Photons | Class 12 NEB Physics | Complete notes | Short questions and numerical problem solutions
Photons
Quantum Nature of Radiation & Photoelectric Effect
Quantum Nature of Radiation
Light exhibits a dual nature, possessing both wave character and particle character. Depending on the physical situation observed, one nature becomes dominant over the other:
When light passes through a double slit, it produces an interference pattern that can strictly be interpreted only via the wave theory of light. Conversely, when light of a sufficiently low wavelength strikes a metallic surface, electrons are ejected immediately. This phenomenon is called the Photoelectric Effect and can only be understood in terms of the particle nature of light.
According to Planck's quantum theory of radiation, energy from a radiating entity is emitted in discrete, individual packets called quanta. Each quantum carries a distinct, fixed quantity of energy and is identified as a Photon. The energy of an individual photon is expressed by:
\[E = hf \hspace{0.1cm} \text{... (i)}\]Where:
f = frequency of the radiation
h = Planck's constant \((6.625 \times 10^{-34} \text{ J}\cdot\text{s})\)
Equation (i) can be modeled alongside wave variables (\(c = f\lambda\)) to yield:
\[E = \frac{hc}{\lambda} \hspace{0.1cm} \text{... (ii)}\]From these relations, it is evident that photons with a higher frequency and shorter wavelength carry greater energy properties. Conversely, those with lower frequencies and longer wavelengths present reduced energy parameters.
- A photon always travels at a constant velocity \(c \approx 3.0 \times 10^8 \text{ m/s}\) in a vacuum.
- The invariant rest mass of a photon equals identically \(0\).
- Each single photon possesses a definitive linear momentum vector and discrete energy value.
- During encounters with material particles, total energy and momentum indicators are strictly conserved.
Consider a parallel beam of light of wavelength \(600 \text{ nm}\). Find the corresponding energy of an individual photon in electron-volts.
Speed of light (\(c\)) = \(3 \times 10^8 \text{ m/s}\) | Planck's constant (\(h\)) = \(6.62 \times 10^{-34} \text{ J}\cdot\text{s}\)
Applying the Planck-Einstein energy relation:
\[\begin{align*} E &= \frac{hc}{\lambda} = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{600 \times 10^{-9}}\\ &= 3.31 \times 10^{-19} \text{ J} \end{align*}\]Converting Joules to electron-volts using the scale factor \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\):
\[E = \frac{3.31 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.069 \text{ eV}\]Photoelectric Experiment Summary
To evaluate the structural formulation of Einstein's equations, the primary empirical markers observed during classical photoelectric experiments are summarized below:
- Instantaneous Emission: Ejection of electrons occurs almost instantaneously (\(< 10^{-9} \text{ s}\)) after light hits the metal surface, provided the wavelength criteria are satisfied.
- Threshold Wavelength (\(\lambda_0\)): A distinct boundary parameter exists for every metal. If incoming light features a wavelength greater than \(\lambda_0\) (or frequency below \(f_0\)), no emissions take place regardless of the source's intensity.
- Stopping Potential (\(V_0\)): The inverse negative potential configuration on the anode required to fully terminate the photo-current circuit. This is directly proportional to the maximum kinetic energy profile of the ejections.
- Intensity Dynamics: The stopping potential is independent of the intensity of the incident light. Elevating intensity indices solely expands the net volume of current (flows of photoelectrons), not their specific kinetic velocity values.
Work Function (\(\phi\)): The minimum energetic value needed to separate an electron cleanly from its host metal boundary matrix. It scales uniquely across diverse elemental substrates.
Einstein's Photoelectric Equation
In 1905, Albert Einstein successfully resolved the anomalous contradictions of classical electrodynamics by extending Planck's blackbody cavity concepts directly to light radiation fields. He modeled light as an active assembly of independent corpuscular energy pockets.
When an explicit photon strikes an electron embedded inside a metallic system, it surrenders its complete energetic quantum profile in a localized collision. Part of this acquired parameter is consumed overcoming the surface bounding forces (\(\phi\)), whereas any surplus translates directly into forward kinetic motion vector indicators:
\[\begin{align*} E &= \phi + \text{K.E.}_{\text{max}}\\ hf &= \phi + \text{K.E.}_{\text{max}}\\ \text{K.E.}_{\text{max}} &= hf - \phi \hspace{0.1cm} \text{... (iii)} \end{align*}\]Equation (iii) stands as Einstein's Photoelectric Equation, an achievement which earned him the Nobel Prize in Physics for 1921.
By defining threshold limits where velocity indices approach zero (\(f = f_0\)), we state \(\phi = hf_0\). This transforms the primary structural framework equation into:
\[\frac{1}{2}m v_{\text{max}}^2 = hc\left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right)\]- If \(\lambda > \lambda_0\): Energetic supply yields values below work parameters; zero emissions occur.
- If \(\lambda = \lambda_0\): Surface separation finishes successfully, but \(\text{K.E.}_{\text{max}} = 0\).
- If \(\lambda \le \lambda_0\): Free structural photo-emissions are cleanly produced.
Find the maximum wavelength of light capable of triggering a photoelectric response inside a sodium target. (Work function of sodium \(\phi = 2.3 \text{ eV}\)).
Planck's Constant (\(h\)) = \(6.62 \times 10^{-34} \text{ J}\cdot\text{s}\) | Speed of Light (\(c\)) = \(3 \times 10^8 \text{ m/s}\)
The maximum wavelength restriction equals the critical threshold limit (\(\lambda_0\)):
\[\begin{align*} \lambda_0 &= \frac{hc}{\phi} = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{3.68 \times 10^{-19}}\\ &= 5.396 \times 10^{-7} \text{ m} \end{align*}\]Millikan's Verification & Planck's Constant
Alkali metals were placed on a rotatable wheel \(W\) within an evacuated chamber to prevent surface oxidation. A clean scraping knife \(K\) refreshed the metal targets in-situ.
By applying a retarding bias voltage, the maximum kinetic energy components translate directly into electrical potential metric units:
\[\text{K.E.}_{\text{max}} = \frac{1}{2}m v_{\text{max}}^2 = eV_0\]Substituting this directly into Einstein's fundamental statement yields:
\[eV_0 = hf - hf_0 \implies V_0 = \left(\frac{h}{e}\right)f - \left(\frac{h}{e}\right)f_0\]This matches the slope intercept profile of a standard straight line equation (\(y = mx + c\)).
By plotting the measured values of \(V_0\) against \(f\), the slope of the line explicitly evaluates as:
\[m = \tan\theta = \frac{h}{e} \implies h = e \cdot \tan\theta \approx 6.625 \times 10^{-34} \text{ J}\cdot\text{s}\]Millikan's measurements validated Einstein's equations completely, earning him the 1923 Nobel Prize.
Conceptual Question Bank
In what ways do photons resemble other material entities (like electrons), and how do they differ? Do they carry mass, electric charges, or accelerate under field pressures?
Resemblances: Photons share discrete localized momentum features with material elements and strictly obey dynamic conservation laws during localized impact collisions.
Differences: Photons lack a rest-mass profile (\(m_0 = 0\)), present no electric net charge attributes, and showcase polarization parameters unique to electromagnetic wave entities.
Acceleration & Charge: Photons carry zero net electric charge and cannot undergo traditional linear velocity acceleration because their speed remains fixed at the speed of light \(c\) across any single medium index.
Why don't humans witness continuous discrete flashing effects when viewing everyday environments under classic photon models?
The explicit emission density rates reaching human visual receptor tracks are exceptionally vast. Because the eye's cognitive frame processing thresholds cannot isolate these high-frequency discrete transits, the brain integrates them into a single, continuous visual stream.
Why don't household metal fittings lose their electrons via photoelectric activation whenever domestic lights are turned on?
Standard household ambient bulbs emit lower spectral frequencies (visible light ranges). The energy parameter of these photons fall below the necessary threshold work function constraints (\(hf < \phi\)) required to emancipate bound states from household structural metals.
Human skin remains mostly unaffected by visible light, but ultraviolet (UV) radiation causes rapid burns. Explain this using quantum principles.
Ultraviolet bands process substantially higher operational frequencies relative to visible bands. Since \(E = hf\), UV photons carry enough energetic capacity to mutate, break chemical bonds, and ionize cellular structures within skin tissue layer matrices.
Why is monochromatic black-and-white photographic film highly sensitive to blue light rays but mostly unresponsive to red or infrared bands?
Because blue light features a shorter wavelength component (\(E = \frac{hc}{\lambda}\)), its individual photon energy exceeds that of red or infrared light. This allows blue photons to successfully initiate the localized chemical reductions on the film emulsion layer.
Can an external electric or magnetic field vector deflect a photon path?
No. Because a photon is entirely neutral and carries no net electric charge, it passes through magnetic and electric force fields without experiencing any Lorentz path deflection.
A heated object rests inside an isolated, cold chamber. Does the net volume of photons inside the room expand?
Yes. The thermal energy differential causes the hot body to continuously radiate electromagnetic energy into the cold space. This thermal radiation expands the net number of photons occupying the room.
Photosynthesis activates cleanly under standard solar exposure fields but fails under pure infrared illumination. Clarify the root cause.
The visible components of sunlight carry sufficient photon energy to induce key electronic transitions within chlorophyll protein receptors. Infrared photons, conversely, lack the necessary energy quantum to trigger these vital metabolic steps.
If we gradually increase the wavelength of incident light striking a metal target, how do the total volume and kinetic profiles of the emissions respond?
Increasing the wavelength lowers the energy parameter per photon. Consequently, the maximum kinetic energy of the emitted electrons decreases linearly. The total volume of ejected electrons remains unchanged until the wavelength crosses the threshold value \(\lambda_0\), at which point all emissions cease completely.
What changes occur to the maximum kinetic energy values of photoelectrons if the source illumination intensity is doubled?
There is absolutely no change. The maximum kinetic energy of photoelectrons is determined solely by the incident frequency and the target's work function. Doubling the intensity only doubles the number of photons striking the surface per second, increasing the net current volume.
Does the threshold frequency parameter for a given metal display any dependence on source illumination intensity fields?
No. The threshold frequency is defined as \(f_0 = \frac{\phi}{h}\). It depends strictly on the electronic work function, which is an intrinsic property of the target material, and is entirely independent of external light intensity.
In a photoelectric experiment, the stopping potential decreases from \(1.85 \text{ V}\) to \(0.82 \text{ V}\) as the wavelength shifts from \(300 \text{ nm}\) to \(400 \text{ nm}\). Calculate Planck's constant from these data.
\(V_{02} = 0.82 \text{ V}\) at \(\lambda_2 = 400 \times 10^{-9} \text{ m}\)
\(e = 1.6 \times 10^{-19} \text{ C}\) | \(c = 3 \times 10^8 \text{ m/s}\)
Setting up the energy conservation equations for both wavelengths:
\[\frac{hc}{\lambda_1} = \phi + eV_{01} \quad \text{and} \quad \frac{hc}{\lambda_2} = \phi + eV_{02}\]Subtracting the second equation from the first eliminates the work function (\(\phi\)):
\[\begin{align*} hc\left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right) &= e(V_{01} - V_{02})\\ h \times (3 \times 10^8) \times \left(\frac{1}{300 \times 10^{-9}} - \frac{1}{400 \times 10^{-9}}\right) &= 1.6 \times 10^{-19} \times (1.85 - 0.82)\\ h \times (3 \times 10^8) \times \left(8.333 \times 10^5\right) &= 1.6 \times 10^{-19} \times 1.03\\ h \times (2.5 \times 10^{14}) &= 1.648 \times 10^{-19}\\ h &= \frac{1.648 \times 10^{-19}}{2.5 \times 10^{14}} = 6.592 \times 10^{-34} \text{ J}\cdot\text{s} \end{align*}\]Light of wavelength \(4.0 \times 10^{-7} \text{ m}\) strikes a sodium surface (\(\phi = 2.3 \text{ eV}\)). What is the maximum kinetic energy of the emitted electrons in electron-volts?
First, evaluate total input photon energy in Joules:
\[E = \frac{hc}{\lambda} = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{4.0 \times 10^{-7}} = 4.965 \times 10^{-19} \text{ J}\]Converting input energy value to electron-volts:
\[E = \frac{4.965 \times 10^{-19}}{1.6 \times 10^{-19}} = 3.103 \text{ eV}\]Applying Einstein's photoelectric relation to solve for kinetic energy:
\[\text{K.E.}_{\text{max}} = E - \phi = 3.103 \text{ eV} - 2.3 \text{ eV} = 0.803 \text{ eV}\]Light of frequency \(5.0 \times 10^{14} \text{ Hz}\) liberates electrons with kinetic energy \(2.31 \times 10^{-19} \text{ J}\) from a metal. What is the wavelength of a UV light beam that liberates electrons with kinetic energy \(8.93 \times 10^{-19} \text{ J}\) from this same surface?
\(\text{K.E.}_{2} = 8.93 \times 10^{-19} \text{ J}\) | \(h = 6.625 \times 10^{-34} \text{ J}\cdot\text{s}\)
First, find the work function (\(\phi\)) using the initial configuration data:
\[\begin{align*} hf_1 &= \phi + \text{K.E.}_{1}\\ (6.625 \times 10^{-34} \times 5.0 \times 10^{14}) &= \phi + 2.31 \times 10^{-19}\\ 3.3125 \times 10^{-19} &= \phi + 2.31 \times 10^{-19}\\ \phi &= 1.0025 \times 10^{-19} \text{ J} \end{align*}\]Now, solve for the incident UV wavelength (\(\lambda_2\)) using the second configuration:
\[\begin{align*} \frac{hc}{\lambda_2} &= \phi + \text{K.E.}_{2}\\ \frac{hc}{\lambda_2} &= 1.0025 \times 10^{-19} + 8.93 \times 10^{-19} = 9.9325 \times 10^{-19} \text{ J}\\ \lambda_2 &= \frac{6.625 \times 10^{-34} \times 3 \times 10^8}{9.9325 \times 10^{-19}} \approx 2.001 \times 10^{-7} \text{ m} \end{align*}\]The work function of potassium is \(2 \text{ eV}\). If the surface is illuminated with light of wavelength \(350 \text{ nm}\), what stopping potential must be applied to prevent electron collection, and what is the maximum kinetic energy of the emitted electrons?
Calculate total input photon energy in eV:
\[\begin{align*} E &= \frac{hc}{\lambda} = \frac{6.625 \times 10^{-34} \times 3 \times 10^8}{350 \times 10^{-9}} = 5.678 \times 10^{-19} \text{ J}\\ E &= \frac{5.678 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.55 \text{ eV} \end{align*}\]Using the photoelectric equation to determine maximum kinetic energy:
\[\text{K.E.}_{\text{max}} = E - \phi = 3.55 \text{ eV} - 2.0 \text{ eV} = 1.55 \text{ eV}\]Since \(\text{K.E.}_{\text{max}} = eV_0\), the corresponding stopping potential is:
\[V_0 = 1.55 \text{ V}\]Converting kinetic energy to Joules yields:
\[\text{K.E.}_{\text{max}} = 1.55 \times 1.6 \times 10^{-19} \text{ J} = 2.48 \times 10^{-19} \text{ J}\]Calculate the energy in electron-volts of a quantum of X-radiation featuring a wavelength component of exactly \(0.15 \text{ nm}\). [Take \(h = 6.5 \times 10^{-34} \text{ J}\cdot\text{s}\)].
Applying the standard wavelength to energy equation:
\[\begin{align*} E &= \frac{hc}{\lambda} = \frac{6.5 \times 10^{-34} \times 3 \times 10^8}{0.15 \times 10^{-9}}\\ &= 1.30 \times 10^{-15} \text{ J} \end{align*}\]Converting the value into electron-volt metrics:
\[E = \frac{1.30 \times 10^{-15}}{1.6 \times 10^{-19}} = 8125 \text{ eV}\]
Comments
Post a Comment