201. Consider the following statements about radioactive emissions.
I. Alpha particles have relatively high ionising power.
II. Gamma rays have high penetrating power.
III. Beta minus particles are helium nuclei.
Which statements are valid?
ⓐ. II and III only
ⓑ. I only
ⓒ. I, II, and III
ⓓ. I and II only
Correct Answer: I and II only
Explanation: Statement I is valid because alpha particles are massive and carry charge \(+2e\), so they ionise matter strongly over a short path. Statement II is valid because gamma rays are neutral high-energy photons and are highly penetrating. Statement III is not valid because beta minus particles are electrons, not helium nuclei. Helium nuclei are alpha particles. Beta radiation has moderate penetration and ionising ability compared with alpha and gamma. The three emissions must be distinguished by identity, charge, and interaction with matter.
202. Assertion: Alpha particles have low penetrating power but high ionising power.
Reason: Alpha particles are relatively massive and carry charge \(+2e\).
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Both Assertion and Reason are true, and Reason explains Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Alpha particles are helium nuclei, so they are much more massive than beta particles and carry charge \(+2e\). Their large charge makes them interact strongly with atoms of the material through which they pass. This produces strong ionisation but also causes them to lose energy quickly. As a result, their range and penetrating power are low. The Reason correctly gives the particle properties responsible for the Assertion. The same properties explain why alpha radiation can be stopped by a thin sheet of paper.
203. Gamma rays can pass through matter more effectively than alpha particles mainly because gamma rays:
ⓐ. Have charge \(+2e\)
ⓑ. Are heavier than helium nuclei
ⓒ. Are slow-moving neutral atoms
ⓓ. Neutral high-energy photons
Correct Answer: Neutral high-energy photons
Explanation: Gamma rays are photons and carry no electric charge. Because they are uncharged, they do not continuously lose energy through strong direct ionisation in the same way as alpha particles. Their high energy also helps them penetrate matter deeply. Alpha particles are massive and charged, so they interact strongly and are stopped over short distances. Gamma rays still interact with matter, but their penetration is much greater. This is why dense shielding such as lead or concrete is used for gamma radiation.
204. The line between nuclear radiation and atomic-shell radiation is clearest in which comparison?
ⓐ. Alpha particles are outer-shell electrons, while beta particles are reflected light
ⓑ. Radioactivity is controlled mainly by chemical colour
ⓒ. Nuclear radiation always has zero energy
ⓓ. Nuclear rather than electronic energy changes
Correct Answer: Nuclear rather than electronic energy changes
Explanation: Nuclear radiation originates from changes in the nucleus. Gamma rays are emitted when an excited nucleus moves to a lower energy state. Ordinary visible light usually comes from changes in electronic energy levels of atoms or molecules. This comparison separates nuclear processes from atomic-shell processes. Alpha and beta emissions also involve nuclear changes, not simple outer-shell light emission. Nuclear radiation can have very high energy, so it is not a zero-energy process.
205. A radioactive nucleus emits a gamma ray after an alpha or beta decay has left it in an excited state. The gamma emission mainly:
ⓐ. Changes every neutron into a proton
ⓑ. Removes two protons and two neutrons
ⓒ. Lower energy; same \(A\) and \(Z\)
ⓓ. Adds an electron to the nucleus permanently
Correct Answer: Lower energy; same \(A\) and \(Z\)
Explanation: Gamma emission is the release of a high-energy photon from an excited nucleus. It reduces the energy of the nucleus but does not remove a nucleon. Therefore, the mass number \(A\) remains unchanged. It also does not change the number of protons, so \(Z\) remains unchanged. This is different from alpha decay, which changes both \(A\) and \(Z\), and beta decay, which changes \(Z\). Gamma emission is best understood as nuclear de-excitation.
206. A source emits particles with charge \(-e\). The radiation is deflected in an electric field toward the positive plate. The most suitable identification is:
ⓐ. Alpha radiation
ⓑ. Gamma radiation
ⓒ. Beta minus radiation
ⓓ. Neutral neutron radiation
Correct Answer: Beta minus radiation
Explanation: A particle with charge \(-e\) is attracted toward the positive plate in an electric field. Beta minus radiation consists of electrons emitted in nuclear decay, and each such electron has charge \(-e\). Alpha particles carry charge \(+2e\), so they would be deflected toward the negative plate. Gamma rays are neutral photons and would not be deflected by the electric field in this way. The direction of deflection therefore helps identify the sign of charge of the emitted radiation.
207. A source emits radiation that is not deflected by an electric field and has very high penetrating power. The most likely radiation is:
ⓐ. Alpha radiation
ⓑ. Gamma radiation
ⓒ. Beta minus radiation
ⓓ. A stream of singly positive ions
Correct Answer: Gamma radiation
Explanation: Gamma rays are electrically neutral photons. Since they have no charge, they are not deflected by an electric field. They also have very high penetrating power compared with alpha and beta radiations. Alpha and beta particles are charged, so they can be deflected in electric and magnetic fields. Singly positive ions would also be deflected because they carry charge. The combination of no electric deflection and high penetration points to gamma radiation.
208. A sample gives off radiation that blackens a photographic plate even when kept in the dark. This observation is closest to the early evidence for:
ⓐ. Spontaneous radioactive emission
ⓑ. Ordinary reflection of sunlight
ⓒ. Cooling by evaporation
ⓓ. Formation of chemical bonds only
Correct Answer: Spontaneous radioactive emission
Explanation: Radiation from radioactive substances can affect photographic plates without ordinary visible light exposure. This was part of the evidence that certain materials emit radiation spontaneously. The effect is not ordinary reflection of sunlight because the plate can be affected even in darkness. It is also not explained simply by chemical bond formation. The radiation comes from unstable nuclei in the material. Such observations helped establish radioactivity as a real nuclear phenomenon.
209. The relative ionising power of alpha, beta, and gamma radiations is best arranged as:
ⓐ. Alpha \(\gt\) beta \(\gt\) gamma
ⓑ. Gamma \(\gt\) beta \(\gt\) alpha
ⓒ. Beta \(\gt\) alpha \(\gt\) gamma
ⓓ. Alpha \(=\) beta \(=\) gamma
Correct Answer: Alpha \(\gt\) beta \(\gt\) gamma
Explanation: Alpha particles are massive and carry charge \(+2e\), so they produce strong ionisation over a short path. Beta particles are lighter and singly charged, so their ionising power is moderate. Gamma rays are neutral photons and are usually less ionising than alpha and beta radiations, although they are much more penetrating. Thus the usual ionising-power order is alpha greater than beta greater than gamma. The penetrating-power order is roughly the opposite. Confusing these two orders leads to wrong shielding and safety reasoning.
210. A compact comparison says:
Radiation P has the greatest ionising power but the least penetration. Radiation Q has moderate ionising and penetrating power. Radiation R has the greatest penetration but comparatively low ionising power.
The sequence P, Q, R is:
ⓐ. Gamma, beta, alpha
ⓑ. Beta, alpha, gamma
ⓒ. Alpha, gamma, beta
ⓓ. Alpha, beta, gamma
Correct Answer: Alpha, beta, gamma
Explanation: Alpha radiation has the highest ionising power because alpha particles are massive and carry charge \(+2e\). It also has the least penetration because it loses energy quickly in matter. Beta radiation is intermediate in both ionising power and penetrating power. Gamma radiation has the greatest penetration because it consists of neutral high-energy photons. It has comparatively lower ionising power than alpha radiation. The description therefore matches the sequence alpha, beta, gamma.
211. In alpha decay, the emitted particle is:
ⓐ. \({}^{0}_{-1}e\)
ⓑ. \({}^{0}_{0}\gamma\)
ⓒ. \({}^{4}_{2}He\)
ⓓ. \({}^{1}_{0}n\)
Correct Answer: \({}^{4}_{2}He\)
Explanation: An alpha particle is a helium-\(4\) nucleus. It contains \(2\) protons and \(2\) neutrons, so its mass number is \(4\) and its atomic number is \(2\). This is why it is represented as \({}^{4}_{2}He\). A beta minus particle is represented as \({}^{0}_{-1}e\), and gamma radiation is written as \({}^{0}_{0}\gamma\). A neutron is \({}^{1}_{0}n\), which is not the usual alpha particle. In alpha decay, the parent nucleus loses a compact group of four nucleons.
212. When a nucleus undergoes alpha decay, its daughter nucleus has:
ⓐ. Mass number decreased by \(2\) and atomic number decreased by \(4\)
ⓑ. Mass number decreased by \(4\) and atomic number decreased by \(2\)
ⓒ. Mass number unchanged and atomic number increased by \(1\)
ⓓ. Mass number unchanged and atomic number unchanged
Correct Answer: Mass number decreased by \(4\) and atomic number decreased by \(2\)
Explanation: Alpha decay means emission of an alpha particle \({}^{4}_{2}He\). The emitted alpha particle carries away \(4\) nucleons, so the parent mass number decreases by \(4\). It also carries away \(2\) protons, so the atomic number decreases by \(2\). The daughter nucleus is therefore different from the parent element. This change is not like gamma emission, where \(A\) and \(Z\) remain unchanged. Alpha decay must conserve both total nucleon number and total charge number.
213. A parent nucleus \({}^{238}_{92}U\) emits an alpha particle. The daughter nucleus has:
ⓐ. \(A=234\), \(Z=90\)
ⓑ. \(A=236\), \(Z=88\)
ⓒ. \(A=238\), \(Z=91\)
ⓓ. \(A=242\), \(Z=94\)
Correct Answer: \(A=234\), \(Z=90\)
Explanation: \( \textbf{Given parent nucleus:} \) \({}^{238}_{92}U\).
\( \textbf{Alpha particle:} \) \({}^{4}_{2}He\).
\( \textbf{Mass-number conservation:} \)
\[
238=A_d+4
\]
\( \textbf{Daughter mass number:} \)
\[
A_d=238-4=234
\]
\( \textbf{Atomic-number conservation:} \)
\[
92=Z_d+2
\]
\( \textbf{Daughter atomic number:} \)
\[
Z_d=92-2=90
\]
\( \textbf{Final answer:} \) The daughter nucleus has \(A=234\) and \(Z=90\). The loss of two protons changes the element identity of the daughter nucleus.
214. The alpha decay equation \({}^{A}_{Z}X\rightarrow{}^{A-4}_{Z-2}Y+{}^{4}_{2}He\) is mainly based on conservation of:
ⓐ. Mass and charge numbers
ⓑ. Atomic radius and density
ⓒ. Electron shell energy only
ⓓ. Chemical valency only
Correct Answer: Mass and charge numbers
Explanation: In a nuclear equation, the total mass number must be the same on both sides. The total atomic number, which represents charge number, must also be conserved. In alpha decay, the emitted \({}^{4}_{2}He\) takes away mass number \(4\) and atomic number \(2\). Therefore, the daughter nucleus must have \(A-4\) and \(Z-2\). This balancing does not come from chemical valency or electron shells. Nuclear equations are balanced by nucleon number and charge number first.
215. A nucleus \({}^{226}_{88}Ra\) decays by alpha emission. The equation is best written as:
ⓐ. \({}^{226}_{88}Ra\rightarrow{}^{226}_{89}Ac+{}^{0}_{-1}e\)
ⓑ. \({}^{226}_{88}Ra\rightarrow{}^{226}_{88}Ra+{}^{0}_{0}\gamma\)
ⓒ. \({}^{226}_{88}Ra\rightarrow{}^{230}_{90}Th+{}^{4}_{2}He\)
ⓓ. \({}^{226}_{88}Ra\rightarrow{}^{222}_{86}Rn+{}^{4}_{2}He\)
Correct Answer: \({}^{226}_{88}Ra\rightarrow{}^{222}_{86}Rn+{}^{4}_{2}He\)
Explanation: In alpha decay, the emitted particle is \({}^{4}_{2}He\). The daughter mass number is \(226-4=222\). The daughter atomic number is \(88-2=86\). The element with \(Z=86\) is radon, represented by \(Rn\). Changes that keep \(A\) unchanged are beta-like rather than alpha decay, gamma emission changes neither \(A\) nor \(Z\), and increasing both \(A\) and \(Z\) is opposite to alpha emission.
216. A row in a decay table is missing the daughter nucleus:
| Parent | Emission | Daughter |
|---|
| \({}^{210}_{84}Po\) | \({}^{4}_{2}He\) | ? |
The missing daughter is:
ⓐ. \({}^{206}_{82}Pb\)
ⓑ. \({}^{214}_{86}Rn\)
ⓒ. \({}^{210}_{83}Bi\)
ⓓ. \({}^{210}_{84}Po\)
Correct Answer: \({}^{206}_{82}Pb\)
Explanation: \( \textbf{Parent nucleus:} \) \({}^{210}_{84}Po\).
\( \textbf{Emission:} \) Alpha particle \({}^{4}_{2}He\).
\( \textbf{Daughter mass number:} \)
\[
A_d=210-4=206
\]
\( \textbf{Daughter atomic number:} \)
\[
Z_d=84-2=82
\]
\( \textbf{Element identification:} \) The element with \(Z=82\) is lead, \(Pb\).
\( \textbf{Balanced equation:} \)
\[
{}^{210}_{84}Po\rightarrow{}^{206}_{82}Pb+{}^{4}_{2}He
\]
\( \textbf{Final answer:} \) The missing daughter is \({}^{206}_{82}Pb\). Both the upper and lower nuclear numbers must balance separately.
217. A claim says, “In alpha decay, the daughter nucleus has the same chemical element as the parent because only energy is emitted.” The best evaluation is:
ⓐ. The claim is valid because alpha decay does not change \(Z\)
ⓑ. The claim is not valid because alpha decay decreases \(Z\) by \(2\)
ⓒ. The claim is valid only if the parent is heavy
ⓓ. The claim is not valid because alpha decay increases \(A\) by \(4\)
Correct Answer: The claim is not valid because alpha decay decreases \(Z\) by \(2\)
Explanation: The chemical element is determined by atomic number \(Z\). In alpha decay, the nucleus emits \({}^{4}_{2}He\), which removes \(2\) protons from the parent nucleus. Therefore, the daughter has atomic number \(Z-2\), so it is a different element. The mass number also decreases by \(4\). Only gamma emission leaves both \(A\) and \(Z\) unchanged. The claim confuses alpha emission with a pure energy de-excitation process.
218. An alpha decay has \(Q>0\). The positive \(Q\)-value means:
ⓐ. Energy must be supplied for the decay to occur
ⓑ. Energy is released in the decay
ⓒ. The daughter nucleus has the same \(Z\) as the parent
ⓓ. The alpha particle has zero mass
Correct Answer: Energy is released in the decay
Explanation: The \(Q\)-value of a nuclear process represents the net energy released or absorbed. If \(Q>0\), the final products have lower total mass-energy than the initial parent nucleus. The difference is released as kinetic energy of the products and possibly radiation. In alpha decay, this energy is often shared mainly between the alpha particle and the recoiling daughter nucleus. A positive \(Q\)-value does not mean the daughter keeps the same atomic number. It indicates an energetically allowed release of nuclear energy.
219. In an alpha decay, the alpha particle is usually observed with a definite energy rather than a broad continuous range. This is mainly because:
ⓐ. The alpha particle is an outer atomic electron
ⓑ. Alpha particles have no kinetic energy
ⓒ. Two-body breakup to definite nuclear states
ⓓ. Charge conservation is not obeyed in alpha decay
Correct Answer: Two-body breakup to definite nuclear states
Explanation: In a simple alpha decay to a definite daughter state, the parent nucleus breaks into two main products: the daughter nucleus and the alpha particle. Conservation of energy and momentum then fixes how the available \(Q\)-value is shared. This gives alpha particles with definite energies for a given transition. The alpha particle is not an atomic electron; it is a helium nucleus. It certainly carries kinetic energy after emission. Charge and nucleon number conservation remain central to writing the alpha decay equation.
220. In beta minus decay, the nuclear change at the nucleon level is best represented as:
ⓐ. \(p\rightarrow n+e^++\nu\)
ⓑ. \({}^{4}_{2}He\rightarrow2p+2n\)
ⓒ. \(n\rightarrow p+e^-+\bar{\nu}\)
ⓓ. \(\gamma\rightarrow p+n\)
Correct Answer: \(n\rightarrow p+e^-+\bar{\nu}\)
Explanation: Beta minus decay occurs when a neutron in the nucleus changes into a proton. An electron and an antineutrino are emitted in the process. This is represented as \(n\rightarrow p+e^-+\bar{\nu}\). Since a neutron becomes a proton, the atomic number of the daughter nucleus increases by \(1\), while the mass number remains unchanged. The emitted beta electron is produced in the decay process and is not simply an orbital electron. The antineutrino helps account for energy, momentum, and other conservation requirements.
