Is it correct to say: Assuming photoelectric emission depends on light intensity?

Emission depends on *frequency* (must exceed ν_0). Intensity affects only the *number* of ejected electrons, not their KE.

Is it correct to say: Using KE_max = hν (missing the work function φ)?

Correct form: KE_max = hν − φ. Otherwise, KE would be non-zero even for ν < ν_0, which contradicts experiments.

Is it correct to say: Writing energy of Bohr level as positive?

E_n = −13.6/n² eV. Sign is negative — bound state. Energy = 0 means the electron just escapes.

Is it correct to say: Confusing decay constant λ with wavelength λ?

Same symbol, different quantity. In decay: units of s⁻¹. In waves: units of m. Context decides.

Is it correct to say: Using T_(1/2) = 1/λ for half-life?

T_(1/2) = 0.693/λ (with ln 2). τ = 1/λ is *mean* life. Different!

Is it correct to say: Forgetting 1 u = 931.5 MeV/c² in mass-defect problems?

Binding energy = Δm × 931.5 MeV when Δm is in u. Common source of factor-931 errors.

Formula sheet (PDF) →

Physics · Chapter 11

Dual Nature of Radiation and Matter / Atoms & Nuclei

Q: Is it correct to say: Writing energy of Bohr level as positive?

E_n = −13.6/n² eV. Sign is negative — bound state. Energy = 0 means the electron just escapes.

Q: Is it correct to say: Confusing decay constant λ with wavelength λ?

Same symbol, different quantity. In decay: units of s⁻¹. In waves: units of m. Context decides.

Q: Is it correct to say: Using T_(1/2) = 1/λ for half-life?

T_(1/2) = 0.693/λ (with ln 2). τ = 1/λ is *mean* life. Different!

Q: Is it correct to say: Forgetting 1 u = 931.5 MeV/c² in mass-defect problems?

Binding energy = Δm × 931.5 MeV when Δm is in u. Common source of factor-931 errors.

Photoelectric effect, de Broglie wavelength, Bohr model of hydrogen, radioactive decay — NCERT Class 12 Physics Ch 11, 12, 13

14 formulas · 3 solved examples · 6 practice questions · Board weight: 7 marks · CBSE Class 12 · 2025–26 syllabus

Board Exam Tips

→Einstein's photoelectric equation derivation + graph (stopping potential vs frequency) is a repeat 3-mark question every year.
→Bohr's postulates + derivation of E_n = −13.6/n² eV is a 5-mark HOTS favourite.
→de Broglie wavelength calculation for electron and comparison with photon of same energy — classic 2-mark question.
→Half-life and mean-life formulas — do NOT confuse T_(1/2) = 0.693/λ with τ = 1/λ.
→Energy conversions: 1 eV = 1.6×10⁻¹⁹ J; hc ≈ 1240 eV·nm — memorise both.

Diagram

Photon striking a metal surface ejecting a photoelectron

Formulas(14)

Photon EnergyBoard favourite

E = h ν = \frac{h c}{λ}

Symbol	Meaning
$h$	Planck's constant = 6.626×10⁻³⁴ J·s
$ν$	Frequency (Hz)
$λ$	Wavelength (m)
$c$	Speed of light = 3×10⁸ m/s

Einstein's Photoelectric EquationBoard favourite

K E_{ma x} = h ν - ϕ = h (ν - ν_{0})

Symbol	Meaning
$ϕ$	Work function of metal (J or eV)
$ν_{0}$	Threshold frequency (Hz)

Stopping Potential

e V_{0} = K E_{ma x} = h (ν - ν_{0})

de Broglie WavelengthBoard favourite

λ = \frac{h}{p} = \frac{h}{m v}

de Broglie Wavelength of Electron via Voltage

λ = \frac{h}{2 m e V} \approx \frac{1.227}{V} nm

Bohr's Quantisation of Angular MomentumBoard favourite

m v r = \frac{nh}{2 π}

Radius of nth Bohr Orbit

r_{n} = \frac{n ^{2} h ^{2} ε _{0}}{π m e ^{2}} = 0.529 n^{2} A ˚

Energy of nth Bohr LevelBoard favourite

E_{n} = - \frac{13.6}{n ^{2}} eV

Photon Frequency for Transition

h ν = E_{n_{2}} - E_{n_{1}}

Rydberg Formula

\frac{1}{λ} = R (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})

Mass–Energy Equivalence

E = m c^{2}

Radioactive Decay Law

N = N_{0} e^{- λ t}

Half-Life and Mean Life

T_{1/2} = \frac{l n 2}{λ} = \frac{0.693}{λ}, τ = \frac{1}{λ}

Activity

A = λ N = A_{0} e^{- λ t}

Solved Examples

1Solved Exampleeasy3 steps

Light of wavelength 400 nm falls on a metal of work function 2.0 eV. Find (a) energy of one photon in eV, (b) maximum KE of photoelectrons, (c) stopping potential.

Photon energy using hc/λ (with hc ≈ 1240 eV·nm)

E = \frac{h c}{λ} = \frac{1240 eV\cdotnm}{400 nm} = 3.1 eV

2Solved Exampleboard4 steps

An electron in the hydrogen atom jumps from n = 3 to n = 2. Find (a) the energy of the emitted photon in eV, (b) its wavelength, (c) identify the spectral series.

Energy levels

E_{2} = - \frac{13.6}{4} = - 3.4 eV, E_{3} = - \frac{13.6}{9} \approx - 1.51 eV

3Solved ExampleHOTS5 steps

A radioactive sample has a half-life of 20 minutes. Initial activity is 8000 disintegrations/s. Find (a) the decay constant, (b) the activity after 1 hour, (c) the number of nuclei present initially.

Decay constant from half-life

λ = \frac{0.693}{T _{1/2}} = \frac{0.693}{20 \times 60} \approx 5.78 \times 1 0^{- 4} s^{- 1}

Traps & Common Mistakes

Common Mistakes6

1
Assuming photoelectric emission depends on light intensity
✓Emission depends on *frequency* (must exceed ν_0). Intensity affects only the *number* of ejected electrons, not their KE.
2
Using KE_max = hν (missing the work function φ)
✓Correct form: KE_max = hν − φ. Otherwise, KE would be non-zero even for ν < ν_0, which contradicts experiments.
3
Writing energy of Bohr level as positive
✓E_n = −13.6/n² eV. Sign is negative — bound state. Energy = 0 means the electron just escapes.
4
Confusing decay constant λ with wavelength λ
✓Same symbol, different quantity. In decay: units of s⁻¹. In waves: units of m. Context decides.
5
Using T_(1/2) = 1/λ for half-life
✓T_(1/2) = 0.693/λ (with ln 2). τ = 1/λ is *mean* life. Different!
6
Forgetting 1 u = 931.5 MeV/c² in mass-defect problems
✓Binding energy = Δm × 931.5 MeV when Δm is in u. Common source of factor-931 errors.

Practice Yourself

Practice Yourself6 questions

Q1
Threshold frequency of a metal is 5×10¹⁴ Hz. Light of 8×10¹⁴ Hz falls on it. Stopping potential?
Q2
de Broglie wavelength of an electron accelerated through 100 V?
Q3
Wavelength of the H-α line (Balmer, n=3→2)?
Q4
Energy required to ionise a hydrogen atom from ground state?
Q5
A radioactive sample decays to 1/8 of its initial value in 30 minutes. Half-life?
Q6
Mass defect of a nucleus is 0.05 u. Binding energy in MeV?

Notes

Dual Nature, Atoms and Nuclei — key concepts

Three chapters combined: light behaves as photons, matter behaves as waves, atoms have discrete energy levels, and unstable nuclei decay exponentially.

The photon picture

Light is quantised into photons of energy $h\nu$ and momentum $h/\lambda$ .
Photoelectric emission is instantaneous, has a threshold frequency, and stopping potential depends only on $\nu$ — three facts that classical wave theory cannot explain.
Einstein's equation $KE_{max} = h\nu - \phi$ encapsulates all three.

The matter wave picture

de Broglie: every particle of momentum p has a wavelength $\lambda = h/p$ .
For electrons, $\lambda \approx 1.227/\sqrt{V}$ nm (V in volts). Confirmed by Davisson–Germer diffraction from a nickel crystal.
Wave nature is significant only when $\lambda$ is comparable to the system size (electrons: yes; cricket ball: no).

Bohr's atomic model — three postulates

Electrons orbit only in certain "allowed" orbits without radiating.
Angular momentum is quantised: $mvr = nh/(2\pi)$ .
Radiation is emitted or absorbed only when the electron jumps between orbits: $h\nu = E_2 - E_1$ .

From these, r_n ∝ n² and E_n = −13.6/n² eV for hydrogen.

Spectral series of hydrogen

Lyman (UV): transitions to n = 1.
Balmer (visible): transitions to n = 2. Includes red H- $\alpha$ at 656 nm.
Paschen, Brackett, Pfund (IR): to n = 3, 4, 5.

Radioactivity

$N = N_0 e^{-\lambda t}$ , $A = \lambda N$ .
Half-life $T_{1/2} = 0.693/\lambda$ ; mean life $\tau = 1/\lambda$ .
After n half-lives, fraction surviving = 1/2ⁿ.

Quick sanity checks

Photon of visible light ( $\lambda \approx 500$ nm) has energy ≈ 2.5 eV — comparable to work functions of common metals.
If stopping potential is not linear in $\nu$ , something is wrong with the experiment.
E_n gets less negative as n increases; ionisation from ground state needs 13.6 eV.
Activity is always positive; N always decreases with time. Any answer violating this is wrong.
Units: 1 eV = 1.6×10⁻¹⁹ J; 1 u = 1.66×10⁻²⁷ kg ≈ 931.5 MeV/c².