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NEET UG Β· Physics

Dual nature of radiation and matter

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NEET UG Β· Physics

Continue studying Dual Nature Of Radiation And Matter

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Separate photon energy from intensity, test emission thresholds and compare matter wavelengths. Dual nature: check the threshold and the momentum

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Dual nature: check the threshold and the momentum

Separate photon energy from intensity, test emission thresholds and compare matter wavelengths.

What you will understand

For NEET UG Physics students: test the photoelectric threshold, distinguish intensity from photon energy, interpret stopping potential and choose the correct momentum for a de Broglie wavelength.

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Quick answer: photon energy, photon rate and electron momentum differ

In ordinary one-photon photoemission, one absorbed photon supplies energy hΞ½. The maximum emitted-electron kinetic energy is hΞ½βˆ’Ο† only when the threshold permits emission. At fixed frequency and surface, increasing illumination can increase the emission rate under suitable collection conditions without raising that maximum energy. Matter wavelength instead follows Ξ»=h/p, so the momentum relation matters. Do not replace every energy with pc: a photon and a slow electron obey different energy–momentum relations.

Photoelectric threshold: no negative emitted kinetic-energy branch

Original ideal one-photon photoelectric energy plot for a supplied surface work function 2.00 eV. The horizontal axis is photon energy in eV and the vertical axis is maximum emitted-electron kinetic energy in eV. The green line starts at photon energy 2.00 eV and follows Kmax=photon energy minus 2.00 eV only on the allowed side. The region below threshold has no emitted-electron kinetic-energy branch, rather than a negative Kmax. A 400 nm photon has 3.10 eV and gives maximum kinetic energy 1.10 eV; an 800 nm photon has 1.55 eV and cannot eject an electron in this ordinary one-photon model. Supplied hc is 1240 eV nm. This is an illustrative surface/energy model, not measured detector data or a statement about intense-field multiphoton processes.
The plot uses photon energy on the horizontal axis and maximum emitted-electron kinetic energy on the vertical axis. For the supplied 2.00 eV work function, the emission branch begins at the 2.00 eV threshold. Below it, the ordinary one-photon model gives no emission; a negative subtraction result is not the kinetic energy of an emitted electron.

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Check wavelength against the threshold before using stopping potential

For a supplied surface take Ο†=2.00 eV and hc=1240 eV nmhc=1240\,\mathrm{eV\,nm}. At Ξ»=400 nm, photon energy is 1240/400=3.10 eV, so Kmax⁑=3.10βˆ’2.00=1.10 eVK_{\max}=3.10-2.00=1.10\,\mathrm{eV}. The magnitude of the ideal stopping potential is 1.10 V because eVs=Kmax⁑eV_s=K_{\max}. The threshold wavelength is 1240/2.00=620 nm. At 800 nm the photon has only 1.55 eV: no one-photon emission occurs, rather than an emitted electron with βˆ’0.45 eV. Contact potentials, detailed surface states and intense-field multiphoton processes are outside this elementary model.

Intensity and stopping potential answer different questions

At fixed above-threshold frequency, a higher incident photon rate can increase photocurrent when surface response and collection conditions are unchanged. Saturation current also depends on collection and the available emitted-electron rate. The stopping potential reflects the maximum kinetic energy and is ideally independent of illumination intensity at that fixed frequency. Increasing intensity below threshold does not rescue the ordinary one-photon energy deficit. The reverse-bias voltage sign depends on the stated electrode convention; Vs here denotes its positive stopping magnitude.

Electron acceleration: four times voltage halves wavelength

Original de Broglie scaling diagram for an electron accelerated from rest through an ideal potential difference V. In the nonrelativistic approximation K=eV and p=sqrt(2meV), so lambda=h/p is proportional to one over sqrt(V). The supplied 100 V and 400 V cases give approximately 0.123 nm and 0.0613 nm with the rounded constants in the lesson. The shorter-wavelength sketch has half the period on the same horizontal scale. The sinusoidal sketches illustrate a relative wavelength, not a literal electron trajectory, particle size or surrounding electric-field oscillation. Charge magnitude, electron mass and initially-at-rest conditions are fixed; high energies require a relativistic momentum relation.
The supplied electron starts from rest and gains kinetic energy eV in an ideal accelerating potential. At these low energies, p=√(2meV), so four times V gives twice the momentum and half the de Broglie wavelength. The wave sketches show relative wavelength, not an electron trajectory, particle radius or oscillating electric field around the particle.

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From accelerating voltage to electron wavelength

Use supplied h=6.626Γ—10βˆ’34 J sh=6.626\times10^{-34}\,\mathrm{J\,s}, me=9.109Γ—10βˆ’31 kgm_e=9.109\times10^{-31}\,\mathrm{kg} and e=1.602Γ—10βˆ’19 Ce=1.602\times10^{-19}\,\mathrm{C}. For an electron accelerated from rest through 100 V, K=eV=1.602Γ—10βˆ’17 JK=eV=1.602\times10^{-17}\,\mathrm J. The nonrelativistic momentum p=2meKp=\sqrt{2m_eK} is about

5.402Γ—10βˆ’24 kg m sβˆ’15.402\times10^{-24}\,\mathrm{kg\,m\,s^{-1}}
, giving
Ξ»=h/pβ‰ˆ1.227Γ—10βˆ’10 m=0.123 nm\lambda=h/p\approx1.227\times10^{-10}\,\mathrm m=0.123\,\mathrm{nm}
. At 400 V, momentum doubles and Ξ»β‰ˆ0.0613 nm. The approximation is appropriate here because the gained energies are much smaller than the electron rest energy, about 511 keV. High-voltage cases may require relativistic momentum.

Same momentum and same kinetic energy are different comparisons

Two particles with the same momentum have the same de Broglie wavelength h/p. In a nonrelativistic comparison at equal kinetic energy, however, momentum is √(2mK), so the heavier particle has the shorter wavelength. At equal speed, momentum is mv in that approximation, giving a different mass dependence. Read whether the problem fixes momentum, kinetic energy, speed or accelerating voltage. Equal voltage implies equal gained energy only for equal charge magnitude and the stated initial conditions.

A photon and an electron at 100 eV do not have equal wavelength

Take a photon of energy 100 eV and an electron of kinetic energy 100 eV. With supplied hc=1240 eV nm, the photon wavelength is 12.4 nm because photon energy is pc. The electron, using the previous nonrelativistic calculation, has wavelength about 0.1227 nm because its kinetic energy is p2/(2me)p^2/(2m_e), not pc. The photon wavelength is about 101 times larger. The electron also has rest energy; comparing its kinetic energy with the photon total energy does not make their momenta equal. The general wavelength relation h/p is shared, while the energy–momentum relations differ.

Diffraction supports a wave description without a drawn particle path

Electron diffraction patterns show why a matter-wave description is useful. A sinusoid used to mark a wavelength is not a literal zigzag electron trajectory. A detector records local events, and an interference pattern can build over many events. The conditions of preparation, propagation and detection matter; a classical particle sketch alone does not describe the full experiment.

Common mistakes: do the threshold and momentum checks first

Shorter wavelength means higher photon energy. Do not report negative emitted kinetic energy below threshold, or treat greater intensity at fixed frequency as greater maximum electron energy. Stopping potential has units of volts; electron energy may be written in eV, but they are different quantities related by charge. For matter wavelength, choose p from the correct nonrelativistic or relativistic model and check what the comparison holds fixed.

Try it: can brightness overcome this threshold?

For the supplied Ο†=2.00 eV surface, 800 nm photons have 1.55 eV each. Under the ordinary one-photon model, does doubling intensity produce emission? What changes for 400 nm light if frequency and collection conditions stay fixed?

Reveal the answer

The 800 nm photons remain below threshold, so doubling intensity does not produce ordinary one-photon emission. At 400 nm, each photon remains 3.10 eV and Kmax remains 1.10 eV. Greater photon rate can increase the photocurrent under unchanged response and collection conditions; it does not raise the ideal stopping magnitude.

Try it: quarter the electron wavelength

In the same initially-at-rest nonrelativistic electron model, how must accelerating voltage change to reduce de Broglie wavelength to one quarter? Why is a factor of four in voltage insufficient?

Reveal the answer

Wavelength is proportional to V raised to βˆ’1/2. A fourfold increase in voltage only halves it. A sixteenfold increase is needed for one quarter, provided the final energy still lies within the nonrelativistic regime.

Come back tomorrow: identify which energy and which momentum

Explain the threshold before calculating, then compare photon and electron wavelength at one supplied energy. Revisit NEET Atoms for discrete photon gaps, or NEET Electromagnetic Waves to separate field amplitude, energy flux and photon energy.

Chapter practice opens the broad Dual Nature pool.

Sources and review

Technical review: Codex (automated). Primary textbook sources checked on 23 October 2026. Diagrams, questions and numerical examples are original and use supplied rounded constants with explicit models. Human subject review is not claimed. These explanations and chapter catalogues do not establish complete coverage of every current exam format or syllabus.

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