Dual Nature Of Matter And Radiation – JEE Main PhysicsPractice Questions & PYQs

Einstein's photoelectric equation describes the relationship between the maximum kinetic energy of photoelectrons and the incident photon energy, formulated as $K = hn - \phi = \frac{1}{2}mv^2$, where $\phi$ represents the work function of the metal surface. When the incident frequency increases to $3n$, the new maximum kinetic energy $K'$ becomes $3hn - \phi$. By substituting the initial equation $hn = K + \phi$ into this new expression, the energy becomes $K' = 3(K + \phi) - \phi$, which simplifies to $K' = 3K + 2\phi$. Since the work function $\phi$ is always a positive value, $K'$ must be strictly greater than $3K$. Substituting the velocity terms, the relationship $\frac{1}{2}m(v')^2 > 3(\frac{1}{2}mv^2)$ implies that the new maximum velocity $v'$ is greater than $\sqrt{3}v$.

The number of photons $n$ in a laser pulse is determined by dividing the total energy of the pulse, $E_{total} = P \times t$, by the energy of a single photon, $E_{photon} = \frac{hc}{\lambda}$, which yields the formula $n = \frac{P t \lambda}{hc}$.

Substituting the given values where power $P = 500 \text{ W}$, pulse width $t = 0.06 \text{ s}$, wavelength $\lambda = 660 \times 10^{-9} \text{ m}$, Planck's constant $h = 6.62 \times 10^{-34} \text{ Js}$, and speed of light $c \approx 3 \times 10^8 \text{ m/s}$:

$n = \frac{500 \times 0.06 \times 660 \times 10^{-9}}{6.62 \times 10^{-34} \times 3 \times 10^8}$

$n = \frac{30 \times 660 \times 10^{-9}}{19.86 \times 10^{-26}}$

$n = \frac{19800 \times 10^{-9}}{19.86 \times 10^{-26}} \approx 997 \times 10^{17} \approx 10^{20}$

Einstein's photoelectric equation states that the maximum kinetic energy of emitted photoelectrons is determined by the difference between the incident photon energy and the material's work function, represented as $K_{max} = \frac{hc}{\lambda} - W$. In the first scenario, the incident light has a wavelength $\lambda$, resulting in a kinetic energy of $K_1 = \frac{hc}{\lambda} - W$. When the wavelength is halved to $\lambda/2$, the incident photon energy doubles to $\frac{hc}{\lambda/2} = \frac{2hc}{\lambda}$, yielding a new kinetic energy of $K_2 = \frac{2hc}{\lambda} - W$. Given that the second kinetic energy is three times the first, we substitute the relationship $K_2 = 3K_1$ into the equations to get $\frac{2hc}{\lambda} - W = 3\left(\frac{hc}{\lambda} - W\right)$. Distributing the factor of 3 on the right side produces $\frac{2hc}{\lambda} - W = \frac{3hc}{\lambda} - 3W$. By adding $3W$ to both sides and subtracting $\frac{2hc}{\lambda}$ from both sides, we isolate the terms to find $2W = \frac{hc}{\lambda}$, which simplifies to the final work function value of $W = \frac{hc}{2\lambda}$.

Einstein's photoelectric equation states that the energy of an incident photon is partitioned between the work function of the material and the maximum kinetic energy of the emitted photoelectrons, expressed as $eV = \frac{hc}{\lambda} - \phi_0$. Applying this principle to the first two scenarios gives the equations $eV = \frac{hc}{\lambda_1} - \phi_0$ and $3eV = \frac{hc}{\lambda_2} - \phi_0$. Subtracting the first equation from the second eliminates the constant work function, resulting in $2eV = \frac{hc}{\lambda_2} - \frac{hc}{\lambda_1}$, which simplifies to $eV = \frac{hc}{2} \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \right)$. By substituting this expression for $eV$ back into the first equation, the work function can be determined as $\phi_0 = \frac{hc}{\lambda_1} - \frac{hc}{2} \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \right)$, which simplifies to $\phi_0 = hc \left( \frac{3}{2\lambda_1} - \frac{1}{2\lambda_2} \right)$. For the third wavelength $\lambda_3$, the stopping potential $V'$ satisfies $eV' = \frac{hc}{\lambda_3} - \phi_0$. Substituting the derived value of $\phi_0$ into this equation yields $eV' = \frac{hc}{\lambda_3} - hc \left( \frac{3}{2\lambda_1} - \frac{1}{2\lambda_2} \right)$, ultimately leading to $V' = \frac{hc}{e} \left[ \frac{1}{\lambda_3} + \frac{1}{2\lambda_2} - \frac{3}{2\lambda_1} \right]$.

Einstein's photoelectric equation dictates that the maximum kinetic energy of an emitted electron is the difference between the incident photon energy and the work function, which is expressed as $\frac{hc}{\lambda} = \frac{1}{2}mv^2 + \phi$. When the wavelength is adjusted to $\frac{3\lambda}{4}$, the new incident energy becomes $\frac{4hc}{3\lambda}$, which can be rewritten as $\frac{4}{3}(\frac{1}{2}mv^2 + \phi)$. Equating this to the new kinetic energy $\frac{1}{2}mv_1^2$ plus the work function $\phi$ results in the equation $\frac{1}{2}mv_1^2 = \frac{4}{3}(\frac{1}{2}mv^2 + \phi) - \phi$. Distributing the terms yields $\frac{1}{2}mv_1^2 = \frac{4}{3}(\frac{1}{2}mv^2) + \frac{1}{3}\phi$. Since the work function $\phi$ is a positive quantity, the new kinetic energy must be strictly greater than $\frac{4}{3}(\frac{1}{2}mv^2)$, leading to the following relationship for the new speed:

$v_1 > v \left( \frac{4}{3} \right)^{1/2}$

The Franck-Hertz experiment confirmed the existence of discrete energy levels in atoms by demonstrating that electrons only lose energy in specific, quantized amounts during collisions with gas atoms. The photo-electric effect provides empirical evidence for the particle nature of light, showing that electromagnetic radiation interacts with matter as discrete energy quanta known as photons. The Davison-Germer experiment established the wave nature of electrons by observing diffraction patterns, which experimentally verified the de Broglie hypothesis that particles exhibit wave-like characteristics.

The emission of photoelectrons occurs only when the energy of the incident photons $E = \frac{hc}{\lambda}$ is greater than or equal to the work function $\phi$ of the cathode material, which defines a specific threshold wavelength $\lambda_0 = \frac{hc}{\phi}$. Since the photoelectric current $I$ depends on the number of photons incident per unit time, and this intensity remains constant, the current remains constant for all wavelengths $\lambda \leq \lambda_0$ where emission is possible. For any wavelength exceeding this threshold $\lambda > \lambda_0$, the individual photon energy is insufficient to overcome the work function, resulting in the cessation of photoelectron emission and a current of zero.

The Davisson-Germer experiment empirically validated the wave nature of electrons, which is described by the de Broglie relation $\lambda = \frac{h}{p}$, by demonstrating that electron beams undergo diffraction after scattering from atomic crystal structures. Since diffraction and interference are hallmark properties associated exclusively with wave motion, observing these patterns confirms that electrons possess intrinsic wave-like characteristics. Consequently, the second statement provides the essential theoretical justification for the first, as the observed diffraction in the experiment serves as proof of wave nature precisely because waves are known to exhibit such phenomena.

The behavior of photoelectrons is governed by Einstein's photoelectric equation, $K_{\max} = h v - h v_0$, which establishes that both the maximum kinetic energy and the stopping potential are linear functions of the incident frequency, confirming the second statement is true. Because the equation includes a constant threshold energy $h v_0$, doubling the incident frequency to $2v$ yields a new kinetic energy $K_{\max}' = 2hv - hv_0$, which is greater than twice the original kinetic energy of $2(hv - hv_0)$. Consequently, doubling the frequency leads to a value more than double the initial maximum kinetic energy and stopping potential, proving the first statement false.

Einstein's photoelectric equation, $K_{max} = hv - \phi$, establishes that the maximum kinetic energy of emitted photoelectrons is determined by the frequency $v$ of the incident radiation and the work function $\phi$ of the material. Because X-rays have a significantly higher frequency than ultraviolet light, replacing the source with X-rays directly increases both the maximum kinetic energy $K_{max}$ and the stopping potential $V_0$, which is defined by the relation $eV_0 = K_{max}$. The second statement is false because the distribution of photoelectron speeds from zero to the maximum value arises from the energy lost by electrons as they escape from various depths within the metal or undergo collisions before exiting, not because of the incident light's frequency distribution. This range of speeds is observed even with monochromatic light, meaning the variation is an internal property of the emission process rather than a result of having multiple incident frequencies.

The total power emitted by a source is the product of the number of photons emitted per second and the energy of each individual photon, which follows the relationship $P = \frac{n h \nu}{t}$. By rearranging this equation to isolate the frequency, we obtain the expression $\nu = \frac{P \times t}{n h}$. Substituting the power of 4 x 10^3 W, the emission rate of 10^20 photons per second, and Planck's constant of 6.63 x 10^-34 J s into this formula yields a frequency of 6 x 10^16 Hz. This specific frequency is characteristic of the radiation known as X-rays.

The photoelectric effect describes how incident light transfers energy to electrons in a metal, where the total energy of each incoming photon is either used to liberate an electron, known as the work function, or contributes to the kinetic energy of the ejected photoelectron. This conservation of energy principle is defined by the following relationship:

$E = W + K.E.$

where $E = \frac{hc}{\lambda}$ represents the energy of the incident photon. Calculating the energy of the incoming photon gives 3.1 eV. Using the rearranged equation $W = E - K.E.$ with the provided kinetic energy, the work function is determined to be 1.42 eV, matching the first option provided.

Passing electrons through a narrow slit of width $d$ forces them to exhibit wave-like characteristics, resulting in a single-slit diffraction pattern determined by their de Broglie wavelength. As a consequence of this wave behavior, the detected number of electrons $N$ peaks at the center position $y=0$, where constructive interference is strongest. Because the slit width is comparable to the de Broglie wavelength, the central diffraction peak is quite broad, spreading out such that the distance between the first minima on either side of the central maximum is wider than $d$.

Constructive interference occurs when waves reflected from parallel crystal planes separated by distance $d$ have a path difference equal to an integer multiple of the de Broglie wavelength. Because the incident angle $i$ is defined relative to the normal of the crystal planes, the glancing angle used in the standard diffraction condition is the complement, $90^{\circ} - i$. Applying this geometric relationship to the path difference equation leads to the condition $2d \cos i = n \lambda_{d B}$, which defines the requirement for observing a diffraction peak.

The interaction of an electron beam with the atomic planes of a crystal creates a diffraction pattern that is fundamentally governed by Bragg's law. Because electrons exhibit wave-like behavior characterized by a de Broglie wavelength, their diffraction is described by the relation $2d \sin \theta = n\lambda$, where $d$ is the spacing between the atomic planes and $\theta$ is the glancing angle of incidence.

In this specific setup, the glancing angle $\theta$ is the angle between the incident beam and the crystal planes rather than the crystal surface. Since the angle of incidence relative to the crystal surface is $30^{\circ}$, the glancing angle is $90^{\circ} - 30^{\circ} = 60^{\circ}$. By setting the order of diffraction $n=1$, Bragg's equation simplifies to $2d \sin 60^{\circ} = \lambda$. Substituting the plane spacing $d = 1 \times 10^{-10} \text{ m}$ and the value of $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$, the required wavelength is $\lambda = 2 \times 1 \times 10^{-10} \times \frac{\sqrt{3}}{2} = \sqrt{3} \times 10^{-10} \text{ m}$.

To determine the necessary accelerating potential $V$, we use the relation $\lambda = \frac{12.27}{\sqrt{V}} \times 10^{-10} \text{ m}$. Equating this to the calculated wavelength gives $\sqrt{3} \times 10^{-10} = \frac{12.27}{\sqrt{V}} \times 10^{-10}$, which simplifies to $\sqrt{V} = \frac{12.27}{\sqrt{3}}$. Squaring both sides yields $V = \frac{(12.27)^2}{3}$, resulting in a value of approximately 50 V.

To determine the momentum of a photon, we link its energy to its effective mass using Einstein's mass-energy equivalence. Since the energy of a photon is expressed as $E = hv$ and Einstein's relationship provides $E = mc^2$, we equate these values as follows:
$hv = mc^2$
By rearranging this equation to isolate the product $mc$, which represents the momentum $p$, we divide both sides by the velocity of light $c$ to obtain the final expression $p = \frac{hv}{c}$.

The photoelectric current is fundamentally determined by the kinetic energy of emitted electrons, which is directly linked to the energy of incoming photons expressed as $E = \frac{hc}{\lambda}$. When the wavelength $\lambda$ of the light is reduced, the frequency and energy of the photons increase, imparting greater velocity to the photoelectrons. These more energetic electrons possess a higher probability of successfully reaching the anode, which leads to a direct increase in the plate current $I$.

The photoelectric effect relies on the immediate transfer of energy from an incident photon to an electron, which contradicts classical wave theory predictions of a time-delayed energy accumulation. Since this interaction involves a direct collision, the emission of a photoelectron is essentially instantaneous, occurring within approximately $10^{-10} \mathrm{s}$ after the photon strikes the surface.

The photoelectric effect dictates that the energy of an incident photon is shared between overcoming the material's threshold work function and providing kinetic energy to the ejected electron, expressed by the equation $E = \phi + eV_0$. Given a threshold energy of 6.2 eV and a stopping potential of 5 V, the incident photon energy equals 11.2 eV.

This energy correlates to a specific wavelength through the relation $E = \frac{hc}{\lambda}$. Solving for $\lambda$ with these parameters yields a wavelength on the order of $10^{-7}$ m, which is approximately 100 nm. Because electromagnetic radiation with wavelengths in the range of 10 nm to 400 nm is classified as ultraviolet, the incident light belongs to the ultraviolet region.

The intensity of light radiating from a point source follows the inverse square law, which states that intensity is inversely proportional to the square of the distance from the source, represented as $I \propto \frac{1}{r^2}$. When the light source moves from a distance of 1 m to a distance of 0.5 m, the intensity increases by a factor of four according to the calculation
$\frac{I_2}{I_1} = \left( \frac{r_1}{r_2} \right)^2 = \left( \frac{1}{0.5} \right)^2 = 4$
Since the number of photoelectrons emitted per unit time is directly proportional to the intensity of the incident light, increasing the intensity by a factor of four causes the number of emitted photoelectrons to increase by that same factor.