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Structure of Atom: one-page cheat sheet

NEET Chemistry · Unit 2 · Physical Chemistry

ScoreChemby CSJ Sir
Weightage: 1–2 questions · 4–8 marks

1 · Particles & rays

Cathode rayselectrons;

e / m = 1.76 \times 1 0^{11} C kg^{- 1}

— same for every gas and electrode

Canal rayspositive ions;

e / m

depends on the gas (highest for H

^{+}

= proton)

Constants

e = 1.602 \times 1 0^{- 19} C

(Millikan);

m_{e} = 9.109 \times 1 0^{- 31} kg

NeutronChadwick 1932:

_{4}^{9} Be +_{2}^{4} He \to_{6}^{12} C +_{0}^{1} n

2 · Radiation & Planck

c = ν λ, ν ˉ = \frac{1}{λ} = \frac{ν}{c}

E = h ν = \frac{h c}{λ} = h c ν ˉ, h = 6.626 \times 1 0^{- 34} J s

Order of ERadio < Microwave < IR < Visible < UV < X-ray <

γ

n photons

E = nh ν

; one mole of photons:

N_{A} h c / λ

3 · Photoelectric effect

h ν = h ν_{0} + KE_{m a x} = W_{0} + \frac{1}{2} m_{e} v^{2}, W_{0} = \frac{h c}{λ _{0}}

KE of electronsrises with frequency

ν

(needs

ν \geq ν_{0}

); not with intensity

Number of electronsproportional to intensity (brightness)

More intensity = more electrons per second, same KE and stopping potential.

4 · Bohr model (H, He $^{+}$ , Li $^{2 +}$ , Be $^{3 +}$ only)

m_{e} v r = \frac{nh}{2 π}

r_{n} = \frac{52.9 n ^{2}}{Z} pm, E_{n} = - 13.6 \frac{Z ^{2}}{n ^{2}} eV = - 2.18 \times 1 0^{- 18} \frac{Z ^{2}}{n ^{2}} J

\frac{1}{λ} = R_{H} Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}), R_{H} = 1.097 \times 1 0^{7} m^{- 1}

Series	$n_{1}$	Region
Lyman	1	UV
Balmer	2	Visible ( $n_{2}$ = 3 to 6)
Paschen	3	Near IR
Brackett	4	IR
Pfund	5	Far IR

Total lines

\frac{n ( n - 1 )}{2}

from level

n

Lines

\frac{( n _{2} - n _{1} ) ( n _{2} - n _{1} + 1 )}{2}

between

n_{2}

and

n_{1}

Hydrogen energy levels

5 · Matter waves & uncertainty

λ = \frac{h}{m v} = \frac{h}{2 m KE} = \frac{h}{2 m q V}

Δ x \cdot Δ p \geq \frac{h}{4 π}, Δ x \cdot Δ v \geq \frac{h}{4 π m}

“Velocity known to

x %

” means

Δ v = v \times \frac{x}{100}

, not

v

itself.

6 · Quantum numbers

Number	Values	Tells you
$n$	1, 2, 3 …	shell, size, energy
$l$	0 to $n - 1$	shape (0 s, 1 p, 2 d, 3 f)
$m_{l}$	$- l$ … 0 … $+ l$	orientation; $2 l + 1$ orbitals
$m_{s}$	$+ \frac{1}{2}, - \frac{1}{2}$	spin

Capacitysubshell

2 (2 l + 1)

electrons; shell

2 n^{2}

Orbital angular momentum

l (l + 1) \frac{h}{2 π}

(zero for s)

Spin-only moment

μ = n (n + 2)

BM,

n

= unpaired electrons

7 · Filling rules

Aufbaufill by increasing

(n + l)

; same

(n + l)

→ lower

n

first

Paulino two electrons share all four quantum numbers; max 2 per orbital, opposite spins

Hundfill degenerate orbitals singly, parallel spins, before pairing

One-electron species (H, He

^{+}

): energy depends on

n

only, so

3 s = 3 p = 3 d

. In many-electron atoms it follows

(n + l)

8 · Nodes

Radial $n - l - 1$

Angular $l$

Total $n - 1$

Example 3p

n = 3, l = 1

→ radial

= 1

, angular

= 1

, total

= 2

9 · Extra stability

Cr (Z = 24)[Ar]

3 d^{5} 4 s^{1}

(not

3 d^{4} 4 s^{2}

)

Cu (Z = 29)[Ar]

3 d^{10} 4 s^{1}

(not

3 d^{9} 4 s^{2}

)

Whysymmetry and maximum exchange energy

Exchanges in

d^{5}

\frac{n ( n - 1 )}{2} = \frac{5 \times 4}{2} = 10

Seen in NEET 2025

Ratio of wavelengths absorbed by H for $2 \to 3$ and $4 \to 6$ :

\frac{λ _{1}}{λ _{2}} = \frac{\frac{1}{16} - \frac{1}{36}}{\frac{1}{4} - \frac{1}{9}} = \frac{5/144}{5/36} = \frac{1}{4}

Wavelength is inverse to the energy gap, so flip the ratio.

Six traps to remember

Cathode-ray $e / m$ is fixed; canal-ray $e / m$ changes with the gas.
Intensity changes the number of photoelectrons, frequency changes their KE.
Bohr formulas work only for one-electron species.
For “ $x %$ accuracy”, use $Δ v = v x /100$ .
Radial nodes $= n - l - 1$ , angular $= l$ , total $= n - 1$ .
Cr and Cu are exceptions: half- or fully-filled $d$ wins.

Structure of Atom: free one-page cheat sheet