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

The phenomenon follows Einstein's photoelectric equation, which states that the energy of an incident photon is split between the metal's work function and the maximum kinetic energy of the emitted photoelectrons. First, determine the energy of the incident UV light using the equation $E = \frac{hc}{\lambda}$. Substituting the given values results in an energy of 4.13 eV. Subtracting the work function of 2.13 eV from this photon energy provides the maximum kinetic energy, $K_{max} = 4.13 - 2.13 = 2$ eV. Because the stopping potential $V_s$ is related to this kinetic energy by the equation $K_{max} = eV_s$, the potential required to halt the electrons is 2 V.

The total energy output of a laser per unit time is determined by the number of photons emitted per second multiplied by the energy of an individual photon. Because light exists as discrete packets known as quanta, the relationship between power, frequency, and photon count is expressed as $P = n h \nu$.

Rearranging this equation yields the expression $n = \frac{P}{h \nu}$ for the rate of emission. Substituting the power of $2 \times 10^{-3}$ W and dividing by the product of Planck's constant $6.63 \times 10^{-34}$ Js and the frequency $6 \times 10^{14}$ Hz results in the equation $n = \frac{2 \times 10^{-3}}{6.63 \times 10^{-34} \times 6 \times 10^{14}}$. Evaluating this quotient leads to a total of $5 \times 10^{15}$ photons emitted each second.

Photodiodes function based on the principle that an incident photon must possess energy at least equal to the band gap energy to promote an electron from the valence band to the conduction band, thereby enabling electrical conduction. The threshold wavelength represents the specific limit where the incident photon energy is exactly equal to the band gap, defined by the relationship $E_g = \frac{hc}{\lambda}$.

To determine the band gap energy in electron-volts, we use the provided constants h = 6.6 x 10^-34 Js, c = 3 x 10^8 m/s, and a wavelength of 660 x 10^-9 m. By incorporating the elementary charge e = 1.6 x 10^-19 C into the energy equation, we get $E_g = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{660 \times 10^{-9} \times 1.6 \times 10^{-19}}$ eV. Simplifying this expression, the powers of 10 cancel out, leaving $E_g = \frac{19.8}{10.56}$ eV, which calculates to 1.875 eV. Expressing 1.875 as a fraction with 8 in the denominator yields $\frac{15}{8}$ eV. Matching this result with the provided form $\frac{X}{8}$ eV confirms that X = 15.

Einstein's photoelectric equation dictates that the energy of an incoming photon is equal to the sum of the work function of a photosensitive material and the maximum kinetic energy of the emitted photoelectrons. Because the stopping potential is 0.5 V, the maximum kinetic energy of the electrons equates to 0.5 eV. This relationship is represented by the formula:

$E = \varphi + eV_s$

By substituting the incident photon energy of 2.48 eV and the energy equivalent of the stopping potential, 0.5 eV, into the equation, the work function can be determined. Rearranging the formula to $\varphi = E - eV_s$ and subtracting 0.5 eV from 2.48 eV yields a final work function of 1.98 eV.

Light exhibits a dual nature, behaving as both a wave and a stream of particles depending on the context of the interaction. While interference, diffraction, polarization, and reflection are successfully described by wave mechanics, the photoelectric effect uniquely demonstrates the particle nature of light through the interaction of discrete energy packets, known as photons, with electrons. This interaction is governed by the relation $E = h\nu$, where $E$ is the energy of a photon, $h$ is Planck's constant, and $\nu$ is the frequency of light, which confirms that this specific phenomenon cannot be accounted for by wave theory alone.

The de-Broglie wavelength relates to momentum through the expression $\lambda = \frac{h}{p}$, which implies that particles sharing the same wavelength must possess identical momentum. Because kinetic energy is linked to momentum via the equation $K = \frac{p^2}{2m}$, it follows that kinetic energy is inversely proportional to mass when momentum remains constant. This inverse relationship determines the ratio between the kinetic energies of the proton and the electron:
$\frac{K_p}{K_e} = \frac{m_e}{m_p}$
Substituting the given fact that the proton's mass is $1836$ times that of the electron, the ratio becomes 1:1836.

Electromagnetic radiation carries momentum proportional to its energy, a fundamental property of light. When a beam strikes a perfectly absorbing surface, it imparts all its momentum to the material. This total momentum transfer is defined by the relationship $p = \frac{E}{c}$, where E represents the incident energy and c is the speed of light.

By substituting the provided values-an energy of 6.48 × 10^5 J and a speed of light of 3 × 10^8 m/s-the magnitude of the momentum delivered to the surface is calculated as follows:

$p = \frac{6.48 \times 10^5}{3 \times 10^8} = 2.16 \times 10^{-3} \text{ kg m/s}$

The threshold frequency of a metal defines the minimum frequency of incident light necessary to cause photoelectric emission, a value determined by the relationship $\phi_0 = h \nu_0$. Calculating this frequency involves converting the work function from eV to Joules by multiplying 6.63 eV by 1.6 x 10^-19 J/eV. Substituting this energy and Planck's constant of 6.63 x 10^-34 J s into the rearranged equation $\nu_0 = \frac{\phi_0}{h}$ yields the numerical expression $\nu_0 = \frac{6.63 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}}$. This operation simplifies to a threshold frequency of $\nu_0 = 1.6 \times 10^{15}$ Hz.

According to the principles of the photoelectric effect, the saturation photoelectric current is directly proportional to the intensity of the incident light, while the stopping potential depends solely on the wavelength of that light. Because the two light beams share the same wavelength, they must exhibit the same stopping potential, meaning the current curves will intersect the horizontal axis at the identical negative voltage. Since $I_2 > I_1$, the curve representing the higher intensity $I_2$ reaches a greater saturation level than the curve for $I_1$, resulting in two curves that originate from the same stopping potential but flatten out at different vertical positions.

Einstein's photoelectric equation establishes that the maximum kinetic energy of a photo-emitted electron is directly determined by the frequency of the incoming light through the relation:
$E_k = hf - \phi$
When plotting this expression on a graph where the vertical axis represents kinetic energy and the horizontal axis represents frequency, the equation mirrors the slope-intercept form of a linear function, $y = mx + c$. In this linear representation, the coefficient of the frequency variable is Planck's constant $h$, which dictates the steepness of the line. Consequently, the physical slope of this graph corresponds directly to Planck's constant, such that the trigonometric tangent of the angle $\theta$ formed by the line with the frequency axis is $\tan \theta = h$.

For the photoelectric effect to occur, the incident light frequency must be equal to or greater than the material's threshold frequency $f_0$. Given that the light starts at $1.5 f_0$ and is subsequently halved to $0.75 f_0$, the new frequency fails to reach the required threshold. Consequently, the photons possess insufficient energy to eject any electrons from the surface regardless of the change in intensity, meaning the emission of photoelectrons drops to zero.

Einstein's photoelectric equation states that incident photon energy equals the sum of the work function and the maximum kinetic energy, modeled as $E = \frac{hc}{\lambda} = \frac{hc}{\lambda_0} + eV_0$. Applying this to the first case with a stopping potential of $8V$ yields the expression
$8e = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$
while for the second case with wavelength $3\lambda$ and stopping potential $2V$, the relationship becomes
$2e = \frac{hc}{3\lambda} - \frac{hc}{\lambda_0}$
Multiplying this second equation by $3$ results in $6e = \frac{hc}{\lambda} - \frac{3hc}{\lambda_0}$. Subtracting this transformed second equation from the first expression eliminates the $\frac{hc}{\lambda}$ term, leaving $2e = \frac{2hc}{\lambda_0}$, which simplifies directly to $e = \frac{hc}{\lambda_0}$. Substituting this identity back into the first equation results in $8e = \frac{hc}{\lambda} - e$, leading to $9e = \frac{hc}{\lambda}$. Since $e = \frac{hc}{\lambda_0}$, the final equation is $\frac{9hc}{\lambda_0} = \frac{hc}{\lambda}$, which confirms that the threshold wavelength for the surface is $9\lambda$.

The total intensity of light is defined by the energy delivered per unit time per unit area, given by the relationship $I = \frac{nh\nu}{A}$, where $n$ represents the number of photons incident on the surface. If intensity remains constant, increasing the frequency $\nu$ necessitates a decrease in the number of photons $n$, which invalidates the claim made in Assertion A. Conversely, Reason R is accurate because the photoelectric equation $k_{max} = h\nu - \phi$ demonstrates that the maximum kinetic energy $k_{max}$ of emitted electrons increases linearly with the frequency of the incident radiation, provided the work function $\phi$ of the metal remains constant.

The relationship between the stopping potential $V_0$ and the frequency $v$ of incident radiation is governed by Einstein's photoelectric equation, expressed as $eV_0 = hv - \phi$. By rearranging this into the linear form $V_0 = (\frac{h}{e})v - (\frac{\phi}{e})$, it becomes clear that the slope of the graph is the constant ratio $\frac{h}{e}$, confirming the validity of the first statement. Because the maximum kinetic energy of emitted photoelectrons is given by $K_{max} = hv - \phi$, a material with a larger work function $\phi$ results in lower kinetic energy for the same incident frequency; since material $M_2$ has a higher work function than $M_1$, it actually produces photoelectrons with less kinetic energy, making the second statement incorrect.

The de Broglie wavelength is related to momentum by the equation $\lambda = \frac{h}{p}$, meaning that particles with the same wavelength must also possess identical momentum. When expressing kinetic energy in terms of momentum using the formula $K = \frac{p^2}{2m}$, it becomes clear that for a fixed momentum, kinetic energy is inversely proportional to mass. Because the mass of a proton is much greater than that of an electron, the proton must have a lower kinetic energy than the electron, resulting in the relationship $K_p < K_e$.

The photoelectric effect dictates that the energy of an incident photon is distributed between the energy required to remove an electron from a metal surface-the work function-and the kinetic energy gained by the ejected electron. This physical balance is expressed as $K.E._{max} = E - \Phi$. Given that the incident ultraviolet light carries 4.13 eV of energy and the metal has a work function of 3.13 eV, subtracting the latter from the former provides the maximum kinetic energy. This calculation yields a final value of 1 eV.

The stopping potential, $V_0$, serves as the retarding potential necessary to halt even the fastest photoelectrons, and it is fundamentally related to incident light via the expression $eV_0 = h\nu - \phi_0$. According to this equation, the maximum kinetic energy and the resulting stopping potential depend strictly on the frequency of the incoming light $\nu$ and the specific work function $\phi_0$ of the emitter material. Intensity is a measure of the number of photons hitting the surface per unit time, which influences the total current rather than the energy per photoelectron. Therefore, the stopping potential remains completely independent of the intensity of incident radiation.

The De Broglie wavelength of a particle is fundamentally related to its momentum and kinetic energy through the equation $\lambda = \frac{h}{\sqrt{2mK}}$. Since Planck's constant $h$ and the kinetic energy $K$ are identical for the electron, proton, and alpha particle, the wavelength depends entirely on the mass $m$ of each particle. Specifically, the relationship $\lambda \propto \frac{1}{\sqrt{m}}$ indicates that lighter particles possess longer wavelengths. Because the mass of an electron is the smallest and the mass of an alpha particle is the largest, with the proton in between, the resulting wavelengths follow the inverse order of $\lambda_\alpha < \lambda_p < \lambda_e$.

Photoelectric emission requires that incident photons possess at least the minimum energy defined by the material's work function. Since photon energy is inversely proportional to wavelength, the longest wavelength capable of ejecting an electron corresponds exactly to the threshold energy where the photon energy equals the work function.

Calculating this threshold involves the relationship $\lambda = \frac{hc}{W}$, where $W$ is the work function and the constant $hc$ is approximately $1240 \text{ eV}\cdot\text{nm}$. Substituting the given work function of $3.0 \text{ eV}$ into the expression $\lambda = \frac{1240 \text{ eV}\cdot\text{nm}}{3.0 \text{ eV}}$ yields a result of approximately 413.33 nm, which aligns with the value of 414 nm.

According to the de Broglie hypothesis, moving particles such as electrons possess wave-like characteristics, including the ability to undergo interference and diffraction, with an associated wavelength defined by $\lambda = \frac{h}{p}$. The Davisson-Germer experiment provided the landmark empirical validation of this theory by observing clear diffraction patterns when a beam of electrons was scattered off a crystalline nickel target. Since this experiment serves as the essential evidence that proves the existence of the wave-like properties mentioned in the assertion, the reason provides the direct scientific basis that justifies the assertion.