Pearson Edexcel IAL Physics Unit 2 Waves and Electricity January 2025 Study Guide

Fundamental Constants and Formula Reference

Physical constants govern all classical and quantum calculations in wave dynamics and circuit theory. The acceleration of free fall close to Earth's surface is $g = 9.81\,m\,s^{-2}$, which is identical to the gravitational field strength $g = 9.81\,N\,kg^{-1}$. An electron carries an elementary charge magnitude of $e = 1.60 \times 10^{-19}\,C$ and has a rest mass of $m_e = 9.11 \times 10^{-31}\,kg$. The conversion factor between energy units is $1\,eV = 1.60 \times 10^{-19}\,J$. Electromagnetic radiation in a vacuum travels at a constant speed of $c = 3.00 \times 10^8\,m\,s^{-1}$, and quantum photon interaction calculations rely on Planck's constant $h = 6.63 \times 10^{-34}\,J\,s$

Classical mechanics relations integrate into electrical and wave energy concepts. Kinematic equations for uniform acceleration include $s = \frac{(u + v)t}{2}$, $v = u + at$, $s = ut + \frac{1}{2}at^2$, and $v^2 = u^2 + 2as$. Dynamics and energy relations dictate that net force is $\Sigma F = ma$, weight is $W = mg$, linear momentum is $p = mv$, and moment of force is $\text{moment} = Fx$. Mechanical work and energy transformations follow $\Delta W = F\Delta s$, kinetic energy $E_k = \frac{1}{2}mv^2$, gravitational potential energy $\Delta E_{\text{grav}} = mg\Delta h$, and mechanical power $P = \frac{E}{t} = \frac{W}{t}$. Efficiency is expressed as the ratio of useful energy output to total energy input, or useful power output to total power input. Material properties include density $\rho = \frac{m}{V}$, Stokes' law for viscous drag $F = 6\pi\eta rv$, Hooke's law $\Delta F = k\Delta x$, elastic strain energy $\Delta E_{\text{el}} = \frac{1}{2}F\Delta x$, and Young modulus $E = \frac{\sigma}{\varepsilon}$ where tensile stress $\sigma = \frac{F}{A}$ and tensile strain $\varepsilon = \frac{\Delta x}{x}$

Wave mechanics equations relate wave speed $v = f\lambda$, transverse wave speed on a stretched string $v = \sqrt{\frac{T}{\mu}}$, radiation intensity $I = \frac{P}{A}$, Snell's law of refraction $n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$, refractive index $n = \frac{c}{v}$, critical angle $\sin(C) = \frac{1}{n}$, and diffraction grating maxima $n\lambda = d \sin(\theta)$. Electrical transport equations define potential difference $V = \frac{W}{Q}$, resistance $R = \frac{V}{I}$, electric power $P = VI = I^2R = \frac{V^2}{R}$, electrical work $W = VIt$, electrical resistivity $R = \frac{\rho l}{A}$, charge flow rate $I = \frac{\Delta Q}{\Delta t}$, and microscopic current density $I = nqvA$. Series resistor combinations sum linearly as $R = R_1 + R_2 + R_3$, whereas parallel combinations sum reciprocally as $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$. Quantum phenomena follow the photon energy relation $E = hf$, Einstein's photoelectric equation $hf = \Phi + \frac{1}{2}mv_{\text{max}}^2$, and the de Broglie wavelength equation $\lambda = \frac{h}{p}$

Section A: Multiple Choice Conceptual Analysis

Microscopic current conduction is defined by the transport equation $I = nqvA$, where $I$ represents electric current, $q$ represents charge per carrier, $v$ represents drift velocity, and $A$ represents cross-sectional area. The variable $n$ represents the charge carrier number density, defined explicitly as the number of conduction electrons contained within a unit volume of $1\,m^3$ of a given material. It does not represent total electrons, nor is it restricted to a specific wire length

The total number of electrons passing a given point in a conductor over a time interval $t$ depends on total charge $Q$ and elementary charge $e$. Since total charge passing a point is $Q = I t$, dividing total charge by the charge of a single electron $e$ yields the total number of electrons $N = \frac{I t}{e}$

Pulse-echo positioning systems utilize the velocity and travel time of reflected sound waves to measure distance. A sensor emits a pulse that travels to a target object and returns after time $t$. The total distance traversed by the pulse during time $t$ at speed $v$ is $2d = v t$. Solving for the distance between the vehicle and the object yields $d = 0.5 v t$

When two identical transverse pulses of wavelength $\lambda$ travel toward each other on a string at equal speeds, their spatial positions determine the interference pattern. When both waves advance by half a wavelength ($0.5\lambda$), their crests align directly in phase at the midpoint. By the principle of superposition, the resultant displacement is the algebraic sum of individual displacements, producing constructive interference with peak displacement equal to the sum of the individual amplitudes

In a circuit containing a negative temperature coefficient (NTC) thermistor, a decrease in ambient temperature causes a reduction in thermal energy supplied to the crystal lattice. This reduces the number of bound electrons promoted to the conduction band, decreasing the conduction electron number density $n$. With fewer available charge carriers, the thermistor's electrical resistance increases, which reduces total circuit current and causes the ammeter reading to decrease

A potential divider circuit containing three identical resistors of resistance $R$ connected to an e.m.f. source $\varepsilon$ with negligible internal resistance distributes voltage according to equivalent resistance. Two resistors connected in parallel yield an equivalent resistance of $R_p = \frac{R}{2} = 0.5R$. Connected in series with the third resistor $R$, the total circuit resistance becomes $R_{\text{total}} = R + 0.5R = 1.5R = \frac{3}{2}R$. Circuit current is $I = \frac{\varepsilon}{1.5R} = \frac{2\varepsilon}{3R}$. The potential difference across the single series resistor measured by a voltmeter is $V = I R = \left(\frac{2\varepsilon}{3R}\right) R = \frac{2}{3}\varepsilon$

Stationary waves on a string feature fixed spatial variations in displacement amplitude and phase. All points oscillating within the same anti-nodal loop between two adjacent nodes move in phase, resulting in a phase difference of $0^\circ$. However, amplitude varies continuously along the loop from zero at the nodes to a maximum value $A$ at the antinode. Therefore, any two arbitrary points $P$ and $Q$ on the string within a loop (not both at antinodes) have a phase difference of $0^\circ$ and an oscillation amplitude strictly less than $A$

Electrical resistance of a uniform conductor is directly proportional to its length $L$ for a constant cross-sectional area and material composition. If a segment of length $d$ exhibits resistance $R$, the resistance per unit length is $\frac{R}{d}$. Multiplying this unit resistance by the total length $L$ yields the total resistance of the copper wire as $R_{\text{total}} = \frac{L}{d} \times R$

Atomic energy absorption and subsequent photon emission follow discrete energy quantization. An electron in the ground state ($0\,eV$) of a mercury atom absorbing a photon of energy $E = 8.74 \times 10^{-19}\,J$ gains an equivalent energy in electronvolts calculated by $E = \frac{8.74 \times 10^{-19}\,J}{1.60 \times 10^{-19}\,J\,eV^{-1}} = 5.4625\,eV \approx 5.46\,eV$. This promotes the electron directly to the $5.46\,eV$ excited energy level. From the $5.46\,eV$ level, de-excitation can occur via three distinct radiative decay pathways: a direct transition to the ground state ($5.46\,eV \rightarrow 0\,eV$), a transition to an intermediate state ($5.46\,eV \rightarrow 4.91\,eV$), and subsequent transition from that intermediate state to the ground state ($4.91\,eV \rightarrow 0\,eV$). These three distinct energy transitions produce exactly 3 unique frequencies of emitted light

A cell of e.m.f. $E = 9\,V$ connected to an external load resistor $R = 10\,\Omega$ producing a terminal potential difference $V = 6\,V$ loses potential difference across its internal resistance $r$ equal to lost volts $v = E - V = 9\,V - 6\,V$. The circuit current is $I = \frac{V}{R} = \frac{6}{10}$. Applying internal resistance formula $r = \frac{E - V}{I}$ yields $r = \frac{10}{6} \times (9 - 6)$

Quantum Physics and De Broglie Wavelength Analysis

Moving particles exhibit wave-like properties with a characteristic de Broglie wavelength inversely proportional to their linear momentum. For an electron traveling at a velocity of $v = 3.6 \times 10^7\,m\,s^{-1}$, the linear momentum $p$ is calculated using the electron rest mass $m_e = 9.11 \times 10^{-31}\,kg$

$p = m_e v = (9.11 \times 10^{-31}\,kg) \times (3.6 \times 10^7\,m\,s^{-1}) = 3.2796 \times 10^{-23}\,kg\,m\,s^{-1}$

Applying the de Broglie relation $\lambda = \frac{h}{p}$ with Planck's constant $h = 6.63 \times 10^{-34}\,J\,s$

$\lambda = \frac{6.63 \times 10^{-34}\,J\,s}{3.2796 \times 10^{-23}\,kg\,m\,s^{-1}} = 2.0216 \times 10^{-11}\,m$

Thus, the de Broglie wavelength of the electron is $2.02 \times 10^{-11}\,m$

Solar Cell Intensity and Power Calculations

To determine whether a watch battery will charge, the absolute incident optical power on the surface of a solar cell must be evaluated against the device's operational threshold of $0.80\,W$. The solar cell has a total surface area of $A = 6.4\,cm^2$, which converts to square metres as

$A = 6.4 \times 10^{-4}\,m^2$

When light of intensity $I = 750\,W\,m^{-2}$ strikes the solar cell, total incident power $P_{\text{input}}$ is given by the intensity relation $I = \frac{P}{A}$

$P_{\text{input}} = I \times A = 750\,W\,m^{-2} \times 6.4 \times 10^{-4}\,m^2 = 0.48\,W$

Comparing the calculated power input of $0.48\,W$ to the minimum required threshold of $0.80\,W$ demonstrates that $0.48\,W < 0.80\,W$. Consequently, the incident optical power is insufficient, and the watch battery will not charge under an intensity of $750\,W\,m^{-2}$

Temperature Dependence of Resistance and Wire Resistivity

When electrical power is supplied to a metal wire cutting tool, resistive heating increases the thermal energy of the metallic lattice. As temperature rises, positive metal ions vibrate with greater amplitude and kinetic energy about their fixed lattice positions. Conduction electrons moving through the wire experience more frequent collisions with these rapidly vibrating ions. This increases the rate of carrier scattering, obstructing charge flow and causing the overall electrical resistance of the wire to increase

The resistivity $\rho$ of a conductor with circular cross-section depends on resistance $R$, length $l$, and diameter $d$. For a cutting wire with resistance $R = 0.67\,\Omega$, length $l = 0.14\,m$, and diameter $d = 0.51\,mm = 0.51 \times 10^{-3}\,m$, the radius is $r = \frac{d}{2} = 0.255 \times 10^{-3}\,m$

The cross-sectional area $A$ is calculated as

$A = \pi r^2 = \pi \times (0.255 \times 10^{-3}\,m)^2 = 2.0428 \times 10^{-7}\,m^2$

Rearranging the resistivity formula $R = \frac{\rho l}{A}$ to solve for resistivity $\rho$

$\rho = \frac{R A}{l} = \frac{0.67\,\Omega \times 2.0428 \times 10^{-7}\,m^2}{0.14\,m} = 9.776 \times 10^{-7}\,\Omega\,m$

The resistivity of the metal wire at its operating cutting temperature is $9.78 \times 10^{-7}\,\Omega\,m$

Wave Properties and Diffraction Grating Optics

A transverse wave is defined as a wave in which particle oscillations or field vibrations occur perpendicular ($90^\circ$) to the direction of wave propagation and energy transfer

A diffraction grating experiment utilizes monochromatic light of wavelength $\lambda = 650\,nm = 650 \times 10^{-9}\,m$ incident on a grating positioned at a distance $L = 1.30\,m$ from a screen. The observed spatial separation between the central zero-order maximum and the first-order ($n = 1$) maximum is $x = 0.22\,m$. The angle of diffraction $\theta$ for the first-order peak is calculated using trigonometry

$\tan(\theta) = \frac{x}{L} = \frac{0.22\,m}{1.30\,m} = 0.16923 \implies \theta = \arctan(0.16923) = 9.603^\circ$

Evaluating $\sin(\theta)$

$\sin(9.603^\circ) = 0.16686$

Alternatively, using the exact hypotenuse distance $\sqrt{L^2 + x^2} = \sqrt{1.30^2 + 0.22^2} = 1.3185\,m$

$\sin(\theta) = \frac{x}{\sqrt{L^2 + x^2}} = \frac{0.22}{1.3185} = 0.16686$

Applying the diffraction grating equation $n\lambda = d \sin(\theta)$ for $n = 1$

$d = \frac{\lambda}{\sin(\theta)} = \frac{650 \times 10^{-9}\,m}{0.16686} = 3.8955 \times 10^{-6}\,m$

The number of grating lines per millimetre $N_{\text{mm}}$ is the reciprocal of slit spacing $d$ expressed in millimetres

$d = 3.8955 \times 10^{-3}\,mm$

$N_{\text{mm}} = \frac{1}{d} = \frac{1}{3.8955 \times 10^{-3}\,mm} = 256.7\,\text{lines/mm}$

Thus, the diffraction grating contains 257 lines per mm

A first-order maximum forms on the screen because monochromatic light passes through adjacent narrow slits of the grating and undergoes diffraction, spreading outward. As these diffracted wavefronts overlap on the screen, the path difference between light originating from adjacent slits is equal to exactly one whole wavelength ($1\lambda$). Arriving at the screen with a path difference of $1\lambda$, the waves are completely in phase (phase difference of $0$ or $2\pi\,\text{rad}$) and superpose constructively, generating a high-intensity bright line

Circuit Analysis and Non-Ohmic Component Evaluation

A potential divider circuit containing a filament lamp connected in parallel with a $15\,\Omega$ fixed resistor is driven by a series-connected variable resistor and a $4.5\,V$ battery of negligible internal resistance. When the variable resistor is adjusted so the voltmeter across the parallel combination reads $3.6\,V$, the current through each parallel branch can be determined. From the characteristic current-voltage graph of the filament lamp, at potential difference $V = 3.6\,V$, the lamp current is $I_{\text{lamp}} = 0.48\,A$

The current flowing through the fixed $15\,\Omega$ resistor is calculated using Ohm's law

$I_{\text{resistor}} = \frac{V}{R} = \frac{3.6\,V}{15\,\Omega} = 0.24\,A$

The total current $I_{\text{total}}$ entering the parallel combination is the sum of the branch currents

$I_{\text{total}} = I_{\text{lamp}} + I_{\text{resistor}} = 0.48\,A + 0.24\,A = 0.72\,A$

The combined resistance $R_{\text{comb}}$ of the parallel branch is

$R_{\text{comb}} = \frac{V}{I_{\text{total}}} = \frac{3.6\,V}{0.72\,A} = 5.0\,\Omega$

Alternatively, calculating lamp resistance $R_{\text{lamp}} = \frac{3.6\,V}{0.48\,A} = 7.5\,\Omega$ and applying the parallel resistance formula

$\frac{1}{R_{\text{comb}}} = \frac{1}{7.5\,\Omega} + \frac{1}{15\,\Omega} = \frac{2}{15} + \frac{1}{15} = \frac{3}{15}\,\Omega^{-1} \implies R_{\text{comb}} = 5.0\,\Omega$

To determine total power dissipated in the entire circuit when the battery delivers an e.m.f. of $\varepsilon = 4.5\,V$ with negligible internal resistance, total power is the product of total e.m.f. and total current drawn from the battery. Since total circuit current is $I_{\text{total}} = 0.72\,A$

$P_{\text{total}} = \varepsilon \times I_{\text{total}} = 4.5\,V \times 0.72\,A = 3.24\,W$

Hence, total electrical power dissipated in the circuit is $3.24\,W$

LDR Characteristics and Internal Resistance Dynamics

The electromotive force (e.m.f.) of a cell is defined as the total electrical energy converted from chemical energy per unit charge passing through the cell ($\varepsilon = \frac{W}{Q}$

In a circuit where a Light Dependent Resistor (LDR) is connected in series with a cell possessive of internal resistance $r$, a voltmeter placed across the terminals of the cell measures its terminal potential difference $V_{\text{terminal}}$

Increasing the intensity of light incident on the LDR increases photon absorption within its semiconductor substrate. Photons supply energy to liberate bound valence electrons into the conduction band, increasing the free charge carrier number density $n$. This causes the electrical resistance of the LDR ($R_{\text{LDR}}$) to decrease

Because the LDR is in series with the cell's internal resistance $r$, the total circuit resistance $R_{\text{total}} = R_{\text{LDR}} + r$ decreases. With constant cell e.m.f. $\varepsilon$, the total circuit current $I = \frac{\varepsilon}{R_{\text{total}}}$ increases

The potential difference lost across the cell's internal resistance (lost volts) is given by $v_{\text{lost}} = I r$. As current $I$ increases, lost volts $I r$ increase. The terminal potential difference measured by the voltmeter is $V_{\text{terminal}} = \varepsilon - I r$. Because e.m.f. $\varepsilon$ remains constant while lost volts $I r$ increase, the voltmeter reading across the cell terminals decreases

Stationary Waves on Vibrating Strings

Stationary waves feature distinct spatial positions where wave amplitude is fixed. Nodes (N) are points along the wave pattern where destructive interference causes zero displacement amplitude. Antinodes (A) are points where constructive interference produces maximum displacement amplitude. On a string driven at its third harmonic (3 anti-nodal loops), 4 nodes (N) are located at the fixed ends and intermediate zero-displacement boundary positions, while 3 antinodes (A) are located at peak displacement positions in the center of each loop

A stationary wave forms on a string through wave reflection and superposition. The vibration generator produces progressive transverse waves that travel down the length of the string toward the fixed support. Upon reaching the support, the waves reflect back in the opposite direction along the string with identical frequency, wavelength, and speed. As incident and reflected waves traverse each other, they superpose. At locations where waves arrive in phase, constructive interference forms antinodes. At locations where waves arrive $180^\circ$ out of phase, destructive interference forms nodes, establishing a stationary wave pattern

The relationship governing standing wave frequency $f$ and string tension $T$ derives from wave speed equations. The velocity of a transverse wave on a string is $v = \sqrt{\frac{T}{\mu}}$, where $\mu$ is mass per unit length. Combining this with wave speed $v = f\lambda$ yields $f\lambda = \sqrt{\frac{T}{\mu}}$. For a string of fixed length $l$ vibrating at its fundamental mode, $\lambda = 2l$. Substituting for $\lambda$

$f(2l) = \sqrt{\frac{T}{\mu}} \implies f = \frac{1}{2l}\sqrt{\frac{T}{\mu}}$

Squaring both sides produces the linear relation

$f^2 = \frac{1}{4l^2 \mu} T$

Plotting $f^2$ on the y-axis against $T$ on the x-axis yields a straight line passing through the origin with a constant gradient equal to $\frac{1}{4l^2 \mu}$

To determine string mass per unit length $\mu$ from experimental data with fixed length $l = 0.85\,m$, the line gradient $m$ is calculated from graph coordinates $(T, f^2)$ such as $(0\,N, 0\,Hz^2)$ and $(4.0\,N, 2500\,Hz^2)$

$m = \frac{\Delta f^2}{\Delta T} = \frac{2500\,Hz^2 - 0}{4.0\,N - 0} = 625\,Hz^2\,N^{-1}$

Equating experimental gradient to theoretical gradient $\frac{1}{4l^2 \mu}$

$625 = \frac{1}{4(0.85\,m)^2 \mu} = \frac{1}{4(0.7225)\mu} = \frac{1}{2.89 \mu}$

$\mu = \frac{1}{2.89 \times 625} = \frac{1}{1806.25} = 5.536 \times 10^{-4}\,kg\,m^{-1}$

The mass per unit length of the string is $5.54 \times 10^{-4}\,kg\,m^{-1}$

Optical Fibre Wave Propagation and Signal Transmission

Light entering the flat end face of a glass optical fibre ($n_2 = 1.50$) from air ($n_1 = 1.00$) at an angle of incidence $\theta_1 = 50^\circ$ refracts toward the normal. Applying Snell's law

$n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \implies 1.00 \times \sin(50^\circ) = 1.50 \times \sin(\theta_2)$

$\sin(\theta_2) = \frac{\sin(50^\circ)}{1.50} = \frac{0.766044}{1.50} = 0.51070$

$\theta_2 = \arcsin(0.51070) = 30.71^\circ$

The angle of refraction inside the glass fibre is $30.7^\circ$

To deduce whether a ray inside glass ($n_1 = 1.50$) incident on an air boundary ($n_2 = 1.00$) at an angle $\theta = 45^\circ$ undergoes Total Internal Reflection (TIR), the critical angle $C$ must be calculated

$\sin(C) = \frac{n_2}{n_1} = \frac{1.00}{1.50} = 0.6667 \implies C = \arcsin(0.6667) = 41.81^\circ$

Because light travels from an optically denser medium (glass) toward a less dense medium (air), and the angle of incidence $\theta = 45^\circ$ is greater than the critical angle $C = 41.8^\circ$, the ray will undergo total internal reflection at the boundary

When light pulses propagate along an optical fibre, signal degradation occurs via attenuation and modal dispersion. Light intensity leaving the fibre is lower than initial intensity $I$ because optical energy is absorbed by impurities in glass, scattered by density fluctuations, and partially lost through micro-refractions at surface boundary imperfections. Pulse duration leaving the fibre is longer than initial duration $T$ due to modal dispersion. Light rays enter at different angles and follow different paths: axial rays travel straight along the center (shortest path length), whereas marginal rays bounce via multiple internal reflections (longer path length). Traveling at the same speed $v = \frac{c}{n}$, axial rays arrive earlier than marginal rays, causing the output pulse to broaden over time

Adding cladding with a lower refractive index than glass ($n_{\text{cladding}} < n_{\text{glass}}$) alters the boundary conditions compared to an unclad glass-air boundary ($n_{\text{air}} = 1.00$). Because $n_{\text{cladding}} > n_{\text{air}}$, the refractive index ratio $\frac{n_{\text{cladding}}}{n_{\text{glass}}}$ is larger than $\frac{n_{\text{air}}}{n_{\text{glass}}}$. Since $\sin(C) = \frac{n_{\text{boundary}}}{n_{\text{glass}}}$, a higher numerical ratio increases $\sin(C)$, which increases the critical angle $C$ for light inside the core

Photoelectric Effect and Quantum Nature of Light

In a photocell, monochromatic light delivers photons of uniform individual energy $E = hf$. Einstein's photoelectric equation states that maximum kinetic energy is $E_{k,\text{max}} = hf - \Phi$, where $\Phi$ is the work function of the metal. Photoelectrons are emitted with a spectrum of kinetic energies ranging from zero up to $E_{k,\text{max}}$. Surface electrons bound with minimum energy require only $\Phi$ to escape, leaving them with $E_{k,\text{max}}$. Electrons situated deeper below the metal surface require additional energy to overcome lattice interactions and undergo collisions prior to liberation. Energy losses from these internal collisions reduce kinetic energy upon exit, producing a range of kinetic energies

Photoelectric experiments yield observations that contrast classical wave theory with quantum photon theory:

1. Low-intensity white light produces no current
2. High-intensity white light produces no current
3. Low-intensity ultraviolet light produces a current

Classical wave theory predicts that energy is delivered continuously and absorbed over time proportional to wave intensity. Under wave theory, high-intensity white light should accumulate sufficient energy to eject electrons. The complete absence of current under high-intensity white light disproves classical wave theory

These results are explained by the photon model, where light consists of discrete energy quanta with photon energy $E = hf$. Electron emission requires one-to-one interaction between a single photon and a single bound electron. Emission occurs only if single photon energy exceeds work function ($hf \ge \Phi$), defining a threshold frequency $f_0$. White light consists of lower frequencies ($f < f_0$), so individual photons lack sufficient energy ($hf < \Phi$) to release electrons regardless of intensity. Ultraviolet light has a higher frequency ($f_{\text{UV}} > f_0$), so individual UV photons carry energy $hf > \Phi$. Even at low intensity, single UV photons instantly transfer energy to liberate electrons, producing a current

To test current flow when battery connections are reversed to supply a opposing potential difference of $V_{\text{reverse}} = 1.5\,V$, maximum kinetic energy $E_{k,\text{max}}$ must be compared against the stopping work $E_{\text{stop}} = e V_{\text{reverse}}$. The photocell is illuminated by UV light of wavelength $\lambda = 250\,nm = 250 \times 10^{-9}\,m$, and the metal plate has a work function of $\Phi = 6.9 \times 10^{-19}\,J$

Photon energy $E_{\text{photon}}$ is calculated using $h = 6.63 \times 10^{-34}\,J\,s$ and $c = 3.00 \times 10^8\,m\,s^{-1}$

$E_{\text{photon}} = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\,J\,s \times 3.00 \times 10^8\,m\,s^{-1}}{250 \times 10^{-9}\,m} = 7.956 \times 10^{-19}\,J$

Applying Einstein's photoelectric equation for $E_{k,\text{max}}$

$E_{k,\text{max}} = E_{\text{photon}} - \Phi = 7.956 \times 10^{-19}\,J - 6.900 \times 10^{-19}\,J = 1.056 \times 10^{-19}\,J$

The energy required for an electron to overcome the opposing $1.5\,V$ electric potential difference is

$E_{\text{stop}} = e V_{\text{reverse}} = 1.60 \times 10^{-19}\,C \times 1.5\,V = 2.40 \times 10^{-19}\,J$

Comparing kinetic energy to stopping work demonstrates that $1.056 \times 10^{-19}\,J < 2.40 \times 10^{-19}\,J$ ($E_{k,\text{max}} < E_{\text{stop}}$). No photoelectrons possess sufficient kinetic energy to cross the potential barrier and reach the opposite electrode. Consequently, there is no current in the circuit