Radioactivity and a,b and g rays Questions in English

(C) The initial radionuclide is $_{90}^{234}Th$.
$1$. First $\beta$-decay: $_{90}^{234}Th \rightarrow {}_{91}^{234}X + {}_{-1}^{0}e$.
$2$. Second $\beta$-decay: $_{91}^{234}X \rightarrow {}_{92}^{234}Y + {}_{-1}^{0}e$.
$3$. One $\alpha$-decay: $_{92}^{234}Y \rightarrow {}_{90}^{230}Z + {}_{2}^{4}He$.
Thus,the resulting radionuclide has an atomic number of $90$ and a mass number of $230$.

(C) The given reaction is $_5B^8 \to _4Be^8 + _1e^0$.
In this reaction,the atomic number decreases by $1$ while the mass number remains constant.
This process corresponds to the emission of a positron $(+1e^0)$,which is also known as $\beta^+$-decay.

(C) In a nuclear reaction,the sum of mass numbers and atomic numbers must be conserved on both sides.
For the reaction $_{93}Np^{239} \to _{94}Pu^{239} + _{z}X^{a}$:
Mass number: $239 = 239 + a \implies a = 0$.
Atomic number: $93 = 94 + z \implies z = -1$.
The particle with mass number $0$ and atomic number $-1$ is an electron (beta particle),denoted as $_{ -1}e^{0}$.

(B) The isotope $_{92}U^{235}$ is more radioactive than $_{92}U^{238}$ because it has a shorter half-life,indicating higher instability.
Therefore,the correct option is $(B)$.

(D) In the given nuclear reaction,the mass number remains constant at $64$ on both sides.
For the atomic number,the reactant has $29$ and the product $Ni$ has $28$.
To balance the atomic number,$x$ must have an atomic number of $29 - 28 = +1$.
Thus,$x$ is a positron,represented as $_{1}e^{0}$ or $\beta^+$.
The reaction is: $_{29}Cu^{64} \to _{28}Ni^{64} + _{+1}e^{0}$.

(D) $\gamma$-rays are high-energy electromagnetic radiation,which are neutral and do not possess mass. They are considered packets of energy,also known as photons.

(A) The order of penetrating power of radioactive rays is $\alpha < \beta < \gamma$.
$\alpha$-particles are heavy and have low speed,so they have the least penetrating power.
$\gamma$-rays are electromagnetic radiations with high energy and speed,so they have the highest penetrating power.
Therefore,the statement that $\alpha$-rays have more penetrating power than $\beta$-rays is incorrect.

(C) $(c)$ $\gamma$-rays are electromagnetic radiations with no charge and no mass.
Due to the absence of charge,they are not deflected by electrical or magnetic fields.
They possess the highest penetrating power among $\alpha$,$\beta$,and $\gamma$ radiations.

(B) . Isotopes are not always radioactive; many are stable.
$B$. $\beta$-rays consist of high-energy electrons,which are negatively charged particles. This statement is correct.
$C$. $\alpha$-rays consist of helium nuclei $(He^{2+})$,which are positively charged.
$D$. $\gamma$-rays are electromagnetic radiation with no charge,so they are not deflected by magnetic or electric fields.

(D) When a radioactive substance is inside the human body,the radiation emitted is absorbed by the surrounding tissues.
$\alpha$-particles have the highest ionizing power among the given options.
Because they are heavy and carry a $+2$ charge,they cause intense localized damage to biological cells and tissues when emitted internally.
Therefore,$\alpha$-emitters are the most harmful when ingested or inhaled.

(A) Henry Becquerel discovered radioactivity in $1896$ by observing the emission of penetrating rays from potassium uranyl sulphate. Later,Madam Curie coined the term radioactivity.

(B) Radioactivity is the process by which an unstable atomic nucleus loses energy by radiation. Among the given options,$Polonium$ ($Po$,atomic number $84$) is a well-known radioactive element discovered by Marie Curie. Sulphur,Tellurium,and Selenium are non-radioactive elements belonging to Group $16$ of the periodic table.

(C) The penetrating power of radioactive radiations follows the order: $\alpha$-rays $< \beta$-rays $< \gamma$-rays.
Since $\alpha$-particles are heavy and carry a $+2$ charge,they have the lowest penetrating power compared to $\beta$-particles and $\gamma$-rays.
Therefore,the penetrating power of an $\alpha$-particle is less than that of $\beta$-rays.
Hence,the correct option is $(C)$.

(C) In $\beta$-decay,a neutron in the nucleus is converted into a proton and an electron (the $\beta$-particle) along with an antineutrino. The process is represented as: $^1_0n \rightarrow ^1_1p + ^0_{-1}e + \bar{\nu}_e$. Thus,the emission of a $\beta$-particle is caused by the conversion of a neutron to a proton.

(A) The penetrating power of radiations follows the order: $\alpha < \beta < X\text{-rays} < \gamma$.
Since $\alpha$-rays have the lowest penetrating power,they are the most easily stopped by air.

(B) . $\alpha$-rays have the least penetrating power because they are heavy,positively charged particles with a high ionization potential,which causes them to lose energy rapidly when passing through matter.

(D) Radioactive decay involves the emission of particles like $\alpha$-particles (helium nuclei),$\beta$-particles (electrons or positrons),and $\gamma$-rays (electromagnetic radiation).
Protons are not typically emitted during natural radioactive decay processes.
Therefore,the correct option is $(D)$.

(D) The radiations emitted from a radioactive element consist of $\alpha$,$\beta$,and $\gamma$ rays.
$\gamma$-rays are neutral and do not deflect in a magnetic field.
Both $\alpha$-particles (positively charged) and $\beta$-particles (negatively charged) are deflected by a magnetic field,but they deflect in opposite directions.
If a radiation is observed to deflect in a specific direction,it could be either an $\alpha$-ray or a $\beta$-ray,depending on the charge of the particle and the orientation of the magnetic field.
Therefore,the correct option is $D$.

(C) The isotope $_{88}Ra^{226}$ is radioactive.
For this nucleus,the number of protons $(p)$ is $88$ and the number of neutrons $(n)$ is $226 - 88 = 138$.
The $\frac{n}{p}$ ratio is $\frac{138}{88} \approx 1.568$.
Since the $\frac{n}{p}$ ratio is significantly greater than $1.5$,the nucleus is unstable and undergoes radioactive decay.

(C) During $\beta^-$-decay,a neutron $(n)$ inside the nucleus transforms into a proton $(p)$,an electron $(e^-)$,and an antineutrino $(\bar{\nu}_e)$.
The process is represented as: $n \rightarrow p + e^- + \bar{\nu}_e$.
Thus,the electron is created at the moment of decay and then ejected from the nucleus.

(B) An $\alpha$-particle is a helium nucleus,represented as $_2^4He$.
When a radioactive nucleus emits an $\alpha$-particle,its mass number $(A)$ decreases by $4$ and its atomic number $(Z)$ decreases by $2$.
The nuclear reaction is represented as: $_Z^AX \rightarrow _{Z-2}^{A-4}Y + _2^4He$.

(D) In $Alpha$ emission,the atomic number decreases by $2$ units.
In $Electron$ capture,a proton is converted into a neutron,decreasing the atomic number by $1$ unit.
In $Positron$ emission,a proton is converted into a neutron,decreasing the atomic number by $1$ unit.
Therefore,all these processes result in a decrease in atomic number.

(C) Let the reaction be $_{84}X^{218} \to _{84}Y^{214} + x_{2}\alpha^{4} + y_{-1}\beta^{0}$.
Number of $\alpha$-particles emitted is given by the change in mass number: $x = \frac{218 - 214}{4} = \frac{4}{4} = 1$.
For atomic number conservation: $84 = 84 + 2x - y$.
Substituting $x = 1$: $84 = 84 + 2(1) - y$,which gives $y = 2$.
Therefore,the number of $\alpha$-particles is $1$ and $\beta$-particles is $2$.

(A) When an $\alpha$-particle $(_{2}He^{4})$ is emitted by a nucleus,the mass number decreases by $4$ units and the atomic number decreases by $2$ units.
The nuclear reaction is represented as:
$_{88}Ra^{224} \rightarrow _{86}X^{220} + _{2}He^{4}$
Thus,the new element has a mass number of $220$ and an atomic number of $86$.

(A) For the nuclear reaction $_{90}^{228}Th \to _{83}^{212}Bi$:
$1$. The number of $\alpha$-particles emitted is calculated by the change in mass number:
$\text{Number of } \alpha = \frac{228 - 212}{4} = \frac{16}{4} = 4$.
$2$. The number of $\beta$-particles emitted is calculated by the change in atomic number,considering the effect of $\alpha$-particles:
$\text{Number of } \beta = (\text{Initial Atomic Number} - \text{Final Atomic Number}) - 2 \times (\text{Number of } \alpha)$
$= (90 - 83) - 2 \times 4 = 7 - 8 = -1$.
Wait,let us re-evaluate the atomic number change: $90 - 2 \times 4 + \beta = 83 \implies 90 - 8 + \beta = 83 \implies 82 + \beta = 83 \implies \beta = 1$.
Therefore,$4$ $\alpha$-particles and $1$ $\beta$-particle are emitted.

(C) The emission of an $\alpha$-particle $(_{2}He^{4})$ from the parent nucleus $_{92}X^{238}$ results in a daughter nucleus with a mass number reduced by $4$ and an atomic number reduced by $2$.
$_{92}X^{238} \rightarrow _{90}Y^{234} + _{2}He^{4}$
The number of neutrons in the daughter nucleus $_{90}Y^{234}$ is calculated as:
$\text{Number of neutrons} = \text{Mass number} - \text{Atomic number}$
$\text{Number of neutrons} = 234 - 90 = 144$.

(D) When a radioactive element emits an electron (beta particle,$_{-1}e^0$),the atomic number $(Z)$ of the daughter element increases by $1$ unit,while the mass number $(A)$ remains unchanged.
The nuclear reaction is represented as: $_Z^A X \to _{Z+1}^A Y + _{-1}e^0$.
Thus,the daughter element has an atomic number one unit more than the parent element.

(A) The emission of $\beta$-particles occurs due to the conversion of a neutron into a proton and an electron within the nucleus. The reaction is: $_0n^1 \to _{+1}p^1 + _{-1}e^0$. Here,the electron $(_{-1}e^0)$ is the $\beta$-particle emitted from the nucleus.

(D) The given reaction $_{11}Na^{24} \to _{12}Mg^{24} + _{-1}e^0$ represents $\beta$-decay.
In $\beta$-decay,a neutron is converted into a proton,which increases the atomic number by $1$ while the mass number remains constant.
This process leads to the emission of $\beta$-radiation,helps in achieving a more stable nuclide,and results in a decrease in the neutron $:$ proton ratio.

(D) During $\beta$-decay,a neutron is converted into a proton and an electron ($\beta$-particle).
Since the total number of nucleons (protons + neutrons) remains constant,the mass number of the atomic nucleus remains unaffected.
Therefore,the correct option is $(D)$.

(B) The initial atom is $_{90}X^{232}$.
Step $1$: Emission of two $\beta$-particles.
Each $\beta$-particle emission increases the atomic number by $1$ but does not change the mass number.
After $2\beta$-emission: $_{90}X^{232} \to _{92}Y^{232} + 2_{-1}e^{0}$.
Step $2$: Emission of $x$ $\alpha$-particles to reach mass number $212$ and atomic number $82$.
Each $\alpha$-particle $(_{2}He^{4})$ decreases the mass number by $4$ and the atomic number by $2$.
Change in mass number: $232 - 212 = 20$.
Number of $\alpha$-particles $(x)$ $= \frac{20}{4} = 5$.
Verification of atomic number: $92 - (5 \times 2) = 92 - 10 = 82$. This matches the target atomic number.
Therefore,the number of $\alpha$-particles emitted is $5$.

(D) The initial nucleus is $_{92}X^{238}$.
After the emission of one $\alpha$-particle $(_{2}He^{4})$,the nucleus becomes $_{90}Y^{234}$.
After the subsequent emission of one $\beta$-particle $(_{-1}e^{0})$,the nucleus becomes $_{91}Z^{234}$.
The number of neutrons is calculated as $A - Z = 234 - 91 = 143$.

(A) Let the element be $X$. The initial state is $_{84}X^{218}$.
When an element emits one $\alpha$-particle $(_{2}He^{4})$,the atomic number decreases by $2$ and the mass number decreases by $4$.
After $1$ $\alpha$-emission: $_{84-2}X^{218-4} = _{82}X^{214}$.
When an element emits one $\beta$-particle $(_{-1}e^{0})$,the atomic number increases by $1$ and the mass number remains unchanged.
After $2$ $\beta$-emissions: $_{82+(2 \times 1)}X^{214} = _{84}X^{214}$.
Thus,the final element has atomic number $84$ and mass number $214$.

(B) The emission of an alpha particle $(_{2}He^{4})$ results in a decrease of $2$ in the atomic number and $4$ in the mass number.
For the reaction $_{84}Po^{215} \to _{82}Pb^{211} + x(_{2}He^{4})$:
Change in atomic number = $84 - 82 = 2$.
Change in mass number = $215 - 211 = 4$.
Since one alpha particle causes a change of $2$ in atomic number and $4$ in mass number,the number of alpha particles emitted is $1$.

(A) An $\alpha$-particle is a helium nucleus,represented as $_{2}He^{4}$.
When a radioactive nucleus emits an $\alpha$-particle,its mass number decreases by $4$ and its atomic number decreases by $2$.
The nuclear reaction is: $_{92}U^{238} \to _{90}Th^{234} + _{2}He^{4}$.
Thus,the product has a mass number of $234$ and an atomic number of $90$.

(C) During $\beta$-emission,a neutron is converted into a proton,which increases the atomic number by $1$ while the mass number remains unchanged.
The nuclear reaction is: $_{20}Ca^{42} \to _{21}Sc^{42} + _{-1}e^{0}$.
Therefore,the remaining nucleus is $_{21}Sc^{42}$.

(B) When a radioactive nucleus emits an $\alpha$-particle $(_{2}He^{4})$,the mass number decreases by $4$ units and the atomic number decreases by $2$ units.
The nuclear reaction is represented as: $_{Z}X^{A} \rightarrow _{Z-2}Y^{A-4} + _{2}He^{4}$.
Thus,both the mass number and atomic number decrease.

(C) During $\beta$-emission,a neutron in the nucleus is converted into a proton and an electron ($\beta$-particle).
The emitted electron is ejected from the nucleus,while the proton remains.
This increases the atomic number $(Z)$ by $1$ while the mass number $(A)$ remains unchanged.
The reaction is represented as: $_{40}X \to _{41}Y + _{-1}e^0$.
Therefore,the new atomic number is $40 + 1 = 41$.

(B) The emission of an $\alpha$-particle $(_{2}He^{4})$ from a nucleus results in a decrease of $2$ in the atomic number $(Z)$ and $4$ in the mass number $(A)$.
The nuclear reaction is: $_{92}U^{236} \to _{90}X^{232} + _{2}He^{4}$.
The number of protons $(Z)$ in the remaining nucleus is $92 - 2 = 90$.
The number of neutrons $(N)$ is calculated as $A - Z = 232 - 90 = 142$.
Therefore,the remaining nucleus has $142$ neutrons and $90$ protons.

(D) The correct option is $(D)$.
Uranium minerals contain trace amounts of $Ra$ (Radium) as a decay product.
Due to the high specific activity of $Ra$ compared to $U$,the appreciable radioactivity observed in these minerals is mainly attributed to the presence of $Ra$.

(A) The nuclear reaction is represented as: $_{92}U^{238} \to _{82}Pb^{206} + x(_{2}\alpha^{4}) + y(_{-1}\beta^{0})$.
Number of $\alpha$ particles $(x)$ = $\frac{238 - 206}{4} = \frac{32}{4} = 8$.
Number of $\beta$ particles $(y)$ = $92 - 82 - 2x = 10 - 2(8) = 10 - 16 = -6$. Wait,let us re-calculate: $92 = 82 + 2(8) - y \implies 92 = 82 + 16 - y \implies 92 = 98 - y \implies y = 6$.
Total number of particles lost = $x + y = 8 + 6 = 14$.

(B) In $Beta$ emission,a neutron is converted into a proton and an electron (beta particle). The proton remains in the nucleus,increasing the atomic number $(Z)$ by $1$,while the mass number $(A)$ remains unchanged.
The process is represented as: $^A_Z X \to ^A_{Z+1} Y + ^0_{-1} e$.

(D) When a $\beta $-particle $(^0_{-1}e)$ is emitted from a nucleus,a neutron is converted into a proton.
This process increases the atomic number $(Z)$ by $1$ while the mass number $(A)$ remains unchanged.
The nuclear reaction is represented as: $^A_Z X \rightarrow ^A_{Z+1} Y + ^0_{-1}e$.
Therefore,the atomic number increases by one unit.

(C) Let the number of $\alpha$-particles emitted be $x$ and the number of $\beta$-particles emitted be $y$.
The nuclear reaction is: $_{92}U^{238} \to _{82}Pb^{206} + x(_{2}He^{4}) + y(_{-1}e^{0})$.
Equating the mass number on both sides:
$238 = 206 + 4x + 0y$
$4x = 32 \implies x = 8$.
Equating the atomic number on both sides:
$92 = 82 + 2x - y$
$92 = 82 + 2(8) - y$
$92 = 82 + 16 - y$
$92 = 98 - y$
$y = 98 - 92 = 6$.
Therefore,the number of $\beta$-particles emitted is $6$.

(D) The nuclear transformation is $_{90}^{232}Th \xrightarrow{} _{82}^{208}Pb + x(\alpha) + y(\beta)$.
Step $1$: Calculate the number of $\alpha-$ particles $(x)$: The change in mass number is $232 - 208 = 24$. Since each $\alpha-$ particle has a mass of $4$,$x = \frac{24}{4} = 6$.
Step $2$: Calculate the number of $\beta-$ particles $(y)$: The change in atomic number is $90 - 82 = 8$. Each $\alpha-$ particle decreases the atomic number by $2$,so $6 \alpha-$ particles decrease it by $12$. Each $\beta-$ particle increases the atomic number by $1$. Thus,$90 - (6 \times 2) + y = 82$,which gives $90 - 12 + y = 82$,so $78 + y = 82$,resulting in $y = 4$.
Therefore,the number of $\alpha-$ and $\beta-$ particles emitted are $6$ and $4$ respectively.

(A) The radioactive decay process is given by: $_{90}E^{232} \rightarrow _{86}G^{220} + x(\alpha) + y(\beta)$.
Number of $\alpha$ particles $(x)$ is calculated by the change in mass number: $x = \frac{232 - 220}{4} = \frac{12}{4} = 3$.
Number of $\beta$ particles $(y)$ is calculated using the change in atomic number: $90 = 86 + 2x - y$,which gives $y = 86 + 2(3) - 90 = 86 + 6 - 90 = 2$.
Therefore,$3$ $\alpha$ particles and $2$ $\beta$ particles are emitted.

(A) The nuclear reaction is $_{92}U^{238} \to _{90}Th^{234} \to _{91}Pa^{234}$.
Step $1$: For the emission of $\alpha$-particles,the mass number decreases by $4$ and the atomic number decreases by $2$.
Change in mass number $= 238 - 234 = 4$.
Number of $\alpha$-particles $= \frac{4}{4} = 1$.
Step $2$: For the emission of $\beta$-particles,the atomic number increases by $1$ while the mass number remains constant.
After the emission of $1$ $\alpha$-particle,the atomic number becomes $90$.
To reach $91$ $(Pa)$,the atomic number must increase by $1$,which corresponds to the emission of $1$ $\beta$-particle.
Therefore,the number of $\alpha$ and $\beta$-particles emitted are $1$ and $1$ respectively.

(D) radioactive nucleus does not emit $\gamma$-rays alone. $\gamma$-ray emission is a secondary process that occurs to release excess energy after the emission of $\alpha$ or $\beta$ particles,which leaves the daughter nucleus in an excited state.

(A) The initial nucleus is $_{72}^{180}X$.
Emission of $2\alpha$ particles: Each $\alpha$ particle $(_{2}^{4}He)$ reduces the atomic number by $2$ and mass number by $4$.
After $2\alpha$ emission: $Z = 72 - (2 \times 2) = 68$ and $A = 180 - (2 \times 4) = 172$. The nucleus becomes $_{68}^{172}P$.
Emission of $1\beta$ particle: $\beta$ emission $(_{-1}^{0}e)$ increases the atomic number by $1$ and keeps the mass number constant.
After $\beta$ emission: $Z = 68 + 1 = 69$ and $A = 172$. The nucleus becomes $_{69}^{172}Q$.
Emission of $\gamma$ radiation: $\gamma$ emission does not change the atomic number or mass number.
Final nucleus: $_{69}^{172}X'$.
Therefore,$Z = 69$ and $A = 172$.

(B) The emission of a beta particle occurs when a neutron in the nucleus decays into a proton and an electron (beta particle).
The nuclear reaction is represented as: $n \to p + e^- + \bar{\nu}$.
As a result,the number of neutrons decreases by $1$ and the number of protons increases by $1$.
Therefore,the loss of a beta particle is equivalent to a decrease of one neutron.