Dual Nature of Radiation and Matter

Dual Nature of Radiation and MatterNEET Physics · Class 12 · NCERT Chapter 11

Overview Notes NEET Questions Interactive Learning

7 interactive concept widgets for Dual Nature of Radiation and Matter. Drag any slider, change any number, and watch the formula and the answer update live. Built so you understand how each NEET problem actually works, not just the final number.

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Photon energy and momentum from wavelength

Einstein's photoelectric equation calculator

Stopping potential vs frequency

Three photoelectric graphs

de Broglie wavelength of a moving particle

Electron de Broglie wavelength after V volts

Photon vs particle wavelength at the same energy

Photon energy and momentum

A photon carries energy hν and momentum h/λ. Quick conversions between λ, f and E.

Photon energy

Photon energy and momentum from wavelength

Drag the wavelength slider; energy and momentum follow.

Each photon carries energy E = h f and momentum p = h / λ. Quick mental check: E (in eV) × λ (in nm) ≈ 1240.

Wavelength λ: 550 nm

Frequency f

5.45e+14 Hz

Energy E

2.255 eV

3.61e-19 J

Momentum p = h/λ

1.20e-27 kg·m/s

E = h ν = \frac{h c}{λ}, p = \frac{h}{λ}

Try this

550 nm (green): E ≈ 2.25 eV. About the work function of caesium.
300 nm (UV): E ≈ 4.13 eV. Enough to eject electrons from most common metals.
X-ray (1 nm): E ≈ 1.24 keV. Far above any work function.
For NEET shortcut: E (eV) × λ (nm) = 1240. Memorise this; converts wavelength to energy in seconds.

Photoelectric effect

Einstein's equation, stopping potential vs frequency line, and the three NEET-favourite graphs.

Einstein's photoelectric equation

Einstein's photoelectric equation calculator

Pick a metal and a wavelength; see whether electrons come out and with what energy.

Pick a metal and a light wavelength. If photon energy exceeds the work function, electrons are ejected with KE = h f - W_0. The stopping potential V_0 directly gives KE_max in volts.

Metal:

Caesium (2.14 eV)

Potassium (2.3 eV)

Sodium (2.36 eV)

Calcium (2.87 eV)

Aluminium (4.28 eV)

Copper (4.65 eV)

Platinum (6.35 eV)

Light λ: 300 nm (E_photon = 4.13 eV)

Photoelectric outcome

Electrons ejected

KE_max = 1.77 eV (2.84e-19 J)

Stopping potential V_0 = 1.77 V

Threshold for Sodium

f_0 = 5.71 × 10¹⁴ Hz

λ_0 = 525 nm

h ν = W_{0} + KE_{m a x}, e V_{0} = KE_{m a x}

Try this

Caesium with green light (550 nm, 2.25 eV) just barely ejects electrons. Try sliding to UV.
Aluminium needs at least UV (W_0 = 4.28 eV; threshold ~290 nm).
Below threshold, intensity does not matter: no electrons. Above threshold, brighter light gives more electrons but same KE_max.
Stopping potential V_0 in volts equals KE_max in eV. Easy to convert.

V_0 vs frequency

Stopping potential vs frequency

Set the work function and watch the line shift; the slope stays universal at h/e.

Stopping potential rises linearly with frequency. Slope is h/e (universal). x-intercept is the threshold frequency f_0 = W_0 / h (depends on the metal).

Work function W_0: 2.50 eV

Slope (h/e)

4.136e-15 V·s

Threshold f₀

6.04 × 10¹⁴ Hz

V_{0} = \frac{h}{e} f - \frac{W _{0}}{e}

Slope is universal: same h/e for every metal. Different metals share the same slope but cut the f-axis at different f_0.

Try this

All metals give parallel lines. The slope = h/e ≈ 4.14 × 10⁻¹⁵ V·s.
Different metals (Cs, Na, K, Pt) shift the line right or left, never tilt it.
A famous Millikan experiment measured h directly from the slope of this line.
Below f_0, V_0 = 0 (no emission). The line below f_0 is dashed: V_0 cannot be negative for emission.

Photoelectric graphs

The three NEET-favourite photoelectric graphs

Toggle between I-V, I_sat-Intensity, and KE_max-frequency.

Toggle between the three NEET-favourite photoelectric graphs.

Photoelectric I vs V: at high positive V, current saturates; bigger intensity → bigger saturation. All curves intersect the V-axis at the same -V_0 (V_0 depends on frequency, not intensity).

Try this

Saturation current depends only on intensity, not on the applied voltage above zero.
Stopping potential depends only on frequency, not on intensity.
KE_max line slope (= h Joule·s) is the same for every metal. Only the intercept changes.
Below f_0, no current flows even at very high intensity.

Matter waves: de Broglie

Wavelength of an electron, proton, alpha particle, even a cricket ball.

de Broglie wavelength

de Broglie wavelength of a moving particle

Pick a particle and energy; see how big or small the matter wavelength is.

Every moving particle has a wave nature with λ = h / p. For everyday objects, λ is too small to detect; for electrons it is comparable to atomic spacings.

Particle:

Electron

Proton

Neutron

Alpha particle

Cricket ball (160 g)

Kinetic energy: 100 eV

(slider is log10 scale; use it for any energy from 0.01 eV to 1 MeV)

de Broglie λ

1.23e+2 pm

Momentum p

5.40e-24 kg·m/s

Speed v

5.93e+6 m/s

λ = \frac{h}{p} = \frac{h}{2 m E _{k}}

Try this

Electron, 100 eV: λ ≈ 0.123 nm, atomic-spacing scale. Davisson-Germer used 54 V → λ ≈ 0.165 nm.
Proton, 1 keV: λ ≈ 9 × 10⁻¹³ m, already very small.
Cricket ball at 100 J: λ ≈ 10⁻³⁴ m, undetectable.
For an electron through V volts: λ = 12.27 / √V Å (V in volts).

Electron de Broglie λ

Electron de Broglie wavelength after V volts

The classic NEET formula: λ = 12.27 / √V Å.

An electron starting at rest accelerated through V volts gains KE = eV. Its de Broglie wavelength is: λ = 12.27 / √V Å.

Accelerating potential V: 100 V

de Broglie λ

1.227 Å

0.1227 nm

λ = \frac{12.27}{V} \overset{˚}{A}

Kinetic energy gained

100 eV

Speed v (non-relativistic)

5.93e+6 m/s

= 1.98% of c

Davisson-Germer used 54 V → λ ≈ 1.67 Å. Bragg diffraction off nickel confirmed it; matter waves are real.

Try this

100 V → λ = 1.227 Å. About the size of a hydrogen atom.
10000 V → λ = 0.123 Å. Used in electron microscopes for sub-atomic detail.
For non-relativistic case: speed scales as √V.
For very high V (above ~50 keV), relativistic correction becomes important.

Photon vs particle λ

Photon and electron wavelengths at the same energy

See why electron microscopes outclass light microscopes.

For the same energy, a photon and an electron have very different wavelengths. The photon is much "wider" because it travels at c, while the electron moves much more slowly.

Common energy E: 100 eV

Photon

λ = 1.24e+1 nm

λ_{p} = \frac{h c}{E}

Electron (KE = E)

λ = 1.23e+2 pm

λ_{e} = \frac{h}{2 m _{e} E}

Ratio λ_photon / λ_electron = 101× (electron always shorter for same E)

Try this

1 eV: photon λ ≈ 1240 nm (IR). Electron λ ≈ 1.23 nm.
100 eV: photon λ ≈ 12.4 nm (X-ray). Electron λ ≈ 0.12 nm.
For the same momentum, both have the same λ (since λ = h / p universally). For the same KE, they differ.
Electron microscopes use the small electron λ to image structures finer than light microscopes.

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