JEE Corner — Bohr Model Derivations, Spectral Lines, Photoelectric Graphs and Orbital Fine Print

Beyond the Board Syllabus: The Competitive Toolkit

Sections 1 to 10 cover what the Board exam wants. JEE asks for more: the velocity of the electron in the third orbit of $\mathrm{Li^{2+}}$, the number of spectral lines from a sample excited to $n = 5$, the slope of a stopping-potential graph, the orbital angular momentum of a 3d electron, the magnetic moment of $\mathrm{Fe^{3+}}$, why palladium is $4d^{10}\,5s^0$. Textbooks hand you $r_n$ and $E_n$ as finished formulae; this section builds the machinery behind them and then uses it in the worked questions.

Bohr's two postulates as two equations

Take a nucleus of charge $+Ze$ with a single electron of mass $m$ moving in a circle of radius $r$ at speed $v$. Bohr's model rests on two statements, each one line of algebra.

Postulate 1 — the Coulomb force supplies the centripetal force: $$\frac{m v^2}{r} = \frac{k Z e^2}{r^2} \qquad \left(k = \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\ \mathrm{N\ m^2\ C^{-2}}\right) \qquad \ldots (1)$$

Postulate 2 — angular momentum is quantised: $$m v r = \frac{n h}{2\pi}, \qquad n = 1, 2, 3, \ldots \qquad \ldots (2)$$

Everything else is a consequence of these two.

Deriving the radius

Rewrite (1) as $m v^2 r = k Z e^2$ and square (2) to get $m^2 v^2 r^2 = n^2 h^2/4\pi^2$. Divide the second by the first — the $v^2$ cancels: $$m r = \frac{n^2 h^2}{4\pi^2 k Z e^2} \quad \Rightarrow \quad \boxed{r_n = \frac{n^2 h^2}{4\pi^2 m k Z e^2} = 0.529\,\frac{n^2}{Z}\ \text{\AA} = 52.9\,\frac{n^2}{Z}\ \text{pm}}$$

The constant $a_0 = 0.529$ Å is the Bohr radius; every other radius is $a_0$ scaled by $n^2/Z$.

Deriving the velocity

Divide (1) in the form $m v^2 r = k Z e^2$ by (2), $m v r = nh/2\pi$. Now $r$ cancels: $$\boxed{v_n = \frac{2\pi k Z e^2}{n h} = 2.18 \times 10^6\,\frac{Z}{n}\ \mathrm{m\ s^{-1}}}$$

For hydrogen in the ground state that is $2.18 \times 10^6$ m s$^{-1}$, about $c/137$ — fast, but not relativistic, which is why the non-relativistic model works. That ratio $v_1/c = 1/137$ is the fine-structure constant.

Deriving the energies

Kinetic energy from (1): $\mathrm{KE} = \dfrac{1}{2} m v^2 = \dfrac{k Z e^2}{2r}$.

Potential energy of a charge $-e$ at distance $r$ from $+Ze$ (zero at infinity): $\mathrm{PE} = -\dfrac{k Z e^2}{r}$.

Total energy: $$E_n = \mathrm{KE} + \mathrm{PE} = \frac{kZe^2}{2r} - \frac{kZe^2}{r} = -\frac{k Z e^2}{2 r_n}$$

Substitute $r_n$: $$\boxed{E_n = -\frac{2\pi^2 m k^2 Z^2 e^4}{n^2 h^2} = -13.6\,\frac{Z^2}{n^2}\ \text{eV} = -2.18 \times 10^{-18}\,\frac{Z^2}{n^2}\ \text{J} = -1312\,\frac{Z^2}{n^2}\ \mathrm{kJ\ mol^{-1}}}$$

Key Point (The three energies): In any Bohr orbit $$\mathrm{KE} = -E_n = +13.6\,\frac{Z^2}{n^2}\ \text{eV}, \qquad \mathrm{PE} = 2E_n = -27.2\,\frac{Z^2}{n^2}\ \text{eV}, \qquad \mathrm{PE} = -2\,\mathrm{KE}$$ so $\mathrm{KE} : \mathrm{PE} : E = 1 : -2 : -1$. The total energy is negative because the electron is bound, and its magnitude equals the kinetic energy (the virial theorem for an inverse-square force).

[JEE Main] "The kinetic energy of the electron in the second orbit of $\mathrm{He^+}$" is $+13.6 \times 4/4 = 13.6$ eV; "its potential energy" is $-27.2$ eV; "its total energy" is $-13.6$ eV. Three near-identical sentences, three different answers. Read the noun.

Time period, frequency and revolutions per second

The electron goes once round a circumference $2\pi r_n$ at speed $v_n$: $$T_n = \frac{2\pi r_n}{v_n} = \frac{2\pi \times 0.529 \times 10^{-10}\, n^2/Z}{2.18 \times 10^6\, Z/n} = 1.52 \times 10^{-16}\,\frac{n^3}{Z^2}\ \text{s}$$ $$\boxed{T_n \propto \frac{n^3}{Z^2}, \qquad f_n = \frac{1}{T_n} = 6.56 \times 10^{15}\,\frac{Z^2}{n^3}\ \mathrm{s^{-1}}}$$

The frequency of revolution is also the number of revolutions per second: in $10^{-8}$ s (a typical excited-state lifetime) an electron in $n = 2$ of hydrogen manages $6.56 \times 10^{15} \times 10^{-8}/8 = 8.2 \times 10^6$ revolutions. Angular velocity $\omega = 2\pi f \propto Z^2/n^3$ and centripetal acceleration $v^2/r \propto Z^3/n^4$ follow the same way.

The ratio table — memorise the exponents, not the numbers

Quantity	Depends on	Working value (H, $n = 1$)	$\mathrm{He^+}$, $n=2$	$\mathrm{Li^{2+}}$, $n=3$
Radius $r_n$	$n^2/Z$	0.529 Å	1.058 Å	1.587 Å
Velocity $v_n$	$Z/n$	$2.18 \times 10^6$ m s$^{-1}$	$2.18 \times 10^6$	$2.18 \times 10^6$
Total energy $E_n$	$Z^2/n^2$	$-13.6$ eV	$-13.6$ eV	$-13.6$ eV
Kinetic energy	$Z^2/n^2$	$+13.6$ eV	$+13.6$ eV	$+13.6$ eV
Potential energy	$Z^2/n^2$	$-27.2$ eV	$-27.2$ eV	$-27.2$ eV
Time period $T_n$	$n^3/Z^2$	$1.52 \times 10^{-16}$ s	$3.04 \times 10^{-16}$ s	$4.56 \times 10^{-16}$ s
Angular momentum	$n$	$h/2\pi$	$2h/2\pi$	$3h/2\pi$

The $n$-th orbit of a species with $Z = n$ has the same velocity and energy as hydrogen's ground state, but a bigger radius and a longer period. Questions are built on that coincidence.

The gap between levels shrinks fast: for hydrogen $E_1 \to E_2$ is 10.2 eV, $E_2 \to E_3$ is 1.89 eV, $E_3 \to E_4$ is 0.66 eV. So the gap between $n = 2$ and $n = 3$ of $\mathrm{Li^{2+}}$ is $9 \times 1.89 = 17.0$ eV — scale hydrogen's gap by $Z^2$ instead of rederiving.

Spectral Lines: Counting Them, Ranking Them, Scaling Them

Section 5 gave the Rydberg formula and the five series. JEE asks three kinds of question on top: how many lines, which line is longest or shortest, and what changes when hydrogen becomes $\mathrm{He^+}$ or $\mathrm{Li^{2+}}$.

Counting the lines

Excite a sample of hydrogen atoms so that electrons reach level $n$. Different atoms fall by different routes, and every pair of levels below $n$ contributes one line: $$\boxed{N_{\text{lines}} = \frac{n(n-1)}{2}} \qquad (\text{from level } n \text{ down to } n = 1)$$

For a drop that ends at a level $n_1$ higher than 1, count only the levels between: $$\boxed{N_{\text{lines}} = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}}$$

So $n = 4 \to 1$ gives 6 lines, $n = 5 \to 1$ gives 10, and $n = 6 \to 2$ gives $(4)(5)/2 = 10$. For one particular series, count how many upper levels drain into that series' lower level: from $n = 5$, Balmer gets $5 \to 2$, $4 \to 2$, $3 \to 2$ — 3 visible lines — while Lyman gets 4, Paschen 2 and Brackett 1, total 10 again.

Key Point: A single atom dropping from $n$ can emit at most $n - 1$ photons (one per step), and only 1 in the "all at once" case. The $n(n-1)/2$ formula is for a sample of many atoms. Exam keys almost always intend the sample answer even when the wording says "an electron"; use $n(n-1)/2$ unless the question insists on one atom.

First line, last line, and the series limit

Within any series (lower level $n_1$ fixed), the wavenumber $\bar{\nu} = R Z^2 \left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$ grows with $n_2$. So:

• First line ($\alpha$ line, $n_2 = n_1 + 1$): smallest $\bar{\nu}$, longest wavelength, lowest energy.
• Series limit ($n_2 = \infty$): $\bar{\nu}_{\text{limit}} = R Z^2/n_1^2$, shortest wavelength of the series, and its energy is exactly the energy needed to ionise from level $n_1$.

Using $1/R_H = 1/109{,}677$ cm $= 91.18$ nm as the unit:

Series	$n_1$	First line $\lambda$	Series limit $\lambda$	Region
Lyman	1	$\dfrac{4}{3R} = 121.6$ nm	$\dfrac{1}{R} = 91.2$ nm	UV
Balmer	2	$\dfrac{36}{5R} = 656.5$ nm	$\dfrac{4}{R} = 364.7$ nm	Visible
Paschen	3	$\dfrac{144}{7R} = 1876$ nm	$\dfrac{9}{R} = 820.6$ nm	Near IR
Brackett	4	$\dfrac{400}{9R} = 4052$ nm	$\dfrac{16}{R} = 1459$ nm	IR
Pfund	5	$\dfrac{900}{11R} = 7460$ nm	$\dfrac{25}{R} = 2279$ nm	Far IR

The series limits go as $n_1^2/R$, so they are in the ratio $1 : 4 : 9 : 16 : 25$. Each series occupies the window from its limit to its first line: Lyman 91 to 122 nm, Balmer 365 to 656 nm, Paschen 821 to 1876 nm, Brackett 1459 to 4052 nm. Lyman, Balmer and Paschen sit in separate windows, but the Brackett limit (1459 nm) lies inside the Paschen window, so Paschen and Brackett overlap, and so do all the higher series.

Wavelength ratios without a calculator

Since $\lambda \propto 1/\bar{\nu}$, ratios are ratios of the bracket factors: $$\frac{\lambda_{\text{Balmer }\alpha}}{\lambda_{\text{Lyman }\alpha}} = \frac{36/5}{4/3} = \frac{27}{5} = 5.4, \qquad \frac{\lambda_{\text{Lyman }\alpha}}{\lambda_{\text{Lyman limit}}} = \frac{4/3}{1} = \frac{4}{3}$$

Hydrogen-like ions: everything scales by $Z^2$

For a one-electron species of nuclear charge $Z$, $$\bar{\nu} = R Z^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right), \qquad \Delta E = 13.6\, Z^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\ \text{eV}, \qquad \boxed{\lambda \propto \frac{1}{Z^2}}$$

Every hydrogen line reappears in $\mathrm{He^+}$ at one-quarter the wavelength and in $\mathrm{Li^{2+}}$ at one-ninth. Inverted, the question becomes: which transition in $\mathrm{He^+}$ has the same wavelength as some transition in H? Match the whole bracket: $$Z^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \text{ is unchanged if every } n \text{ is multiplied by } Z.$$

Hydrogen transition	Same $\lambda$ in $\mathrm{He^+}$ ($\times 2$)	Same $\lambda$ in $\mathrm{Li^{2+}}$ ($\times 3$)
$2 \to 1$ (Lyman $\alpha$, 121.6 nm)	$4 \to 2$	$6 \to 3$
$3 \to 2$ (Balmer $\alpha$, 656.5 nm)	$6 \to 4$	$9 \to 6$
$3 \to 1$	$6 \to 2$	$9 \to 3$
$4 \to 2$	$8 \to 4$	$12 \to 6$

Strictly the match is "almost the same wavelength", because the Rydberg constant changes slightly with nuclear mass ($R_\infty = 109{,}737$ cm$^{-1}$ versus $R_{\mathrm{H}} = 109{,}677$). Exams ignore this unless they hand you the reduced-mass formula.

Ionisation energy, excitation energy, and their potentials

Key Point (Definitions):

• Ionisation energy of a H-like species in state $n$: energy to take the electron from $E_n$ to $E_\infty = 0$: $\mathrm{IE} = 13.6\, Z^2/n^2$ eV. From the ground state, $\mathrm{IE} = 13.6\, Z^2$ eV: H 13.6, $\mathrm{He^+}$ 54.4, $\mathrm{Li^{2+}}$ 122.4, $\mathrm{Be^{3+}}$ 217.6 eV (or $1312\, Z^2$ kJ mol$^{-1}$).
• Excitation energy to level $n$: $E_n - E_1 = 13.6\, Z^2\left(1 - \dfrac{1}{n^2}\right)$ eV. For H: first excitation energy (to $n = 2$) $= 10.2$ eV, second (to $n = 3$) $= 12.09$ eV, third (to $n = 4$) $= 12.75$ eV, to $n = 5$: 13.06 eV.
• Excitation potential and ionisation potential are the same numbers in volts: the potential through which an electron must be accelerated to carry that energy. First excitation potential of H $= 10.2$ V; ionisation potential $= 13.6$ V.
• Binding energy (or separation energy) of an electron in level $n$ is just $\lvert E_n \rvert$: 3.4 eV for H in $n = 2$.

The hydrogen ladder you should be able to write blind: $E_1 = -13.6$, $E_2 = -3.40$, $E_3 = -1.51$, $E_4 = -0.85$, $E_5 = -0.54$, $E_6 = -0.38$ eV.

[JEE Main] A photon must match a gap exactly or it is not absorbed at all: a 12.5 eV photon does nothing to a hydrogen atom. An electron of 12.5 eV can hand over any part of its energy, so it excites the atom to the highest level whose excitation energy is at most 12.5 eV — $n = 3$ (12.09 eV), giving 3 lines, with the electron keeping the 0.41 eV change.

De Broglie Inside the Orbit, and the Wavelength Formulae JEE Uses

Why angular momentum is quantised at all

Bohr's second postulate looks arbitrary until de Broglie explains it. Treat the orbiting electron as a wave of wavelength $\lambda = h/mv$. To survive going round the orbit the wave must join up with itself, so the circumference must hold a whole number of wavelengths; otherwise successive turns interfere destructively and the wave cancels: $$2\pi r = n\lambda = n\,\frac{h}{mv} \quad \Rightarrow \quad mvr = \frac{nh}{2\pi}$$

That is Bohr's condition, derived. It also gives a favourite one-liner: the number of de Broglie waves in the $n$-th orbit is $n$.

The de Broglie wavelength of the orbiting electron

Put $r_n$ in: $$\lambda_n = \frac{2\pi r_n}{n} = \frac{2\pi \times 0.529\, n^2/Z}{n}\ \text{\AA} \quad \Rightarrow \quad \boxed{\lambda_n = 3.32\,\frac{n}{Z}\ \text{\AA}}$$

Check via momentum: $\lambda_1 = h/(m v_1) = 6.626 \times 10^{-34}/(9.1 \times 10^{-31} \times 2.18 \times 10^6) = 3.34 \times 10^{-10}$ m, the same. So hydrogen's ground-state electron has a wavelength equal to its orbit's circumference, and in $n = 4$ of $\mathrm{He^+}$ the wavelength is $3.32 \times 4/2 = 6.64$ Å with four waves round the loop.

Wavelength of a charged particle accelerated through a potential

A particle of charge $q$ and mass $m$ accelerated from rest through $V$ volts gains $\mathrm{KE} = qV$, so $p = \sqrt{2mqV}$ and $$\boxed{\lambda = \frac{h}{\sqrt{2 m q V}}}$$

Particle	$m$ (kg)	$q$	$\lambda$
Electron	$9.109 \times 10^{-31}$	$e$	$\dfrac{12.27}{\sqrt{V}}$ Å
Proton	$1.673 \times 10^{-27}$	$e$	$\dfrac{0.286}{\sqrt{V}}$ Å
$\alpha$-particle	$6.64 \times 10^{-27}$	$2e$	$\dfrac{0.101}{\sqrt{V}}$ Å

The electron line is worth memorising: 100 V gives 1.227 Å, 150 V gives 1.00 Å, $10^4$ V gives 0.123 Å. Beyond about 50 kV the electron becomes relativistic and the formula drifts; exam problems stay below that.

Wavelength from kinetic energy, and thermal neutrons

Since $\mathrm{KE} = p^2/2m$, $$\boxed{\lambda = \frac{h}{\sqrt{2 m\,\mathrm{KE}}}}$$

For a particle in thermal equilibrium at temperature $T$, the average kinetic energy is $\dfrac{3}{2}kT$ with $k = 1.38 \times 10^{-23}$ J K$^{-1}$, so $$\lambda = \frac{h}{\sqrt{3 m k T}}$$

A thermal neutron at 300 K ($m = 1.675 \times 10^{-27}$ kg): $3mkT = 3 \times 1.675 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300 = 2.08 \times 10^{-47}$, root $= 4.56 \times 10^{-24}$, so $\lambda = 1.45$ Å — an interatomic spacing, which is why neutron diffraction works. Since $\lambda \propto 1/\sqrt{T}$, doubling the temperature shrinks the wavelength by $\sqrt{2}$.

The comparison table JEE actually tests

Two particles with the same …	Then $\lambda$ goes as	Electron vs proton ($m_p/m_e = 1836$)
momentum $p$	identical, $\lambda = h/p$	$\lambda_e = \lambda_p$
velocity $v$	$\lambda \propto 1/m$	$\lambda_e/\lambda_p = 1836$
kinetic energy	$\lambda \propto 1/\sqrt{m}$	$\lambda_e/\lambda_p = \sqrt{1836} = 42.8$
accelerating potential (same $q$)	$\lambda \propto 1/\sqrt{m}$	42.8 again

Against a photon of the same energy: photon $\lambda = hc/E$, particle $\lambda = h/\sqrt{2mE}$, so their ratio is $\dfrac{1}{c}\sqrt{\dfrac{E}{2m}}$ — about $10^{-3}$ for a 1 eV electron, which is why electron microscopes beat optical ones.

[JEE Main] Two formats that look different but are the same fact: "an electron moving with velocity equal to $\alpha$ times the velocity in Bohr's first orbit" and "an electron accelerated through $V$ volts". Convert both to momentum, then $\lambda = h/p$.

The Photoelectric Effect as JEE Draws It

Section 4 gave Einstein's equation. JEE gives graphs — and every graph is one rearrangement of the same line.

$$\mathrm{KE}_{\max} = h\nu - W_0 = h\nu - h\nu_0 = hc\left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right)$$

Stopping potential

Apply a reverse voltage to the collector; the photocurrent falls to zero at the stopping potential $V_s$, when even the fastest electron is turned back: $$\boxed{e V_s = \mathrm{KE}_{\max} = h\nu - W_0 \quad \Rightarrow \quad V_s = \left(\frac{h}{e}\right)\nu - \frac{W_0}{e}}$$

If $\mathrm{KE}_{\max}$ is 1.2 eV, $V_s$ is 1.2 V: same number, different unit. $V_s$ depends on frequency only, never on intensity.

The four graphs

Plot ($y$ vs $x$)	Shape	Slope	Intercepts	What varies it
$\mathrm{KE}_{\max}$ vs $\nu$	straight line, starts at $\nu_0$	$h$ (same for every metal)	$x$: $\nu_0$; $y$ (extrapolated): $-W_0$	different metals give parallel lines shifted along $\nu$
$V_s$ vs $\nu$	straight line	$h/e = 4.14 \times 10^{-15}$ V s	$x$: $\nu_0$; $y$: $-W_0/e$	parallel lines for different metals; unchanged by intensity
Photocurrent vs intensity	straight line through origin (for $\nu > \nu_0$)	$\propto$ number of photons per second	passes through origin	zero for any intensity if $\nu < \nu_0$
Photocurrent vs collector voltage	rises, then flat at the saturation current	—	cuts the negative-$V$ axis at $-V_s$	higher intensity: higher plateau, same $V_s$; higher frequency (same intensity): same plateau, larger $\lvert V_s \rvert$

Three consequences the examiners lean on:

1. Doubling the intensity doubles the photons per second, so it doubles the photocurrent and the number of photoelectrons, and does nothing to $\mathrm{KE}_{\max}$ or $V_s$.
2. Doubling the frequency does not double $\mathrm{KE}_{\max}$: it goes from $h\nu - W_0$ to $2h\nu - W_0$, so it more than doubles. Halving the wavelength does the same.
3. Two metals in a $V_s$ vs $\nu$ plot give parallel lines: same slope $h/e$, different threshold frequencies. Non-parallel lines mean the graph is wrong.

[JEE Main] For the current-versus-voltage graph the standard convention is that the plateau depends on intensity and the cut-off voltage on frequency. Strictly, at fixed intensity in W m$^{-2}$ a higher frequency means fewer photons and hence a slightly lower plateau — but exam keys follow the convention in the table.

The 1240 shortcut

$$hc = 6.626 \times 10^{-34} \times 3.0 \times 10^8 = 1.988 \times 10^{-25}\ \text{J m} = 1240\ \text{eV nm}$$ $$\boxed{E_{\text{photon}}(\text{eV}) = \frac{1240}{\lambda(\text{nm})}, \qquad W_0(\text{eV}) = \frac{1240}{\lambda_0(\text{nm})}}$$

A 400 nm photon carries 3.10 eV, a 620 nm photon 2.00 eV, a 200 nm photon 6.20 eV. Sodium ($W_0 = 2.3$ eV) has $\lambda_0 = 1240/2.3 = 539$ nm; caesium ($W_0 = 1.9$ eV) has $\lambda_0 = 653$ nm, which is why caesium is the photocell metal. With $h = 6.63 \times 10^{-34}$ and $c = 3 \times 10^8$ you land on 1242 instead; either is fine to two figures.

Counting photons and electrons

A source of power $P$ (in W) at wavelength $\lambda$ emits $$n_{\text{photons/s}} = \frac{P}{h\nu} = \frac{P\lambda}{hc}$$ photons per second. If a fraction $\eta$ of the photons hitting the metal eject an electron (the quantum efficiency, usually given as a percentage), the number of photoelectrons per second is $\eta\, n_{\text{photons/s}}$ and the saturation current is $$I_{\text{sat}} = \eta\,\frac{P\lambda}{hc}\, e$$

The $\lambda$ sits in the numerator: at the same power, a longer wavelength means more photons per second, each carrying less energy. "Number of photoelectrons is proportional to intensity" is true at a fixed wavelength.

Finding $h$ from two data points

Two frequencies, two stopping potentials — the Millikan experiment (1916): $$e V_{s1} = h\nu_1 - W_0, \qquad e V_{s2} = h\nu_2 - W_0$$ Subtract: $$\boxed{h = \frac{e\,(V_{s2} - V_{s1})}{\nu_2 - \nu_1}}, \qquad W_0 = h\nu_1 - eV_{s1}, \qquad \nu_0 = \frac{W_0}{h}$$

With wavelengths instead of frequencies, $\nu = c/\lambda$, so the denominator becomes $c\left(\dfrac{1}{\lambda_2} - \dfrac{1}{\lambda_1}\right)$. The result should come out near $6.6 \times 10^{-34}$ J s — if it does not, you have mixed J with eV somewhere.

Maximum velocity of the photoelectrons

$$v_{\max} = \sqrt{\frac{2\,\mathrm{KE}_{\max}}{m_e}} = \sqrt{\frac{2 e V_s}{m_e}} = 5.93 \times 10^5\sqrt{V_s}\ \mathrm{m\ s^{-1}}$$

A stopping potential of 1 V means the fastest electrons leave at $5.93 \times 10^5$ m s$^{-1}$; 4 V doubles that. The $5.93 \times 10^5$ is the same $\sqrt{2e/m_e}$ that sits inside $12.27/\sqrt{V}$.

Four clauses settle every statement-type question on this topic: KE and $V_s$ answer to frequency; current and electron count answer to intensity; nothing happens below $\nu_0$ however bright the light; and there is no time lag ($< 10^{-9}$ s) however dim it is.

The Uncertainty Principle in Every Form the Paper Uses

Section 7 stated Heisenberg's principle once. JEE writes it four ways and expects you to switch between them freely.

Key Point (The forms): $$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}, \qquad \Delta x \cdot \Delta v \geq \frac{h}{4\pi m}, \qquad \Delta E \cdot \Delta t \geq \frac{h}{4\pi}, \qquad \Delta x \cdot \Delta p \geq \frac{\hbar}{2}\ \left(\hbar = \frac{h}{2\pi}\right)$$ Numerically $h/4\pi = 5.27 \times 10^{-35}$ J s. The first form is the principle; the second is the first with $p = mv$ and the mass pulled out; the third pairs energy with time; the fourth is the first in the physicist's notation.

Some older problems use $\Delta x \cdot \Delta p \geq h/2\pi$ or even $\geq h$ — order-of-magnitude versions. Use $h/4\pi$ unless the question states otherwise; if the options are a factor of 2 apart, the question's stated constant decides.

Working it

For an electron with $\Delta x$ known, $$\Delta v = \frac{h}{4\pi m\,\Delta x} = \frac{5.27 \times 10^{-35}}{9.1 \times 10^{-31}\,\Delta x} = \frac{5.79 \times 10^{-5}}{\Delta x}\ \mathrm{m\ s^{-1}}$$

Situation	$\Delta x$	Minimum $\Delta v$	Meaning
Electron confined to an atom	$10^{-10}$ m	$5.8 \times 10^5$ m s$^{-1}$	comparable to the orbital speed itself ($2.2 \times 10^6$): a definite orbit is meaningless
Electron confined to a nucleus	$10^{-14}$ m	$5.8 \times 10^9$ m s$^{-1}$	about $20c$ — impossible
1 mg dust grain, $\Delta x = 10^{-6}$ m	$10^{-6}$ m	$5.3 \times 10^{-23}$ m s$^{-1}$	unobservable: classical mechanics is safe

Why the electron cannot live in the nucleus

A nucleus has diameter about $10^{-14}$ m. If the electron were inside it, $\Delta x \leq 10^{-14}$ m, so $$\Delta v \geq \frac{6.626 \times 10^{-34}}{4\pi \times 9.1 \times 10^{-31} \times 10^{-14}} = 5.8 \times 10^{9}\ \mathrm{m\ s^{-1}}$$ about twenty times the speed of light. (With $10^{-15}$ m for the nuclear radius the answer is $5.8 \times 10^{10}$ m s$^{-1}$, worse.) The corresponding kinetic energy, done properly, is tens of MeV, while nuclear binding energies for an electron would be at most a few MeV — the nucleus cannot hold it. So electrons sit outside the nucleus, and $\beta$-particles must be created at the moment of decay rather than pre-existing inside.

The special cases JEE likes

• Equal uncertainties in position and momentum, $\Delta x = \Delta p$: then $(\Delta x)^2 = h/4\pi$, $\Delta x = \sqrt{h/4\pi} = 7.26 \times 10^{-18}$ (SI). Follow-up: the uncertainty in velocity is $\Delta p/m = 7.26 \times 10^{-18}/9.1 \times 10^{-31} = 8.0 \times 10^{12}$ m s$^{-1}$.
• Uncertainty in position equal to the de Broglie wavelength, $\Delta x = \lambda = h/mv$: then $\Delta v \geq \dfrac{h}{4\pi m} \cdot \dfrac{mv}{h} = \dfrac{v}{4\pi}$, about 8% of the velocity itself.
• Percentage uncertainty in velocity given: "an electron moves at 300 m s$^{-1}$ known to 0.001%" means $\Delta v = 3 \times 10^{-3}$ m s$^{-1}$, so $\Delta x = 5.79 \times 10^{-5}/3 \times 10^{-3} = 1.9 \times 10^{-2}$ m — nearly 2 cm.
• Same $\Delta x$ for an electron and a proton: $\Delta v_e/\Delta v_p = m_p/m_e = 1836$.

Energy and time

$\Delta E \cdot \Delta t \geq h/4\pi$ is why spectral lines have a natural width. An excited state that lives $\Delta t = 10^{-8}$ s has an energy uncertain by $$\Delta E = \frac{5.27 \times 10^{-35}}{10^{-8}} = 5.3 \times 10^{-27}\ \text{J} = 3.3 \times 10^{-8}\ \text{eV}$$ tiny, but measurable as a line width. The ground state, which lives forever, has a perfectly sharp energy.

[JEE Main] The uncertainty principle does not say Bohr was wrong about the energies — the levels are exactly right for hydrogen. It says the orbit, a simultaneous exact $r$ and $v$, is not a meaningful object. That is the limitation of the Bohr model the principle exposes.

Orbital Fine Print: Radial Probability, Nodes, Angular Momentum, Magnetic Moment

Probability density versus radial probability

$\psi^2$ at a point is the probability per unit volume of finding the electron there. For a 1s orbital $\psi^2$ is largest at the nucleus and falls off exponentially. Asking instead at what distance the electron is most likely to be found needs the probability of lying anywhere in a thin shell between $r$ and $r + dr$, and that shell has volume $4\pi r^2\,dr$: $$\boxed{P(r) = 4\pi r^2 \psi^2} \qquad (\text{radial probability distribution function, sometimes written } r^2 R^2)$$

At $r = 0$ the shell has zero volume, so $P(0) = 0$ for every orbital, including 1s. For 1s, $P(r)$ rises, peaks and falls; the peak — the most probable distance — is at exactly $a_0 = 52.9$ pm for hydrogen and at $a_0/Z$ for a H-like ion, Bohr's first-orbit radius reappearing in the quantum model. The average distance is a little larger, $1.5\,a_0$.

For the node-free orbitals of each shell (1s, 2p, 3d, 4f — those with $l = n - 1$) the single maximum sits at $n^2 a_0/Z$, Bohr's radius again. Orbitals with radial nodes (2s, 3s, 3p, …) have more than one hump; the outermost is always the tallest, and for 2s it lies at about $5.2\,a_0$, beyond the 2p maximum at $4\,a_0$. The small inner hump of 2s, which 2p lacks, is why 2s is more penetrating and lower in energy than 2p in multi-electron atoms.

Node counting — the recipe

Key Point: For an orbital with quantum numbers $n$ and $l$:

• Radial (spherical) nodes $= n - l - 1$ — spheres where $R(r) = 0$
• Angular nodes $= l$ — planes (or cones) where the angular part is zero
• Total nodes $= n - 1$

An orbital with 2 radial nodes and 1 angular node is a 4p; one with 3 angular nodes and no radial node is a 4f; the first orbital of each type (1s, 2p, 3d, 4f) has no radial node.

Orbital	$n$	$l$	Radial	Angular	Total
1s	1	0	0	0	0
2s	2	0	1	0	1
2p	2	1	0	1	1
3s	3	0	2	0	2
3p	3	1	1	1	2
3d	3	2	0	2	2
4s	4	0	3	0	3
4p	4	1	2	1	3
4d	4	2	1	2	3
4f	4	3	0	3	3

The angular nodes of the p and d orbitals are specific surfaces, and JEE asks for them by name.

Orbital	Nodal surfaces	Count
$p_x$	$yz$-plane	1
$p_y$	$xz$-plane	1
$p_z$	$xy$-plane	1
$d_{xy}$	$xz$- and $yz$-planes	2
$d_{yz}$	$xy$- and $xz$-planes	2
$d_{xz}$	$xy$- and $yz$-planes	2
$d_{x^2 - y^2}$	the two vertical planes $x = y$ and $x = -y$ (at $45^\circ$ to the axes)	2
$d_{z^2}$	no nodal plane; two nodal cones about the $z$-axis	2 nodes, 0 planes

The rule "nodal planes $= l$" is exactly right for every d orbital except $d_{z^2}$, where the two angular nodes are cones. The d orbital with no nodal plane is $d_{z^2}$; the one whose nodal planes do not contain the axes is $d_{x^2-y^2}$.

Orbital and spin angular momentum

Bohr said $L = nh/2\pi$. Quantum mechanics says $$\boxed{L = \sqrt{l(l+1)}\,\frac{h}{2\pi}} \qquad \text{s: } 0, \quad \text{p: } \sqrt{2}\,\frac{h}{2\pi}, \quad \text{d: } \sqrt{6}\,\frac{h}{2\pi}, \quad \text{f: } \sqrt{12}\,\frac{h}{2\pi} = 2\sqrt{3}\,\frac{h}{2\pi}$$

So a 3d electron carries $\sqrt{6}\, h/2\pi$, not $3h/2\pi$; hydrogen's ground-state (1s) electron carries zero, not $h/2\pi$; and the value depends on $l$ only, so 2p, 3p and 4p electrons all carry $\sqrt{2}\,h/2\pi$. The component along a chosen axis is $m_l\, h/2\pi$.

Spin angular momentum follows the same pattern with $s = \tfrac{1}{2}$: $$S = \sqrt{s(s+1)}\,\frac{h}{2\pi} = \sqrt{\tfrac{1}{2} \cdot \tfrac{3}{2}}\,\frac{h}{2\pi} = \frac{\sqrt{3}}{2}\,\frac{h}{2\pi}$$ with $z$-component $m_s\, h/2\pi = \pm\tfrac{1}{2}\, h/2\pi$. In numbers, $h/2\pi = 1.055 \times 10^{-34}$ J s.

Spin-only magnetic moment

Each unpaired electron is a tiny magnet. For an atom or ion with $n$ unpaired electrons the spin-only magnetic moment is $$\boxed{\mu = \sqrt{n(n+2)}\ \text{BM}} \qquad (1\ \text{Bohr magneton} = 9.27 \times 10^{-24}\ \mathrm{J\ T^{-1}})$$

Configuration	Unpaired $n$	$\mu$ (BM)	Example ions
$d^1$	1	$\sqrt{3} = 1.73$	$\mathrm{Ti^{3+}}$, $\mathrm{V^{4+}}$
$d^2$	2	$\sqrt{8} = 2.83$	$\mathrm{V^{3+}}$, $\mathrm{Ti^{2+}}$
$d^3$	3	$\sqrt{15} = 3.87$	$\mathrm{Cr^{3+}}$, $\mathrm{V^{2+}}$
$d^4$	4	$\sqrt{24} = 4.90$	$\mathrm{Cr^{2+}}$, $\mathrm{Mn^{3+}}$
$d^5$	5	$\sqrt{35} = 5.92$	$\mathrm{Mn^{2+}}$, $\mathrm{Fe^{3+}}$
$d^6$	4	4.90	$\mathrm{Fe^{2+}}$, $\mathrm{Co^{3+}}$
$d^7$	3	3.87	$\mathrm{Co^{2+}}$
$d^8$	2	2.83	$\mathrm{Ni^{2+}}$
$d^9$	1	1.73	$\mathrm{Cu^{2+}}$
$d^{10}$	0	0	$\mathrm{Cu^{+}}$, $\mathrm{Zn^{2+}}$

The usual question reads the table backwards: "a trivalent ion of a 3d metal has $\mu = 3.87$ BM" means three unpaired electrons, $d^3$, $\mathrm{M^{3+}} = [\mathrm{Ar}]\,3d^3$, so M has $18 + 3 + 3 = 24$ electrons: chromium. Watch the mirror symmetry — $d^4$ and $d^6$ both give 4.90 BM, so a magnetic moment alone cannot distinguish $\mathrm{Cr^{2+}}$ from $\mathrm{Fe^{2+}}$. (The table assumes Hund's-rule free ions; real complexes can differ — Class 12 territory.)

Zeeman and Stark

In a magnetic field each spectral line splits into several closely spaced lines — the Zeeman effect. Orbitals of different $m_l$, degenerate in free space, now have slightly different energies; this is the experimental origin of the magnetic quantum number. The same splitting in an electric field is the Stark effect. Bohr's model, with its single number $n$, explains neither; the quantum model with $l$ and $m_l$ does.

Exceptional Configurations, Exchange Energy, Ions, and Slater's Rules

The exceptions beyond Cr and Cu

Section 10 explained chromium and copper. The $(n+l)$ rule fails the same way — and in a few new ways — further down the d block.

Element	$Z$	Expected by aufbau	Actual configuration	Why
Cr	24	$[\mathrm{Ar}]\,3d^4\,4s^2$	$[\mathrm{Ar}]\,3d^5\,4s^1$	half-filled d
Cu	29	$[\mathrm{Ar}]\,3d^9\,4s^2$	$[\mathrm{Ar}]\,3d^{10}\,4s^1$	full d
Nb	41	$[\mathrm{Kr}]\,4d^3\,5s^2$	$[\mathrm{Kr}]\,4d^4\,5s^1$	4d-5s gap small; more exchange
Mo	42	$[\mathrm{Kr}]\,4d^4\,5s^2$	$[\mathrm{Kr}]\,4d^5\,5s^1$	half-filled d
Ru	44	$[\mathrm{Kr}]\,4d^6\,5s^2$	$[\mathrm{Kr}]\,4d^7\,5s^1$	4d-5s gap small
Rh	45	$[\mathrm{Kr}]\,4d^7\,5s^2$	$[\mathrm{Kr}]\,4d^8\,5s^1$	4d-5s gap small
Pd	46	$[\mathrm{Kr}]\,4d^8\,5s^2$	$[\mathrm{Kr}]\,4d^{10}\,5s^0$	full d — the only $ns^0$ case
Ag	47	$[\mathrm{Kr}]\,4d^9\,5s^2$	$[\mathrm{Kr}]\,4d^{10}\,5s^1$	full d
La	57	$[\mathrm{Xe}]\,4f^1\,6s^2$	$[\mathrm{Xe}]\,5d^1\,6s^2$	5d dips below 4f at $Z = 57$
Gd	64	$[\mathrm{Xe}]\,4f^8\,6s^2$	$[\mathrm{Xe}]\,4f^7\,5d^1\,6s^2$	half-filled f
Pt	78	$[\mathrm{Xe}]\,4f^{14}\,5d^8\,6s^2$	$[\mathrm{Xe}]\,4f^{14}\,5d^9\,6s^1$	5d-6s gap small
Au	79	$[\mathrm{Xe}]\,4f^{14}\,5d^9\,6s^2$	$[\mathrm{Xe}]\,4f^{14}\,5d^{10}\,6s^1$	full d

Three patterns worth fixing in memory. Group 6 is Cr, Mo with $d^5 s^1$, but tungsten is $5d^4\,6s^2$, regular. Group 11 (Cu, Ag, Au) is uniformly $d^{10} s^1$. Palladium is the lone $d^{10} s^0$. In the 4d series five elements (Nb, Mo, Ru, Rh, Ag) end in $5s^1$ and one (Pd) in $5s^0$ — the least regular row of the table, because 4d and 5s energies are nearly equal. Among the lanthanides, La, Ce ($4f^1\,5d^1\,6s^2$), Gd and Lu carry a $5d^1$ electron; the rest do not.

Exchange energy — the counting formula

Electrons of the same spin in the same subshell can swap places, and each possible swap lowers the energy by a fixed amount $K$. The number of exchanges among $p$ electrons of parallel spin is the number of pairs you can choose from them: $$\boxed{\text{Number of exchange pairs} = \frac{p(p-1)}{2} = {}^{p}C_2}$$

3d configuration	Parallel-spin sets	Exchange pairs	Exchange energy
$d^3$	3 up	$3 \times 2/2 = 3$	$3K$
$d^4$	4 up	6	$6K$
$d^5$	5 up	10	$10K$
$d^6$	5 up, 1 down	$10 + 0 = 10$	$10K$
$d^7$	5 up, 2 down	$10 + 1 = 11$	$11K$
$d^8$	5 up, 3 down	$10 + 3 = 13$	$13K$
$d^9$	5 up, 4 down	$10 + 6 = 16$	$16K$
$d^{10}$	5 up, 5 down	$10 + 10 = 20$	$20K$

The chromium argument is now arithmetic. $3d^4\,4s^2$ has 6 exchange pairs within 3d; $3d^5\,4s^1$ has 10. The extra $4K$ (plus the symmetry and lower repulsion of a half-filled set) outweighs the small cost of promoting one electron from 4s to 3d. Copper gains the same $20K - 16K = 4K$ going $d^9 \to d^{10}$. Counting the 4s electron too, if its spin is parallel to the 3d set, adds a further 5 pairs for $3d^5\,4s^1$; either convention is accepted as long as you state it, and the difference argument comes out the same way.

Removing electrons from ions

Key Point: Electrons are removed from the orbital of highest principal quantum number first, then from the highest $l$ within it. For transition metals this means 4s before 3d, even though 4s was filled first: once 3d has electrons in it, the 3d orbitals sit closer to the nucleus and lower in energy than 4s.

Atom	Configuration	Ion	Configuration	Unpaired	$\mu$ (BM)
Fe (26)	$[\mathrm{Ar}]\,3d^6\,4s^2$	$\mathrm{Fe^{2+}}$	$[\mathrm{Ar}]\,3d^6$	4	4.90
		$\mathrm{Fe^{3+}}$	$[\mathrm{Ar}]\,3d^5$	5	5.92
Cu (29)	$[\mathrm{Ar}]\,3d^{10}\,4s^1$	$\mathrm{Cu^{+}}$	$[\mathrm{Ar}]\,3d^{10}$	0	0 (diamagnetic)
		$\mathrm{Cu^{2+}}$	$[\mathrm{Ar}]\,3d^9$	1	1.73
Cr (24)	$[\mathrm{Ar}]\,3d^5\,4s^1$	$\mathrm{Cr^{3+}}$	$[\mathrm{Ar}]\,3d^3$	3	3.87
Ni (28)	$[\mathrm{Ar}]\,3d^8\,4s^2$	$\mathrm{Ni^{2+}}$	$[\mathrm{Ar}]\,3d^8$	2	2.83
Zn (30)	$[\mathrm{Ar}]\,3d^{10}\,4s^2$	$\mathrm{Zn^{2+}}$	$[\mathrm{Ar}]\,3d^{10}$	0	0
Ga (31)	$[\mathrm{Ar}]\,3d^{10}\,4s^2\,4p^1$	$\mathrm{Ga^{+}}$	$[\mathrm{Ar}]\,3d^{10}\,4s^2$	0	0

Two species with the same electron count need not share a configuration: $\mathrm{Fe^{2+}}$ (24 electrons) is $[\mathrm{Ar}]\,3d^6$, while a neutral Cr atom (also 24) is $[\mathrm{Ar}]\,3d^5\,4s^1$. And $\mathrm{Cu^{2+}}$ is paramagnetic while $\mathrm{Cu^{+}}$ is diamagnetic — the standard "which is coloured / paramagnetic" pair.

Slater's rules — a two-minute version (JEE Advanced only)

Section 9 defined $Z_{\text{eff}} = Z - \sigma$. Slater gave a recipe for $\sigma$. Write the configuration in groups: $(1s)(2s\,2p)(3s\,3p)(3d)(4s\,4p)(4d)(4f)(5s\,5p)\ldots$ Then, for the electron of interest:

• For an $ns$ or $np$ electron: every other electron in the same group contributes 0.35 (0.30 if the group is 1s); every electron in the $(n-1)$ shell contributes 0.85; every electron in shell $n - 2$ or lower contributes 1.00.
• For an $nd$ or $nf$ electron: every other electron in the same group contributes 0.35; every electron in any group to the left contributes 1.00.

Electrons to the right of the group contribute nothing.

A 2p electron in nitrogen ($Z = 7$, groups $(1s^2)(2s^2\,2p^3)$). Other electrons in the $(2s\,2p)$ group: 4, at 0.35 each $= 1.40$. The 1s electrons: 2, at 0.85 $= 1.70$. So $\sigma = 3.10$ and $Z_{\text{eff}} = 7 - 3.10 = 3.90$.

4s versus 3d in zinc ($Z = 30$): $(1s^2)(2s^2\,2p^6)(3s^2\,3p^6)(3d^{10})(4s^2)$. For a 4s electron: $1 \times 0.35 + 18 \times 0.85 + 10 \times 1.00 = 25.65$, $Z_{\text{eff}} = 4.35$. For a 3d electron: $9 \times 0.35 + 18 \times 1.00 = 21.15$, $Z_{\text{eff}} = 8.85$. The 3d electron feels twice the pull of the 4s electron — the quantitative reason 4s is ionised first.

[JEE Main] JEE Main needs only the qualitative facts: $\sigma$ rises down a group (more inner shells), $Z_{\text{eff}}$ rises across a period (same-shell electrons shield poorly at 0.35 each), and $s > p > d > f$ in penetration. JEE Advanced has asked for an actual $Z_{\text{eff}}$, so learn the 0.35 / 0.85 / 1.00 triple and the grouping.