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Conclusion?","back":"Most of the atom is empty space.","mcqOptions":[{"text":"Most of the atom is empty space.","correct":true,"explanation":"Most alpha particles pass through the foil without significant deflection, showing that the atom contains a very large region with no concentrated matter or charge: empty space."},{"text":"Most of the atom is positive charge.","correct":false,"explanation":"The scattering pattern does not show that positive charge fills most of the atom. The correct conclusion is that most of the atom is empty space, with positive charge concentrated in the nucleus."},{"text":"Most of the atom is dense matter.","correct":false,"explanation":"Most alpha particles do not collide with electrons or a solid mass. The correct conclusion is that most of the atom is empty space, allowing the particles to pass through."},{"text":"Most of the atom is negative charge.","correct":false,"explanation":"A straight path does not indicate that the atom is mostly neutral charge. The correct conclusion is that most of the atom is empty space, while charge is concentrated in a tiny nucleus."}]},{"id":"fc82144c-1bc1-4da2-b08c-b3236d41f3e3","cardType":"basic","points":2,"front":"Observation: A very small fraction of alpha particles are deflected by more than 90 degrees.
State two conclusions.","back":"

The atom's positive charge is concentrated in a tiny volume (nucleus).
The nucleus contains most of the atom's mass.

","mcqOptions":[{"text":"

The atom's positive charge is concentrated in a tiny volume (nucleus).
The nucleus contains most of the atom's mass.

","correct":true,"explanation":"Large-angle deflections require strong repulsion from a small, positively charged region, while the alpha particle’s momentum shows that this nucleus contains most of the atom’s mass."},{"text":"

The atom's positive charge is concentrated in a tiny volume (nucleus).
The nucleus contains most of the atom's volume.

","correct":false,"explanation":"A tiny fraction of alpha particles being strongly deflected indicates concentrated positive charge, but the nucleus contains most of the atom’s mass, not most of its volume."},{"text":"

The atom's positive charge is spread out throughout the atom.
The nucleus contains most of the atom's mass.

","correct":false,"explanation":"The nucleus is tiny, so it cannot contain most of the atom’s volume. The observations instead show concentrated positive charge and that most of the atom’s mass is in the nucleus."},{"text":"

The atom's positive charge is spread out throughout the atom.
The nucleus contains only a small fraction of the atom's mass.

","correct":false,"explanation":"Strong deflections require a concentrated positive charge, not a diffuse one, and the nucleus contains most of the atom’s mass rather than only a small fraction."}]},{"id":"e0754cf1-9a9a-46db-aeb2-41c05b46135d","cardType":"cloze","points":null,"front":"An atom has a radius of approximately $\\text{[...]}\\text{ m}$, while the nucleus has a radius of roughly $10^{-15}\\text{ to }10^{-14}\\text{ m}$.","back":"An atom has a radius of approximately $\\boxed{10^{-10}}\\text{ m}$, while the nucleus has a radius of roughly $10^{-15}\\text{ to }10^{-14}\\text{ m}$.","mcqOptions":[{"text":"$$10^{-10}$","correct":true,"explanation":"The typical atomic radius is about $10^{-10}\\text{ m}$, whereas the nuclear radius is around $10^{-15}$ to $10^{-14}\\text{ m}$, demonstrating the nucleus is far smaller than the atom."},{"text":"$$10^{-9}$","correct":false,"explanation":"$$10^{-9}\\text{ m}$ is about ten times larger than the accepted atomic radius. The standard estimate is $10^{-10}\\text{ m}$, while the nucleus is much smaller."},{"text":"$$10^{-11}$","correct":false,"explanation":"$$10^{-11}\\text{ m}$ is ten times smaller than the typical atomic radius. The accepted approximate value is $10^{-10}\\text{ m}$, not this near-miss exponent."},{"text":"$$10^{-14}$","correct":false,"explanation":"$$10^{-14}\\text{ m}$ is a nuclear-scale radius, not the radius of the whole atom. The atomic radius is approximately $10^{-10}\\text{ m}$."}]},{"id":"7f00c214-59e9-429c-a8ca-f165d8443350","cardType":"cloze","points":null,"front":"An atom has a radius of approximately $10^{-10}\\text{ m}$, while the nucleus has a radius of roughly $\\text{[...]}\\text{ to }10^{-14}\\text{ m}$.","back":"An atom has a radius of approximately $10^{-10}\\text{ m}$, while the nucleus has a radius of roughly $\\boxed{10^{-15}}\\text{ to }10^{-14}\\text{ m}$.","mcqOptions":[{"text":"$$10^{-15}$","correct":true,"explanation":"The nuclear radius is typically estimated as $10^{-15}$ to $10^{-14}\\text{ m}$. This is much smaller than the atomic radius of approximately $10^{-10}\\text{ m}$."},{"text":"$$10^{-16}$","correct":false,"explanation":"$$10^{-16}\\text{ m}$ is one order of magnitude below the usual lower estimate. The accepted nuclear-radius range begins at approximately $10^{-15}\\text{ m}$."},{"text":"$$10^{-13}$","correct":false,"explanation":"$$10^{-13}\\text{ m}$ is one order of magnitude above the usual upper estimate for a nucleus. The accepted range is approximately $10^{-15}$ to $10^{-14}\\text{ m}$."},{"text":"$$10^{-10}$","correct":false,"explanation":"$$10^{-10}\\text{ m}$ is the approximate radius of the whole atom, not its nucleus. Nuclear radii are much smaller, beginning at about $10^{-15}\\text{ m}$."}]},{"id":"ac75f22f-606c-47e8-a122-bde85282be36","cardType":"basic","points":2,"front":"What is the composition and charge of an alpha particle?","back":"

Composition: 2 protons and 2 neutrons (helium nucleus).
Charge: $+2e$.

","mcqOptions":[{"text":"

Composition: 2 protons and 2 neutrons (helium nucleus).
Charge: $+2e$.

","correct":true,"explanation":"An alpha particle is a helium nucleus containing two protons and two neutrons. The two protons give it a net charge of $+2e$."},{"text":"

Composition: 2 protons and 2 electrons (helium atom).
Charge: $-2e$.

","correct":false,"explanation":"Electrons are not part of an alpha particle, and two protons give a charge of $+2e$, not $-2e$. An alpha particle is two protons and two neutrons."},{"text":"

Composition: 2 protons and 1 neutron (lithium nucleus).
Charge: $+2e$.

","correct":false,"explanation":"A neutron has no charge, so two protons and one neutron would give $+2e$, but an alpha particle also contains a second neutron and is a helium nucleus."},{"text":"

Composition: 1 proton and 1 neutron (hydrogen nucleus).
Charge: $+e$.

","correct":false,"explanation":"One proton and one neutron form a deuterium nucleus and have charge $+e$. An alpha particle contains two protons and two neutrons, giving $+2e$."}]},{"id":"889d769c-5b23-4f29-bc04-f1ec9345dd01","cardType":"basic","points":1,"front":"Why must the Rutherford scattering experiment be conducted in a vacuum?","back":"To prevent alpha particles being scattered by air molecules.","mcqOptions":[{"text":"To prevent alpha particles being scattered by air molecules.","correct":true,"explanation":"Air molecules would cause additional collisions and deflections, making it impossible to attribute the observed scattering solely to interactions between alpha particles and gold nuclei."},{"text":"To prevent alpha particles being neutralised by air molecules.","correct":false,"explanation":"A vacuum is not required to prevent alpha particles losing charge. It removes air molecules that could collide with and scatter the particles before they reach the foil."},{"text":"To prevent alpha particles being decayed by air molecules.","correct":false,"explanation":"Alpha particles are not prevented from decaying by removing air. The vacuum prevents extra scattering by air molecules, so the measured deflections are due to the gold nuclei."},{"text":"To prevent alpha particles being absorbed by air molecules.","correct":false,"explanation":"The experiment does not require a vacuum to stop alpha particles becoming radioactive. The purpose is to prevent air molecules causing additional scattering and confusing the results."}]},{"id":"1d5fa2cf-9fe1-463d-bd2b-c1df2bed5bbe","cardType":"basic","points":2,"front":"State two reasons why gold foil was used in the scattering experiment.","back":"

It can be beaten into extremely thin layers (few atomic layers).
It has a large nuclear charge ($Z$), making deflections easier to observe.

","mcqOptions":[{"text":"

It can be beaten into extremely thin layers (few atomic layers).
It has a large nuclear charge ($Z$), making deflections easier to observe.

","correct":true,"explanation":"Gold can be hammered into a foil only a few atomic layers thick, reducing multiple scattering, and its high atomic number gives strong nuclear repulsion that makes deflections easier to detect."},{"text":"

It can be beaten into extremely thin layers (few atomic layers).
It has a small nuclear charge ($Z$), making deflections easier to observe.

","correct":false,"explanation":"Gold’s usefulness comes from being extremely thin and having a large nuclear charge, not from having a small nuclear charge. A small charge would produce weaker deflections."},{"text":"

It can be beaten into extremely thick layers (many atomic layers).
It has a large nuclear charge ($Z$), making deflections easier to observe.

","correct":false,"explanation":"A thick foil would cause repeated scattering and obscure the interpretation. Gold was chosen because it forms extremely thin layers and has a large nuclear charge."},{"text":"

It can be beaten into extremely thick layers (many atomic layers).
It has a small nuclear charge ($Z$), making deflections easier to observe.

","correct":false,"explanation":"Gold’s high density is not the stated reason for using it, and a small nuclear charge would make deflections harder to observe. Its key advantages are thinness and large $Z$."}]},{"id":"92cdd12e-cdac-4f4d-9fcd-e4a28aef7b58","cardType":"basic","points":2,"front":"Why did the plum pudding model predict only tiny deflections of alpha particles?","back":"The positive charge was diffuse (spread out), so the electric field was too weak to cause large deflections.","mcqOptions":[{"text":"The positive charge was diffuse (spread out), so the electric field was too weak to cause large deflections.","correct":true,"explanation":"In the plum pudding model, positive charge is spread diffusely through the atom. This produces only a weak electric field, so alpha particles experience small rather than large deflections."},{"text":"The positive charge was concentrated (localised), so the electric field was too strong to cause large deflections.","correct":false,"explanation":"The model predicted small deflections because its positive charge was diffuse and produced a weak field, not because the charge was concentrated and the field was strong."},{"text":"The positive charge was diffuse (spread out), so the electric field was too strong to cause large deflections.","correct":false,"explanation":"A diffuse positive charge gives a weak electric field, but a weak field causes small deflections rather than large ones. The model therefore predicted only tiny deflections."},{"text":"The positive charge was concentrated (localised), so the electric field was too weak to cause large deflections.","correct":false,"explanation":"The plum pudding model did not concentrate positive charge in a nucleus. Its diffuse charge created a weak field, which explains the prediction of tiny deflections."}]},{"id":"b2f945fc-c985-4763-9b8b-60a4bedfc4e8","cardType":"basic","points":1,"front":"How were alpha particles detected in the Geiger-Marsden experiment?","back":"They hit a zinc sulfide screen, which produced tiny flashes of light called scintillations.","mcqOptions":[{"text":"They hit a zinc sulfide screen, which produced tiny flashes of light called scintillations.","correct":true,"explanation":"The zinc sulfide screen emitted a tiny flash whenever an alpha particle struck it. These flashes, called scintillations, allowed individual alpha particles to be detected and counted."},{"text":"They hit a photographic plate, which produced tiny flashes of light called scintillations.","correct":false,"explanation":"A photographic plate records particle tracks chemically rather than producing visible scintillations. In the Geiger-Marsden experiment, alpha particles struck zinc sulfide, which emitted the observed flashes."},{"text":"They hit a zinc sulfide screen, which produced a continuous beam of light called fluorescence.","correct":false,"explanation":"The screen produced separate flashes when individual alpha particles arrived, not a continuous beam. Zinc sulfide scintillations were used because each flash indicated a detected particle."},{"text":"They hit a metal detector, which produced tiny flashes of light called ionisations.","correct":false,"explanation":"Ionisation is the formation of ions, not the name of the visible flashes used for detection. The particles struck a zinc sulfide screen, producing scintillations."}]},{"id":"a7729ba9-070e-4af1-9937-9a8f6457eae9","cardType":"cloze","points":null,"front":"At the distance of closest approach for a head-on collision, the alpha particle's initial $\\text{[...]}$ is entirely transferred into $electrostatic potential energy$.","back":"At the distance of closest approach for a head-on collision, the alpha particle's initial $\\boxed{kinetic energy}$ is entirely transferred into $electrostatic potential energy$.","mcqOptions":[{"text":"kinetic energy","correct":true,"explanation":"At closest approach in a head-on collision, the alpha particle has momentarily stopped, so its initial kinetic energy has been converted into electrostatic potential energy."},{"text":"momentum","correct":false,"explanation":"Momentum is not transferred into electrostatic potential energy, and momentum is conserved for the complete interacting system. The alpha particle’s initial kinetic energy provides the potential energy increase."},{"text":"gravitational potential energy","correct":false,"explanation":"Gravity is negligible at nuclear distances and cannot account for the energy change. The alpha particle’s initial kinetic energy is converted into electrostatic potential energy during repulsion."},{"text":"electric charge","correct":false,"explanation":"The alpha particle does not lose its electric charge during the approach. Its initial kinetic energy is converted into electrostatic potential energy as repulsion brings it to rest momentarily."}]},{"id":"d86db4e6-0a21-4d0a-8b7d-95f7339563d0","cardType":"cloze","points":null,"front":"At the distance of closest approach for a head-on collision, the alpha particle's initial $kinetic energy$ is entirely transferred into $\\text{[...]}$.","back":"At the distance of closest approach for a head-on collision, the alpha particle's initial $kinetic energy$ is entirely transferred into $\\boxed{electrostatic potential energy}$.","mcqOptions":[{"text":"electrostatic potential energy","correct":true,"explanation":"The positively charged alpha particle is repelled by the positively charged nucleus. At closest approach, its kinetic energy has been converted into electrostatic potential energy."},{"text":"gravitational potential energy","correct":false,"explanation":"Gravitational effects are far too weak at nuclear distances to determine the turning point. The relevant stored energy is electrostatic potential energy from the two positive charges."},{"text":"nuclear binding energy","correct":false,"explanation":"The alpha particle does not enter or bind to the nucleus in this situation. Its kinetic energy becomes electrostatic potential energy as electrical repulsion slows it."},{"text":"electric current","correct":false,"explanation":"Electric current describes moving charge through a conductor, not stored energy in this collision. The kinetic energy is converted into electrostatic potential energy at closest approach."}]},{"id":"5c61d961-d237-4427-8812-7c92fea6eb56","cardType":"basic","points":1,"front":"Why are alpha particles not significantly deflected by collisions with electrons?","back":"Alpha particles are much more massive than electrons.","mcqOptions":[{"text":"Alpha particles are much more massive than electrons.","correct":true,"explanation":"An alpha particle is much heavier than an electron, so an electron collision produces only a very small change in the alpha particle’s direction and momentum."},{"text":"Alpha particles are much less massive than electrons.","correct":false,"explanation":"The mass relationship is reversed here. Alpha particles contain two protons and two neutrons, making them much more massive than electrons and therefore difficult to deflect."},{"text":"Alpha particles are electrically neutral compared with electrons.","correct":false,"explanation":"Alpha particles are positively charged, so neutrality is not the reason for their small deflection. Their much greater mass means electron collisions have little effect on their motion."},{"text":"Alpha particles are much faster than electrons.","correct":false,"explanation":"Speed alone does not explain why electron collisions cause little deflection. The important factor is that alpha particles are much more massive than electrons."}]},{"id":"b52d2d22-c38f-4e35-b18d-2970fdbd5b28","cardType":"basic","points":1,"front":"Define the 'impact parameter' in scattering.","back":"The perpendicular distance between the initial path of the particle and the centre of the nucleus.","mcqOptions":[{"text":"The perpendicular distance between the initial path of the particle and the centre of the nucleus.","correct":true,"explanation":"The impact parameter measures the initial sideways offset of the incident particle’s path from the nucleus’s centre, before electrostatic repulsion curves its trajectory."},{"text":"The shortest distance between the deflected path of the particle and the surface of the nucleus.","correct":false,"explanation":"This confuses impact parameter with distance of closest approach. The impact parameter is the perpendicular distance between the particle’s initial path and the centre of the nucleus."},{"text":"The parallel distance between the initial path of the particle and the centre of the nucleus.","correct":false,"explanation":"A parallel separation would be measured along the particle’s direction of travel. The impact parameter is the perpendicular distance from its initial path to the nucleus’s centre."},{"text":"The perpendicular distance between the final path of the particle and the surface of the nucleus.","correct":false,"explanation":"This incorrectly uses the outgoing path and nuclear surface. The impact parameter is defined using the initial undeflected path and the centre of the nucleus."}]},{"id":"d0d63f4e-1543-4c43-a634-b9b89a8256d2","cardType":"cloze","points":null,"front":"The electrostatic force $F$ between the alpha particle and nucleus follows Coulomb's Law: $$F = \\frac{1}{\\text{[...]}} \\frac{Q_1 Q_2}{r^2}$$","back":"The electrostatic force $F$ between the alpha particle and nucleus follows Coulomb's Law: $$F = \\frac{1}{\\boxed{4\\pi\\varepsilon_0}} \\frac{Q_1 Q_2}{r^2}$$","mcqOptions":[{"text":"$$4\\pi\\varepsilon_0$","correct":true,"explanation":"Coulomb’s law in SI units is $F = \\frac{1}{4\\pi\\varepsilon_0}\\frac{Q_1Q_2}{r^2}$, where $\\varepsilon_0$ is the permittivity of free space."},{"text":"$$4\\pi\\varepsilon_0^2$","correct":false,"explanation":"Squaring the permittivity incorrectly changes the dimensions and gives the wrong electrostatic constant. Coulomb’s law contains one factor of $\\varepsilon_0$ in the denominator."},{"text":"$$4\\pi k$","correct":false,"explanation":"The constant $k$ is commonly defined as $\\frac{1}{4\\pi\\varepsilon_0}$, so placing $4\\pi k$ in the denominator does not reproduce Coulomb’s law. The required factor is $4\\pi\\varepsilon_0$."},{"text":"$$4\\pi r\\varepsilon_0$","correct":false,"explanation":"Including an extra factor of $r$ in the denominator would make the force vary as $1/r^3$. Coulomb’s law already contains the required $r^2$ dependence separately."}]},{"id":"b2c697b6-c897-4bab-9ac4-8b8de1585e9f","cardType":"basic","points":1,"front":"Why is the distance of closest approach only an estimate of the nuclear radius?","back":"The alpha particle is repelled and turns around before it actually reaches or touches the nuclear surface.","mcqOptions":[{"text":"The alpha particle is repelled and turns around before it actually reaches or touches the nuclear surface.","correct":true,"explanation":"Electrostatic repulsion converts the alpha particle’s kinetic energy into electric potential energy, stopping it before contact. Its closest distance therefore provides only an upper estimate of nuclear radius."},{"text":"The alpha particle is attracted and speeds up before it actually reaches or touches the nuclear surface.","correct":false,"explanation":"An alpha particle and nucleus are both positively charged, so they repel rather than attract. The particle turns around before touching the surface, making the inferred radius an estimate."},{"text":"The alpha particle is absorbed and stops moving when it actually reaches or touches the nuclear surface.","correct":false,"explanation":"The closest-approach calculation assumes electrostatic repulsion, not absorption at the surface. The alpha particle turns around before contact, so the calculated distance only estimates the nuclear radius."},{"text":"The alpha particle is deflected and accelerates after it actually passes through the nuclear surface.","correct":false,"explanation":"This assumes the alpha particle enters the nucleus, but Coulomb repulsion turns a head-on particle around beforehand. Consequently, closest approach is not a direct measurement of the surface."}]},{"id":"f17f6a0c-f2b3-466e-8a98-cb29d0c39fdd","cardType":"basic","points":1,"front":"Observation: Some alpha particles were deflected through small angles.
Conclusion?","back":"There is a positive charge inside the atom causing repulsion.","mcqOptions":[{"text":"There is a positive charge inside the atom causing repulsion.","correct":true,"explanation":"Small-angle deflections show that alpha particles experience a repulsive force. Since alpha particles are positive, the deflecting charge inside the atom must also be positive."},{"text":"There is a negative charge inside the atom causing repulsion.","correct":false,"explanation":"A negative charge would attract positive alpha particles rather than repel them. Repulsive deflection therefore indicates a positive charge within the atom."},{"text":"There is a positive charge inside the atom causing attraction.","correct":false,"explanation":"Positive charges repel one another, so attraction cannot explain the observed deflection of positive alpha particles. The conclusion is a positive internal charge causing repulsion."},{"text":"There is no charge inside the atom causing repulsion.","correct":false,"explanation":"A repulsive force requires an electrical interaction, so a neutral region cannot cause this effect. The small deflections indicate positive charge producing repulsion."}]},{"id":"2ef31298-f91f-420a-a7f5-973566b99792","cardType":"cloze","points":null,"front":"A $\\text{[...]}$ impact parameter leads to a larger angle of deflection because the particle passes $closer$ to the nucleus.","back":"A $\\boxed{smaller}$ impact parameter leads to a larger angle of deflection because the particle passes $closer$ to the nucleus.","mcqOptions":[{"text":"$$smaller$","correct":true,"explanation":"A smaller impact parameter means the alpha particle’s initial path passes nearer the nucleus. The stronger electrostatic force experienced there produces a larger change in momentum and deflection angle."},{"text":"$$larger$","correct":false,"explanation":"This reverses the relationship: a larger impact parameter takes the particle farther from the nucleus, weakening repulsion. A smaller impact parameter produces the larger angle of deflection."},{"text":"$$constant$","correct":false,"explanation":"The impact parameter varies between particle trajectories and determines how closely they approach the nucleus. A smaller impact parameter causes stronger repulsion and therefore a larger deflection angle."},{"text":"$$maximum$","correct":false,"explanation":"A maximum impact parameter corresponds to a distant pass and weak deflection, not a large angle. The required term is $smaller$, because closer trajectories experience stronger electrostatic repulsion."}]},{"id":"149fe542-10fe-4d92-a961-2026f5338ddb","cardType":"cloze","points":null,"front":"A $smaller$ impact parameter leads to a larger angle of deflection because the particle passes $\\text{[...]}$ to the nucleus.","back":"A $smaller$ impact parameter leads to a larger angle of deflection because the particle passes $\\boxed{closer}$ to the nucleus.","mcqOptions":[{"text":"$$closer$","correct":true,"explanation":"A smaller impact parameter directs the particle nearer the positively charged nucleus. At this reduced separation, electrostatic repulsion is stronger and produces a larger angle of deflection."},{"text":"$$further$","correct":false,"explanation":"This reverses the geometry: a smaller impact parameter makes the initial trajectory pass closer, not further away. The reduced separation increases electrostatic repulsion and hence the deflection."},{"text":"$$parallel$","correct":false,"explanation":"Parallel describes the orientation of paths, not their proximity to the nucleus. The correct term is $closer$, because a smaller impact parameter reduces the trajectory’s sideways offset."},{"text":"$$tangential$","correct":false,"explanation":"Tangential motion does not express the required change in distance from the nucleus. The particle passes $closer$, where stronger electrostatic repulsion causes a greater deflection."}]},{"id":"96ca65b9-6b69-444d-90d2-33c90a6432dd","cardType":"basic","points":2,"front":"To calculate the distance of closest approach for a gold nucleus ($Z=79$), what are the two charges ($Q_1$ and $Q_2$) used in the potential energy formula?","back":"

Alpha particle: $+2e$.
Gold nucleus: $+79e$.

","mcqOptions":[{"text":"

Alpha particle: $+2e$.
Gold nucleus: $+79e$.

","correct":true,"explanation":"An alpha particle contains two protons and has charge $+2e$. Gold has atomic number 79, so its nucleus contains 79 protons and has charge $+79e$."},{"text":"

Alpha particle: $+4e$.
Gold nucleus: $+79e$.

","correct":false,"explanation":"The alpha particle’s nucleon number is 4, but two of those nucleons are neutral neutrons, so its charge is $+2e$. The gold nucleus has charge $+79e$."},{"text":"

Alpha particle: $+2e$.
Gold nucleus: $+197e$.

","correct":false,"explanation":"The value 197 is gold’s nucleon number, not its proton number. Nuclear charge depends on proton number, giving $+79e$ for gold and $+2e$ for an alpha particle."},{"text":"

Alpha particle: $-2e$.
Gold nucleus: $+79e$.

","correct":false,"explanation":"An alpha particle contains two protons and no electrons, so its charge is positive, not negative. The required charges are $+2e$ and $+79e$."}]},{"id":"d4766184-32f5-4868-aba2-e3dc4047fa72","cardType":"cloze","points":null,"front":"In the nuclear model, an atom is electrically neutral because the number of $\\text{[...]}$ in the nucleus equals the number of $electrons$ outside it.","back":"In the nuclear model, an atom is electrically neutral because the number of $\\boxed{protons}$ in the nucleus equals the number of $electrons$ outside it.","mcqOptions":[{"text":"$$protons$","correct":true,"explanation":"Each proton has charge $+e$, while each electron has charge $-e$. Equal numbers of protons and electrons therefore give equal and opposite total charges, making the atom neutral."},{"text":"$$neutrons$","correct":false,"explanation":"Neutrons carry no electric charge, so their number does not balance the electrons’ negative charge. Neutrality requires equal numbers of positively charged protons and negatively charged electrons."},{"text":"$$nucleons$","correct":false,"explanation":"Nucleons include both protons and neutrons, but neutrons are uncharged. Only the number of protons must equal the number of electrons for the atom to be neutral."},{"text":"$$positrons$","correct":false,"explanation":"Positrons are positively charged antiparticles and are not the normal positive constituents of an atomic nucleus. The nucleus contains protons, whose number equals the electrons in a neutral atom."}]},{"id":"36144ff3-78f3-4ff0-baaf-a4d3def74642","cardType":"cloze","points":null,"front":"In the nuclear model, an atom is electrically neutral because the number of $protons$ in the nucleus equals the number of $\\text{[...]}$ outside it.","back":"In the nuclear model, an atom is electrically neutral because the number of $protons$ in the nucleus equals the number of $\\boxed{electrons}$ outside it.","mcqOptions":[{"text":"$$electrons$","correct":true,"explanation":"Protons carry charge $+e$ and electrons carry charge $-e$. Equal numbers give equal positive and negative charge magnitudes, so the atom has zero net charge."},{"text":"$$neutrons$","correct":false,"explanation":"Neutrons are located in the nucleus and have no charge, so they cannot balance proton charge. Neutrality requires an equal number of negatively charged electrons outside the nucleus."},{"text":"$$positrons$","correct":false,"explanation":"Positrons have positive charge, so they would add to rather than cancel the protons’ charge. Negatively charged electrons outside the nucleus balance the positive nuclear charge."},{"text":"$$nucleons$","correct":false,"explanation":"Nucleons are protons and neutrons within the nucleus, not particles orbiting outside it. The number of protons must equal the number of external electrons for neutrality."}]},{"id":"c77cae4f-3530-4eaa-a107-8af56f32af40","cardType":"basic","points":1,"front":"Which subatomic particle was missing from Rutherford's original nuclear model?","back":"The neutron.","mcqOptions":[{"text":"The neutron.","correct":true,"explanation":"Rutherford’s model established a small, positively charged nucleus but predated Chadwick’s discovery of the neutron. It therefore did not include neutral particles within the nucleus."},{"text":"The proton.","correct":false,"explanation":"Rutherford’s model already required concentrated positive charge in the nucleus, associated with protons. The missing particle was the neutron, which Chadwick discovered later in 1932."},{"text":"The electron.","correct":false,"explanation":"Electrons were already known from Thomson’s experiments and were included outside Rutherford’s nucleus. The neutron was missing because it had not yet been discovered."},{"text":"The positron.","correct":false,"explanation":"Positrons are antiparticles of electrons and are not ordinary permanent constituents of atoms. Rutherford’s original model lacked neutrons, the neutral nuclear particles discovered later by Chadwick."}]},{"id":"7cff5f9f-b1b4-4c10-b205-d4b6b0482390","cardType":"basic","points":1,"front":"What is the formula for the distance of closest approach ($r$) rearranged from the energy conservation equation?","back":"$$$r = \\frac{1}{4\\pi\\varepsilon_0} \\frac{2Ze^2}{E_k}$$","mcqOptions":[{"text":"$$$r = \\frac{1}{4\\pi\\varepsilon_0} \\frac{2Ze^2}{E_k}$$","correct":true,"explanation":"At closest approach, the alpha particle’s initial kinetic energy equals the electrostatic potential energy between charges $+2e$ and $+Ze$. Rearranging $E_k = \\frac{1}{4\\pi\\varepsilon_0}\\frac{2Ze^2}{r}$ gives this expression."},{"text":"$$$r = \\frac{1}{4\\pi\\varepsilon_0} \\frac{Ze^2}{E_k}$$","correct":false,"explanation":"This omits the factor of 2 from the alpha particle’s charge, incorrectly treating it as $+e$. An alpha particle has charge $+2e$, so the numerator must contain $2Ze^2$."},{"text":"$$$r = \\frac{1}{4\\pi\\varepsilon_0} \\frac{2Ze^2}{E_k^2}$$","correct":false,"explanation":"This incorrectly squares the kinetic energy when rearranging the energy equation, making the dimensions inconsistent. Since electrostatic potential energy varies as $1/r$, the correct distance is inversely proportional to $E_k$."},{"text":"$$$r = 4\\pi\\varepsilon_0 \\frac{2Ze^2}{E_k}$$","correct":false,"explanation":"This inverts Coulomb’s constant, replacing $\\frac{1}{4\\pi\\varepsilon_0}$ with its reciprocal. The electrostatic potential energy contains Coulomb’s constant, so that factor remains unchanged when solving for $r$."}]},{"id":"ee18a502-80cc-4114-a8fa-83e44a18fdd1","cardType":"cloze","points":null,"front":"Large-angle scattering was surprising because it proved that positive charge was not $\\text{[...]}$ but was instead concentrated in a $tiny volume$.","back":"Large-angle scattering was surprising because it proved that positive charge was not $\\boxed{spread evenly}$ but was instead concentrated in a $tiny volume$.","mcqOptions":[{"text":"spread evenly","correct":true,"explanation":"Frequent large-angle deflections could only occur if an alpha particle occasionally encountered a highly concentrated positive charge. This disproved the model in which positive charge was distributed uniformly throughout the atom."},{"text":"carried by electrons","correct":false,"explanation":"Electrons carry negative charge, so they cannot carry the atom’s positive charge. Alpha scattering showed that positive charge is concentrated in a tiny central nucleus rather than being associated with electrons."},{"text":"located at surface","correct":false,"explanation":"Locating positive charge at the atom’s surface would not produce the intense electric field needed for occasional large-angle deflections. The results instead showed that positive charge occupies a tiny central volume."},{"text":"absent entirely","correct":false,"explanation":"Positive charge cannot be entirely absent because alpha particles experienced electrostatic repulsion. The large deflections demonstrated that substantial positive charge is concentrated within a very small central region."}]},{"id":"6c461240-8d2c-400e-9f67-361357a5b4ea","cardType":"cloze","points":null,"front":"Large-angle scattering was surprising because it proved that positive charge was not $spread evenly$ but was instead concentrated in a $\\text{[...]}$.","back":"Large-angle scattering was surprising because it proved that positive charge was not $spread evenly$ but was instead concentrated in a $\\boxed{tiny volume}$.","mcqOptions":[{"text":"$$tiny volume$","correct":true,"explanation":"Rutherford scattering showed that the atom’s positive charge and most of its mass are concentrated in the nucleus, which occupies a tiny fraction of the atom’s volume."},{"text":"$$large volume$","correct":false,"explanation":"A large volume would resemble the diffuse positive charge proposed by the plum pudding model. Large-angle deflections instead require charge concentrated in a tiny volume."},{"text":"$$diffuse cloud$","correct":false,"explanation":"A diffuse cloud would exert only weak forces on passing alpha particles and could not cause large-angle scattering. The positive charge is concentrated in a tiny volume."},{"text":"$$thin shell$","correct":false,"explanation":"A thin shell would place positive charge around the atom’s exterior, rather than at its centre. Alpha scattering showed that it is concentrated in a tiny volume."}]}]}],"examCountdown":{"examDate":"2027-05-12","paperLabel":"Paper 1"},"heading":"Nuclear atom and alpha scattering","title":"Nuclear atom and alpha scattering Flashcards","topbar":{"backHref":"/as-level/physics/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"8e68888c-0163-515a-8603-43a923838935","walled":"$L39"}],"$L3a","$L3b"]