83}^{214} Bi \rightarrow { }_{84}^{214} Po + { }_{-1}^{0} e$
Emission of an $\alpha$-particle $({ }_{2}^{4} He)$ decreases the mass number by $4$ and the atomic number by $2$:
${ }_{84}^{214} Po \rightarrow { }_{82}^{210} Pb + { }_{2}^{4} He$
The final nucleus is ${ }_{82}^{210} Pb$.
The number of neutrons $N = A - Z = 210 - 82 = 128$.

(A) In positron emission,the atomic number of the nucleus decreases by $1$ while the mass number remains constant.
The decay reaction for $F^{18}$ is:
${ }_{9}F^{18} \longrightarrow { }_{8}O^{18} + { }_{+1}e^{0}$
Since the $F^{18}$ atom is part of the $C_{6}H_{5}F^{18}$ molecule,the $F^{18}$ nucleus transforms into an $O^{18}$ nucleus.
Therefore,the resulting decay product is $C_{6}H_{5}O^{18}$.

(C) During $\beta$-emission,a neutron is converted into a proton,an electron ($\beta$-particle),and an antineutrino $(\bar{\nu})$.
Often,the daughter nucleus is formed in an excited state,which then releases energy in the form of $\gamma$-rays to reach the ground state.
The nuclear reaction is: $_0n^1 \rightarrow _1H^1 + _{-1}e^0 + \bar{\nu} + \gamma$-ray.

(D) The initial nucleus is ${}_{82}A^{210}$ and the final nucleus is ${}_{82}D^{206}$.
The change in mass number is $210 - 206 = 4$ units,and the change in atomic number is $82 - 82 = 0$.
Emission of one $\alpha$-particle decreases the mass number by $4$ and the atomic number by $2$.
Emission of one $\beta$-particle keeps the mass number constant and increases the atomic number by $1$.
To keep the atomic number constant while decreasing the mass number by $4$,we need one $\alpha$-particle ($-2$ in atomic number) and two $\beta$-particles ($+2$ in atomic number).
The sequence is ${}_{82}A^{210}$ $\xrightarrow{-\beta} {}_{83}B^{210}$ $\xrightarrow{-\beta} {}_{84}C^{210}$ $\xrightarrow{-\alpha} {}_{82}D^{206}$.
Thus,the sequence of emission is $\beta, \beta, \alpha$.

(B) The initial atom is $^X_Y M$.
$1$. Emission of first $\alpha$ particle $(^4_2 He)$: The mass number decreases by $4$ and atomic number decreases by $2$. Product: $^{X-4}_{Y-2} N$.
$2$. Emission of second $\alpha$ particle: The mass number decreases by $4$ and atomic number decreases by $2$. Product: $^{X-8}_{Y-4} O$.
$3$. Emission of one $\beta$ particle $(^0_{-1} e)$: The mass number remains unchanged and atomic number increases by $1$. Product: $^{X-8}_{Y-3} P$.
Number of neutrons = Mass number - Atomic number.
Number of neutrons = $(X-8) - (Y-3) = X - 8 - Y + 3 = X - Y - 5$.

(D) The radioactive decay process follows the conservation of mass number and atomic number.
$1$. For $\beta$-decay: ${ }_{89}^{227} Ac \xrightarrow{-\beta} { }_{90}^{227} Th + { }_{-1}^{0} e$. Thus,$[A] = { }_{90}^{227} Th$.
$2$. For $\alpha$-decay: ${ }_{90}^{227} Th \xrightarrow{-\alpha} { }_{88}^{223} Ra + { }_{2}^{4} He$. Thus,$[B] = { }_{88}^{223} Ra$.
$3$. Further $\alpha$-decay: ${ }_{88}^{223} Ra \xrightarrow{-\alpha} { }_{86}^{219} Rn + { }_{2}^{4} He$. This confirms the sequence.
Therefore,$[A]$ is ${ }_{90}^{227} Th$ and $[B]$ is ${ }_{88}^{223} Ra$.

(B) The emission of one $\alpha$ particle $({}_{2}^{4}He)$ decreases the mass number by $4$ and the atomic number by $2$.
The emission of one $\beta$ particle $({}_{-1}^{0}e)$ does not change the mass number but increases the atomic number by $1$.
For the product ${}_{89}^{227}Ac$ from ${}_{92}^{235}U$:
Change in mass number $= 235 - 227 = 8$,which corresponds to $2 \alpha$ particles $(2 \times 4 = 8)$.
Change in atomic number $= 92 - 89 = 3$.
With $2 \alpha$ particles,the atomic number decreases by $2 \times 2 = 4$. To reach $89$,we need to increase the atomic number by $1$ using one $\beta$ particle $(92 - 4 + 1 = 89)$.
Thus,the decay involves $2 \alpha$ and $1 \beta$ particle,resulting in ${}_{89}^{227}Ac$.