NCERT Solutions · Class 11 Chemistry Structure of Atom

Each quantum number narrows down which electrons you're counting — n fixes the shell, l fixes the subshell (shape), and \(\displaystyle m_s \) fixes the spin. Count only the electrons that satisfy every condition given.(a) \(\displaystyle n = 4,\ m_s = -\tfrac{1}{2} \)For a given principal quantum number \(\displaystyle n \), the number of orbitals available is\[\text{number of orbitals} = n^2 \]where \(\displaystyle n^2 \) counts every combination of \(\displaystyle l = 0, 1, 2, \dots, (n-1) \) together with its \(\displaystyle (2l+1) \) values of \(\displaystyle m_l \). For \(\displaystyle n = 4 \):\[\text{number of orbitals} = 4^2 = 16 \]Each orbital holds exactly two electrons — one with spin quantum number \(\displaystyle m_s = +\tfrac{1}{2} \) and one with \(\displaystyle m_s = -\tfrac{1}{2} \) (this is the Pauli exclusion principle: no two electrons in the same orbital can share the same spin). So the total number of electrons for \(\displaystyle n = 4 \) is\[\text{total electrons} = 2n^2 = 2(16) = 32 \]and this total splits evenly between the two spin values. The easy mistake here is to stop at "$\displaystyle 32$ electrons" — but the question asks only for the ones with \(\displaystyle m_s = -\tfrac{1}{2} \), which is exactly half:\[\text{electrons with } m_s = -\tfrac{1}{2} = \frac{2n^2}{2} = n^2 = 4^2 = 16 \](b) \(\displaystyle n = 3,\ l = 0 \)The azimuthal (subshell) quantum number \(\displaystyle l = 0 \) specifies an s subshell — here, the 3s subshell since \(\displaystyle n = 3 \). For a given \(\displaystyle l \), the number of orbitals is\[\text{number of orbitals} = 2l + 1 \]With \(\displaystyle l = 0 \):\[\text{number of orbitals} = 2(0) + 1 = 1 \]This single orbital (3s) can hold a maximum of $\displaystyle 2$ electrons, one for each value of \(\displaystyle m_s \) (\(\displaystyle +\tfrac{1}{2} \) and \(\displaystyle -\tfrac{1}{2} \)):\[\text{maximum electrons} = 2 \times 1 = 2 \]The step people rush past: \(\displaystyle n = 3, l = 0 \) fixes a subshell, not one orbital's worth of electrons by coincidence — it's because \(\displaystyle l = 0 \) always gives just one orbital, regardless of which shell it sits in.Answer: (a) $\displaystyle 16$ electrons have \(\displaystyle n = 4,\ m_s = -\tfrac{1}{2} \). (b) $\displaystyle 2$ electrons can have \(\displaystyle n = 3,\ l = 0 \).

Bohr's quantization rule fixes the angular momentum; de Broglie's relation fixes the wavelength — put them together and the orbit's circumference falls out as a whole number of wavelengths.Step $\displaystyle 1$ — Bohr's postulate. Bohr proposed that the angular momentum of an electron circling the nucleus is quantized: \[mvr = \frac{nh}{2\pi} \] where \(\displaystyle m\) is the mass of the electron, \(\displaystyle v\) is its speed in the orbit, \(\displaystyle r\) is the radius of that orbit, \(\displaystyle n\) is a positive integer (the principal quantum number: \(\displaystyle n = 1, 2, 3, \ldots\)), and \(\displaystyle h\) is Planck's constant.Rearranging for the circumference \(\displaystyle 2\pi r\): \[2\pi r = \frac{nh}{mv} \]Step $\displaystyle 2$ — de Broglie's relation. de Broglie proposed that a particle of momentum \(\displaystyle mv\) has an associated wavelength \[\lambda = \frac{h}{mv} \] The quantity in the denominator, \(\displaystyle mv\), is the electron's linear momentum (mass times speed) — it is easy to mistake this for just the speed \(\displaystyle v\), but the wavelength depends on the full momentum, not on \(\displaystyle v\) alone.Step $\displaystyle 3$ — Combine the two. The right-hand side of the Step $\displaystyle 1$ equation, \(\displaystyle \dfrac{h}{mv}\), is exactly \(\displaystyle \lambda\) from Step 2. Substituting: \[2\pi r = n\left(\frac{h}{mv}\right) = n\lambda \]So the circumference of the Bohr orbit is \[2\pi r = n\lambda, \qquad n = 1, 2, 3, \ldots \]This says the orbit's circumference is exactly \(\displaystyle n\) times the de Broglie wavelength of the electron moving in it — an integral (whole-number) multiple, never a fractional one. Physically this is the condition for the electron's wave to close on itself smoothly around the orbit without cancelling itself out: after going around once, the wave must return to its starting point in phase, which is only possible if the circumference holds a whole number of wavelengths.**Answer: The circumference of the \(\displaystyle n\)th Bohr orbit satisfies \(\displaystyle 2\pi r = n\lambda\), i.e., it is an integral multiple (\(\displaystyle n = 1, 2, 3, \ldots\)) of the de Broglie wavelength \(\displaystyle \lambda = h/(mv)\) of the electron revolving in that orbit — obtained directly by combining Bohr's quantization condition \(\displaystyle mvr = nh/(2\pi)\) with de Broglie's relation \(\displaystyle \lambda = h/(mv)\).

A hydrogen-like ion is not hydrogen — the nuclear charge \(\displaystyle Z\) enters the Rydberg formula as \(\displaystyle Z^2\), so you cannot compare wavelengths across species without it.For any single-electron species (a neutral hydrogen atom, or an ion stripped to one electron, such as \(\displaystyle \text{He}^+ \)), the wavenumber of a spectral line is given by the Rydberg formula\[\bar\nu = \frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]where \(\displaystyle \bar\nu \) is the wavenumber, \(\displaystyle \lambda \) is the wavelength of the emitted photon, \(\displaystyle R_H = 1.097 \times 10^{7}\ \text{m}^{-1} \) is the Rydberg constant, \(\displaystyle Z \) is the atomic number (nuclear charge) of the emitting species, and \(\displaystyle n_1 < n_2 \) are the principal quantum numbers of the lower and upper levels.Step $\displaystyle 1$: Wavenumber of the given \(\displaystyle \mathrm{He^{+}}\) transition.\(\displaystyle \text{He}^+ \) has one electron but a nucleus of charge \(\displaystyle Z = 2 \). Its Balmer-type transition is \(\displaystyle n_2 = 4 \to n_1 = 2 \):\[\frac{1}{\lambda_{\text{He}^+}} = R_H (2)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R_H \times 4 \times \left( \frac{1}{4} - \frac{1}{16} \right) \]Aside: dropping that factor of \(\displaystyle Z^2 = 4 \) — treating \(\displaystyle \text{He}^+ \) as if it were hydrogen — is the single most common slip in this problem.\[\frac{1}{4} - \frac{1}{16} = \frac{4}{16} - \frac{1}{16} = \frac{3}{16} \]\[\frac{1}{\lambda_{\text{He}^+}} = R_H \times 4 \times \frac{3}{16} = R_H \times \frac{3}{4} \]So this \(\displaystyle \mathrm{He^{+}}\) line has wavenumber exactly \(\displaystyle \frac{3}{4} R_H \) — a clean fraction, independent of the numerical value of \(\displaystyle R_H \).Step $\displaystyle 2$: Find the hydrogen transition with the same wavenumber.For hydrogen, \(\displaystyle Z = 1 \), so\[\frac{1}{\lambda_H} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]Equal wavelengths mean equal wavenumbers:\[R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \frac{3}{4} R_H \quad\Longrightarrow\quad \frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{3}{4} \]Try the smallest possible quantum number, \(\displaystyle n_1 = 1 \):\[1 - \frac{1}{n_2^2} = \frac{3}{4} \quad\Longrightarrow\quad \frac{1}{n_2^2} = \frac{1}{4} \quad\Longrightarrow\quad n_2 = 2 \]This lands exactly on whole numbers, so it is the transition: \(\displaystyle n_2 = 2 \to n_1 = 1 \) — the first line of the Lyman series (Lyman-\(\displaystyle \alpha\)).Aside: it isn't a coincidence that this worked out evenly. \(\displaystyle \text{He}^+ \)'s levels scale as \(\displaystyle Z^2/n^2 = 4/n^2 \), which is the same as \(\displaystyle 1/(n/2)^2 \) — so its \(\displaystyle n = 2 \) and \(\displaystyle n = 4 \) levels line up with hydrogen's \(\displaystyle n = 1 \) and \(\displaystyle n = 2 \) levels. The algebra above is what confirms this, not the shortcut alone.Step $\displaystyle 3$: Confirm with an actual wavelength.\[\frac{1}{\lambda} = \frac{3}{4} \times 1.097 \times 10^{7}\ \text{m}^{-1} = 8.2275 \times 10^{6}\ \text{m}^{-1} \]\[\lambda = \frac{1}{8.2275 \times 10^{6}\ \text{m}^{-1}} = 1.215 \times 10^{-7}\ \text{m} = 121.5\ \text{nm} \]This wavelength sits in the Lyman series (UV region), consistent with the \(\displaystyle n = 2 \to n = 1 \) identification above.Answer: The \(\displaystyle n = 2 \to n = 1 \) transition of the hydrogen atom (the Lyman-\(\displaystyle \alpha\) line, \(\displaystyle \lambda \approx 1.215 \times 10^{-7}\ \text{m} = 121.5\ \text{nm} \)) has the same wavelength as the \(\displaystyle n = 4 \to n = 2 \) Balmer transition of \(\displaystyle \text{He}^+ \).

Energy levels scale as \(\displaystyle Z^2\), not linearly with \(\displaystyle Z\), because the nucleus of a hydrogen-like ion pulls on the single electron with a charge of \(\displaystyle Ze\) instead of \(\displaystyle e\).\(\displaystyle \text{He}^+\) has a nucleus of charge \(\displaystyle +2e\) with just one electron sitting around it — exactly like a hydrogen atom, except the pull on that electron is twice as strong. Species like this (one electron, nuclear charge \(\displaystyle Ze\)) are called hydrogen-like, and their orbit energies follow the same formula as hydrogen but scaled by \(\displaystyle Z^2\):\[E_n = -Z^2 \left(2.18 \times 10^{-18}\ \text{J}\right)\frac{1}{n^2} \]Here \(\displaystyle Z\) is the atomic number (nuclear charge in units of \(\displaystyle e\)), \(\displaystyle n\) is the orbit (principal quantum number), and \(\displaystyle 2.18 \times 10^{-18}\ \text{J}\) is the ionization energy of hydrogen — the same constant reappears here because it is the H formula, just carrying the extra \(\displaystyle Z^2\).A step people skip: forgetting the \(\displaystyle Z^2\) and just reusing the H atom's energy directly. For He\(\displaystyle ^+\), \(\displaystyle Z = 2\), so the pull — and the energy — is four times hydrogen's, not twice.The process asked for is \[\text{He}^+(g) \rightarrow \text{He}^{2+}(g) + e^- \] which is exactly the ionization of the one electron still bound in He\(\displaystyle ^+\): it starts in the ground state (\(\displaystyle n = 1\)) and ends completely free (\(\displaystyle n = \infty\), where \(\displaystyle E_\infty = 0\)).Ground-state energy of the electron in He\(\displaystyle ^+\) (\(\displaystyle Z = 2\), \(\displaystyle n = 1\)):\[E_1 = -(2)^2 \left(2.18 \times 10^{-18}\ \text{J}\right)\frac{1}{(1)^2} = -4 \times 2.18 \times 10^{-18}\ \text{J} = -8.72 \times 10^{-18}\ \text{J} \]Energy required is how far you have to climb to reach \(\displaystyle E_\infty = 0\), not the value of \(\displaystyle E_1\) itself — ionization energy is always \(\displaystyle E_\infty - E_n\), so the sign flips:\[\Delta E = E_\infty - E_1 = 0 - \left(-8.72 \times 10^{-18}\ \text{J}\right) = 8.72 \times 10^{-18}\ \text{J} \]This positive value is the energy that must be supplied per He\(\displaystyle ^+\) ion to strip off its last electron.Since the given constant had three significant figures, the result is kept to three:Answer: \(\displaystyle 8.72 \times 10^{-18}\ \text{J}\) per He\(\displaystyle ^+\) ion (about \(\displaystyle 4\) times the ionization energy of the H atom, as expected from the \(\displaystyle Z^2\) factor with \(\displaystyle Z = 2\)).

The number of atoms that fit in a line is just the total length divided by the width of one atom — but only once both are in the same unit.Here the scale length is given in centimetres and the atomic diameter in nanometres, so the first job is to convert one of them.Step $\displaystyle 1$: Convert the diameter to centimetres.Using \(\displaystyle 1\ \text{m} = 100\ \text{cm} \) and \(\displaystyle 1\ \text{nm} = 10^{-9}\ \text{m} \):\[1\ \text{nm} = 10^{-9}\ \text{m} \times \frac{100\ \text{cm}}{1\ \text{m}} = 10^{-7}\ \text{cm} \]So the diameter of one carbon atom is\[d = 0.15\ \text{nm} = 0.15 \times 10^{-7}\ \text{cm} = 1.5 \times 10^{-8}\ \text{cm} \]A jump straight from nm to cm without going through this factor of \(\displaystyle 10^{-7}\) is the step people get wrong — always insert the metre as the common ground between the two units.Step $\displaystyle 2$: Divide the total length by the diameter of one atom.If \(\displaystyle N\) carbon atoms are placed side by side, touching, in a straight line, they span a length \(\displaystyle L = N \times d\). Rearranging for \(\displaystyle N\):\[N = \frac{L}{d} \]where \(\displaystyle L\) is the length of the scale and \(\displaystyle d\) is the diameter of one carbon atom.Substituting \(\displaystyle L = 20\ \text{cm}\) and \(\displaystyle d = 1.5\times10^{-8}\ \text{cm}\):\[N = \frac{20\ \text{cm}}{1.5 \times 10^{-8}\ \text{cm}} \]\[N = \frac{20}{1.5} \times 10^{8} = 13.33 \times 10^{8} \]Writing this in proper scientific notation (one non-zero digit before the decimal point):\[N = 1.333 \times 10^{9} \]The centimetres cancel top and bottom, leaving \(\displaystyle N\) as a pure (unitless) count of atoms, exactly as it should be for a number of particles.Since the diameter \(\displaystyle 0.15\ \text{nm}\) carries two significant figures, the result should be reported to the same precision: \(\displaystyle 1.3 \times 10^{9}\) atoms; the fuller value \(\displaystyle 1.333 \times 10^{9}\) is kept above only to show the division clearly.Answer: \(\displaystyle N = 1.333 \times 10^{9}\ \text{atoms} \approx 1.3 \times 10^{9}\) carbon atoms

Length of the row is the number of atoms multiplied by the diameter of one atom — not its radius. When atoms are lined up "side by side" like beads on a string, each atom contributes its full diameter to the length, because neighbouring atoms touch at a single point along the line.Step $\displaystyle 1$: Set up the relation.\[\text{Total length} = (\text{number of atoms}) \times (\text{diameter of one atom}) \]Here the number of atoms is \(\displaystyle 2 \times 10^{8}\) and the total length is \(\displaystyle 2.4\ \text{cm}\).Step $\displaystyle 2$: Solve for the diameter.\[\text{Diameter} = \frac{\text{Total length}}{\text{Number of atoms}} = \frac{2.4\ \text{cm}}{2 \times 10^{8}} \]\[\text{Diameter} = 1.2 \times 10^{-8}\ \text{cm} \]Step $\displaystyle 3$: Get the radius.The radius is half the diameter — this is the step people skip, reporting the diameter as if it were the radius.\[\text{Radius} = \frac{\text{Diameter}}{2} = \frac{1.2 \times 10^{-8}\ \text{cm}}{2} = 6.0 \times 10^{-9}\ \text{cm} \]Step $\displaystyle 4$: Express this in the usual atomic-radius unit, picometres (pm).Since \(\displaystyle 1\ \text{cm} = 10^{10}\ \text{pm}\):\[\text{Radius} = 6.0 \times 10^{-9}\ \text{cm} \times 10^{10}\ \frac{\text{pm}}{\text{cm}} = 60\ \text{pm} \]Carrying it in SI units instead: \(\displaystyle 6.0 \times 10^{-9}\ \text{cm} = 6.0 \times 10^{-11}\ \text{m}\), the same value written as metres.The data (\(\displaystyle 2 \times 10^{8}\) and \(\displaystyle 2.4\) cm) each carry two significant figures, so the final answer is rounded to two significant figures — not before.Answer: Radius of the carbon atom \(\displaystyle = 6.0 \times 10^{-9}\ \text{cm} = 6.0 \times 10^{-11}\ \text{m} = 60\ \text{pm}\).

The zinc atoms are packed edge to edge, so each atom's diameter is the "length" you divide the total length by — not the radius.(a) Radius of the zinc atom in pmDiameter is a length, and $\displaystyle 1$ Å (angstrom) equals \(\displaystyle 10^{-10}\,\text{m}\), while $\displaystyle 1$ pm (picometre) equals \(\displaystyle 10^{-12}\,\text{m}\). So\[1\ \text{Å} = 10^{-10}\,\text{m} = \frac{10^{-10}}{10^{-12}}\,\text{pm} = 10^{2}\,\text{pm} = 100\ \text{pm} \]Given diameter \(\displaystyle d = 2.6\ \text{Å}\):\[d = 2.6\ \text{Å} \times \frac{100\ \text{pm}}{1\ \text{Å}} = 260\ \text{pm} \]Radius is half the diameter:\[r = \frac{d}{2} = \frac{260\ \text{pm}}{2} = 130\ \text{pm} \](b) Number of atoms in $\displaystyle 1.6$ cm laid side by sideIf atoms sit lengthwise touching each other, each atom occupies exactly one diameter's worth of length, so\[\text{number of atoms} = \frac{\text{total length}}{\text{diameter of one atom}} \]Both lengths must be in the same unit before dividing — this is the step that trips people up. Convert the diameter from Å to cm:\[1\ \text{Å} = 10^{-10}\,\text{m} = 10^{-10}\,\text{m} \times \frac{100\ \text{cm}}{1\ \text{m}} = 10^{-8}\ \text{cm} \]\[d = 2.6\ \text{Å} = 2.6 \times 10^{-8}\ \text{cm} \]Now divide the given length \(\displaystyle L = 1.6\ \text{cm}\) by this diameter:\[\text{number of atoms} = \frac{1.6\ \text{cm}}{2.6 \times 10^{-8}\ \text{cm}} = \frac{1.6}{2.6} \times 10^{8} \]\[\frac{1.6}{2.6} = 0.6154\ldots \]\[\text{number of atoms} = 0.6154 \times 10^{8} = 6.154 \times 10^{7} \]The input data ($\displaystyle 2.6$ Å, $\displaystyle 1.6$ cm) carry three significant figures, so round to three:\[\text{number of atoms} \approx 6.15 \times 10^{7}\ \text{atoms} \]Answer: (a) radius = $\displaystyle 130$ pm; (b) number of atoms = \(\displaystyle 6.15 \times 10^{7}\) atoms.

Charge is quantized — any charge you can actually measure is a whole-number multiple of the charge on one electron, \(\displaystyle e\). So to find how many electrons are sitting on this particle, divide the total charge by the charge of a single electron.The relation is\[n = \frac{Q}{e} \]where The step people rush past: don't round \(\displaystyle e\) to "$\displaystyle 1.6$" before dividing — carry the full \(\displaystyle 1.6022 \times 10^{-19}\) through the division, and only round the final count.Substituting:\[n = \frac{2.5 \times 10^{-16}\ \text{C}}{1.6022 \times 10^{-19}\ \text{C}} \]Handle the powers of ten and the decimal part separately:\[n = \frac{2.5}{1.6022} \times 10^{-16-(-19)} = \frac{2.5}{1.6022} \times 10^{3} \]\[\frac{2.5}{1.6022} = 1.5604 \]\[n = 1.5604 \times 10^{3} = 1560.4 \]Since \(\displaystyle n\) counts individual electrons, it has to be a whole number — a particle can't carry $\displaystyle 0.4$ of an electron. The "$\displaystyle 0.4$" here is just rounding noise from taking \(\displaystyle e\) to four significant figures, so round \(\displaystyle n\) to the nearest whole electron.\[n \approx 1560 \]Answer: The particle carries about $\displaystyle 1560$ electrons (\(\displaystyle n = \dfrac{2.5 \times 10^{-16}\ \text{C}}{1.6022 \times 10^{-19}\ \text{C}} \approx 1560\)).

The charge on the drop is always a whole-number multiple of the charge on one electron — this is Millikan's key finding, and it's exactly what lets you count electrons from a charge.Step $\displaystyle 1$ — Name the relationship.The total charge on the oil drop is\[q = n \times e \]where \(\displaystyle q\) is the charge on the drop, \(\displaystyle e\) is the charge on a single electron \(\displaystyle \left(1.602 \times 10^{-19}\ \text{C}\right)\), and \(\displaystyle n\) is the number of electrons on the drop (a whole number — you cannot have a fraction of an electron).Step $\displaystyle 2$ — Solve for \(\displaystyle n\).Rearranging,\[n = \frac{q}{e} \]The sign of \(\displaystyle q\) just tells you the charge is negative (electrons were deposited on the drop); for counting electrons you only need the magnitude, so use \(\displaystyle q = 1.282 \times 10^{-18}\ \text{C}\).\[n = \frac{1.282 \times 10^{-18}\ \text{C}}{1.602 \times 10^{-19}\ \text{C}} \]Step $\displaystyle 3$ — Carry out the division carefully with the powers of ten.Write both numbers with the same power of ten so the division is clean:\[n = \frac{12.82 \times 10^{-19}\ \text{C}}{1.602 \times 10^{-19}\ \text{C}} = \frac{12.82}{1.602} \]\[n = 8.0025 \]The aside people miss here: \(\displaystyle n\) must come out to (very nearly) a whole number, because charge is quantized — it only exists in integer multiples of \(\displaystyle e\). The \(\displaystyle 0.0025\) beyond $\displaystyle 8$ is just rounding in the given data (the drop's charge and \(\displaystyle e\) are both quoted to $\displaystyle 3$–$\displaystyle 4$ significant figures), not a sign of a fractional electron. So you round to the nearest integer, not to the nearest tenth.\[n = 8 \]Answer: The oil drop carries $\displaystyle 8$ electrons (\(\displaystyle n = 8\)).

The scattering angle in Rutherford's experiment is controlled by two things: how strongly the nucleus repels the alpha particle (its charge) and how much the nucleus itself recoils during the collision (its mass) — both of these fall sharply when you go from a heavy nucleus like gold to a light one like aluminium.In the original experiment the foil is gold ( \(\displaystyle Z = 79\), atomic mass \(\displaystyle \approx 197\ \text{u}\) ). An incoming alpha particle carries charge \(\displaystyle +2e\) and mass \(\displaystyle \approx 4\ \text{u}\). Two features of gold made the classic result possible, and both change if aluminium ( \(\displaystyle Z = 13\), atomic mass \(\displaystyle \approx 27\ \text{u}\) ) is used instead.1. Weaker electrostatic repulsion. The repulsive Coulomb force between the nucleus and the alpha particle at any given separation is proportional to the product of their charges, \[F \propto \frac{Z e \cdot 2e}{r^{2}} \] where \(\displaystyle Z\) is the nuclear charge (atomic number) and \(\displaystyle r\) is the distance from the nucleus. Since aluminium's \(\displaystyle Z = 13\) is nearly six times smaller than gold's \(\displaystyle Z = 79\), an alpha particle passing at the same distance \(\displaystyle r\) feels a much weaker push. To be turned through a large angle it would now have to approach far closer to the aluminium nucleus than it ever needed to with gold — so, statistically, far fewer alpha particles get deflected through large angles.2. The lighter nucleus recoils instead of standing still. This is the step that is easy to miss: Rutherford's large-angle (and $\displaystyle 180$°, "rebound") scattering depended on the gold nucleus being so much heavier than the alpha particle that it acted, for the purposes of the collision, like an immovable wall. In a two-body elastic collision between a projectile of mass \(\displaystyle m\) and a stationary target of mass \(\displaystyle M\), the target's recoil — and hence the maximum angle the projectile itself can be thrown through — depends on the mass ratio \(\displaystyle m/M\). Putting the two effects together. With an aluminium foil, the vast majority of alpha particles would still pass through with little or no deflection, exactly as with gold, because the atom is still mostly empty space — but among the particles that do come close to a nucleus, the fraction scattered through large angles would drop sharply, and no particles at all would rebound straight back at angles near \(\displaystyle 180^\circ\), because both the weaker charge and the smaller mass ratio work against it. The experiment would therefore give much weaker evidence for a small, dense, heavily charged nucleus than the gold-foil result did — which is exactly why Rutherford chose a heavy element like gold in the first place.Answer: with a light-atom (aluminium) foil, far fewer alpha particles would be deflected through large angles, and none would be scattered back near \(\displaystyle 180^\circ\), because aluminium's smaller nuclear charge gives weaker Coulomb repulsion and its mass is not large enough compared to the alpha particle's mass to stay essentially fixed during the collision.