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

The total power $P$ delivered by a source is the product of the number of photons emitted per second, $n$, and the energy of each individual photon, which is determined by the equation $P = \frac{nhc}{\lambda}$ where $c$ is the speed of light and $h$ is Planck's constant. Because the energy of a single photon is inversely proportional to its wavelength, photons with shorter wavelengths carry more energy, meaning fewer are needed to produce the same total power. Given that both light sources operate at an identical power of 200 W, the number of photons emitted per second is directly proportional to the wavelength of the emitted light. Consequently, the ratio of the number of photons for the X-ray source to the visible light source is simply the ratio of their wavelengths, calculated as 1 nm divided by 500 nm. This results in a final ratio of 1/500.

The phenomenon of electron diffraction is a direct consequence of the wave-particle duality of matter, where high-energy electrons behave as waves with a characteristic wavelength known as the De Broglie wavelength. For diffraction to occur, the incident electron beam must interact with the periodic structure of a crystal lattice, satisfying Bragg's Law for constructive interference, which is given by $2 d \sin \theta = \lambda$, where $d$ is the atomic spacing, $\theta$ is the scattering angle, and $\lambda$ is the wavelength. Because the wavelength of an electron relates to its kinetic energy $E$ through the expression $\lambda = \frac{h}{\sqrt{2 m E}}$, these principles can be combined to determine the energy of the electrons required to produce a specific diffraction pattern.

Equating the two expressions for the wavelength results in the fundamental relationship $2 d \sin \theta = \frac{h}{\sqrt{2 m E}}$. Given that the atomic spacing $d$ is $1 \text{ \AA}$ or $1 \times 10^{-10} \text{ m}$ and the scattering angle $\theta$ is $60^{\circ}$, the equation becomes $2 \times 10^{-10} \times \sin 60^{\circ} = \frac{6.64 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times E}}$. Using the value $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$, the left side simplifies to $10^{-10} \sqrt{3}$. Squaring both sides allows us to isolate the energy, leading to $E = \frac{h^2}{2 m (2 d \sin \theta)^2}$. Substituting the provided constants and solving for the numerical value yields an energy of approximately $50 \text{ eV}$ when expressed in electron-volts after dividing by the conversion factor of $1.6 \times 10^{-19} \text{ J/eV}$.

The emission of photoelectrons from a metal surface is governed by the conservation of energy, where the incident photon energy is divided between the metal's work function, denoted as $\phi$, and the kinetic energy of the ejected electron, represented by $K_{max} = \frac{hc}{\lambda} - \phi$. When these electrons enter a uniform magnetic field, they are forced into a circular trajectory with a radius determined by $R = \frac{\sqrt{2 m K_{max}}}{e B}$. By rearranging this equation to isolate the kinetic energy, we obtain $K_{max} = \frac{R^2 e^2 B^2}{2m}$.

Using the provided wavelength of 6561 Å, the energy of the incident photon is approximately 1.89 eV. Substituting the magnetic field of 3 x 10^-4 T and the radius of 10 mm into the rearranged expression for kinetic energy yields a value of approximately 0.79 eV for the electrons. Subtracting this kinetic energy from the incident photon energy gives the work function, which results in a value of 1.1 eV.

Einstein's photoelectric equation dictates that the maximum kinetic energy of an emitted photoelectron is defined by $K_{max} = \frac{hc}{\lambda} - \phi$, where $h$ is Planck's constant, $c$ is the speed of light, $\lambda$ is the wavelength of the incident radiation, and $\phi$ is the work function of the metal. For the initial condition with a wavelength of $500\ \text{nm}$, the maximum kinetic energy is $K_1 = \frac{hc}{500} - \phi$. When the wavelength is reduced to $200\ \text{nm}$, the maximum kinetic energy becomes $K_2 = \frac{hc}{200} - \phi$. Since the problem states that the second kinetic energy is three times the first, we have the relationship $K_2 = 3K_1$.

By substituting the energy constant $hc = 1240\ \text{eV}\cdot\text{nm}$ into the equation $\frac{hc}{200} - \phi = 3\left(\frac{hc}{500} - \phi\right)$, we can determine the work function. Evaluating the terms gives $\frac{1240}{200} - \phi = 3\left(\frac{1240}{500} - \phi\right)$, which simplifies to $6.2 - \phi = 3(2.48 - \phi)$. Expanding the right side results in $6.2 - \phi = 7.44 - 3\phi$. Solving this linear equation for $\phi$ leads to a work function of $0.61\ \text{eV}$.

Determining the change in the de-Broglie wavelength requires finding the final velocity of the particle using the laws of conservation of linear momentum and the restitution condition for elastic collisions. By equating the initial momentum to the total momentum after the collision, we have $\frac{m}{2} v_0 = \frac{m}{2} v_A + \frac{m}{3} v_B$, which simplifies to $3v_0 = 3v_A + 2v_B$. Because the collision is elastic, the relative velocity of separation equals the relative velocity of approach, giving $v_B - v_A = v_0$, or $v_B = v_0 + v_A$.

Substituting this expression for $v_B$ into the momentum equation leads to $3v_0 = 3v_A + 2(v_0 + v_A)$, which results in $3v_0 = 5v_A + 2v_0$, identifying the final velocity of particle A as $v_A = \frac{v_0}{5}$. The de-Broglie wavelength is defined by $\lambda = \frac{h}{p}$, so the initial wavelength is $\lambda_0 = \frac{h}{(m/2)v_0} = \frac{2h}{mv_0}$. Following the collision, the final wavelength is $\lambda_f = \frac{h}{(m/2)(v_0/5)} = \frac{10h}{mv_0}$. The difference is $\Delta \lambda = \lambda_f - \lambda_0 = \frac{10h}{mv_0} - \frac{2h}{mv_0} = \frac{8h}{mv_0}$, which simplifies to $\Delta \lambda = 4 \lambda_0$.

Einstein's photoelectric equation states that the energy of an incident photon equals the sum of the work function of the metal and the maximum kinetic energy of the emitted photoelectrons. Since kinetic energy is proportional to the square of velocity, the ratio of the kinetic energies is the square of the ratio of the maximum speeds, implying that the kinetic energy in the first case is four times greater than in the second case.

Applying this principle, the energy balance for the two cases is expressed as $E_1 = \phi + K_1$ and $E_2 = \phi + K_2$. Given the input energies, these equations become $4 = \phi + K_1$ and $2.5 = \phi + K_2$ in units of eV, while the velocity ratio establishes $K_1 = 4K_2$. Substituting the relationship $K_1 = 4K_2$ into the first equation yields $4 = \phi + 4K_2$. Subtracting the second equation, $2.5 = \phi + K_2$, from this result leads to $1.5 = 3K_2$, so $K_2 = 0.5$ eV. Inserting this kinetic energy value back into the expression for the second case confirms the work function is 2.00 eV.

The de Broglie wavelength of a particle is fundamentally linked to its momentum by the relation $\lambda = \frac{h}{p}$. Since all the particles in this scenario possess identical kinetic energy, it is most effective to express this relationship in terms of mass and kinetic energy using the substitution $p = \sqrt{2m(KE)}$, which transforms the wavelength formula into $\lambda = \frac{h}{\sqrt{2m(KE)}}$. This equation clearly shows that for a constant kinetic energy, the wavelength is inversely proportional to the square root of the particle's mass, or $\lambda \propto \frac{1}{\sqrt{m}}$. Comparing the masses of these particles, we know that the doubly ionized helium ion is the heaviest, followed by the proton, while the electron has the smallest mass, resulting in the mass relationship $m_{\mathrm{He}^{++}} > m_{P} > m_{e}$. Because a smaller mass leads to a larger de Broglie wavelength, the electron exhibits the greatest wavelength and the helium ion exhibits the smallest, establishing the sequence $\lambda_{e} > \lambda_{P} > \lambda_{\mathrm{He}^{++}}$.

The de-Broglie wavelength is fundamentally linked to a particle's momentum through the relation $\lambda = \frac{h}{mv}$. As an electric field exerts a force on an electron, it modifies the particle's velocity, which in turn alters its momentum and wavelength. By determining how the electron's velocity changes over time under the influence of the electric field, one can derive the corresponding evolution of its de-Broglie wavelength.

Initially, the electron possesses a velocity $\vec{v} = v_0 \hat{i} + v_0 \hat{j}$, resulting in an initial speed of $v_i = \sqrt{2}v_0$ and an initial wavelength of $\lambda_0 = \frac{h}{m\sqrt{2}v_0}$. Under the influence of the electric field $\vec{E} = -E_0 \hat{k}$, the electron experiences a constant force $\vec{F} = -e\vec{E} = eE_0 \hat{k}$, which generates an acceleration $\vec{a} = \frac{eE_0}{m} \hat{k}$. Since there is no initial velocity component along the z-axis, the velocity at time $t$ is $\vec{v}(t) = v_0 \hat{i} + v_0 \hat{j} + (\frac{eE_0t}{m}) \hat{k}$.

Calculating the magnitude of this velocity gives the speed at time $t$ as $v = \sqrt{v_0^2 + v_0^2 + (\frac{eE_0t}{m})^2} = \sqrt{2v_0^2 + \frac{e^2E_0^2t^2}{m^2}}$. Inserting this speed into the de-Broglie equation, the wavelength at time $t$ becomes $\lambda' = \frac{h}{m\sqrt{2v_0^2 + \frac{e^2E_0^2t^2}{m^2}}}$. Factoring $2v_0^2$ out from inside the square root in the denominator yields $\lambda' = \frac{h}{m\sqrt{2}v_0 \sqrt{1 + \frac{e^2E_0^2t^2}{2m^2v_0^2}}}$. Since the term $\frac{h}{m\sqrt{2}v_0}$ is equal to $\lambda_0$, the final expression is $\lambda' = \frac{\lambda_0}{\sqrt{1 + \frac{e^2E^2t^2}{2m^2v_0^2}}}$.

Einstein's photoelectric equation states that the maximum kinetic energy of emitted photoelectrons is the difference between the energy of the incident photon and the work function of the metal surface. Since the stopping potential $V$ is directly proportional to this maximum kinetic energy, the relationship is given by $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$, where $\lambda_0$ represents the threshold wavelength of the metal.

Applying this to the first scenario, we have $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$. When the incident wavelength is changed to $3\lambda$ and the stopping potential becomes $V/4$, the equation becomes $\frac{eV}{4} = \frac{hc}{3\lambda} - \frac{hc}{\lambda_0}$. To compare these two conditions, we can multiply the second equation by 4 to isolate $eV$, resulting in $eV = \frac{4hc}{3\lambda} - \frac{4hc}{\lambda_0}$.

Equating these two expressions for $eV$ allows us to solve for the threshold wavelength by writing $\frac{hc}{\lambda} - \frac{hc}{\lambda_0} = \frac{4hc}{3\lambda} - \frac{4hc}{\lambda_0}$. Rearranging the equation to group the terms with $\lambda_0$ on one side gives $\frac{4hc}{\lambda_0} - \frac{hc}{\lambda_0} = \frac{4hc}{3\lambda} - \frac{hc}{\lambda}$. Simplifying both sides leads to $\frac{3hc}{\lambda_0} = \frac{hc}{3\lambda}$, which reduces to $\frac{3}{\lambda_0} = \frac{1}{3\lambda}$. Consequently, $\lambda_0 = 9\lambda$, which identifies $n$ as 9.

The de Broglie wavelength of an electron is inversely proportional to the square root of its maximum kinetic energy, based on the relationship $\lambda = \frac{h}{\sqrt{2mK}}$. Given the condition $\lambda_B = 2\lambda_A$, we can equate the ratios $\frac{h}{\sqrt{2m T_B}} = 2 \left( \frac{h}{\sqrt{2m T_A}} \right)$, which simplifies by squaring both sides to the relationship $T_A = 4T_B$.

The problem states that the kinetic energy of photoelectrons from metal B is $T_B = T_A - 1.5$ eV. Substituting the relationship $T_A = 4T_B$ into this expression yields $T_B = 4T_B - 1.5$ eV. Rearranging this equation to solve for the kinetic energy gives $3T_B = 1.5$ eV, which means the maximum kinetic energy for metal B is $T_B = 0.5$ eV.

Applying the photoelectric equation, the maximum kinetic energy for metal B is the difference between the incident photon energy and its work function, expressed as $T_B = 4.5 - \phi_B$. Substituting the previously determined kinetic energy $T_B = 0.5$ eV into this equation, we get $0.5 = 4.5 - \phi_B$. This leads directly to the conclusion that the work function $\phi_B$ is 4 eV.

The de-Broglie wavelength of a particle is defined by the ratio of the Planck constant to its momentum, expressed as $\lambda = \frac{h}{mv}$. For a nitrogen molecule at a temperature of $400 \text{ K}$, the root-mean-square velocity provides the appropriate velocity term, leading to the substitution $v = \sqrt{\frac{3 k_B T}{m}}$, which simplifies the wavelength formula to $\lambda = \frac{h}{\sqrt{3 m k_B T}}$.

By substituting the provided mass of $4.64 \times 10^{-26} \text{ kg}$, the Boltzmann constant of $1.38 \times 10^{-23} \text{ J/K}$, and the temperature of $400 \text{ K}$ into this expression, the product within the square root evaluates to $7.68 \times 10^{-46}$. Taking the square root of this value yields approximately $2.77 \times 10^{-23}$, and dividing the Planck constant $6.63 \times 10^{-34} \text{ J}\cdot\text{s}$ by this denominator results in a wavelength of roughly $2.39 \times 10^{-11} \text{ meters}$. Converting this distance into angstroms leads to a value of $0.24 \text{ \AA}$.

The de Broglie wavelength $\lambda$ relates to kinetic energy $E$ as $\lambda = \frac{h}{\sqrt{2mE}}$, which establishes that the wavelength is inversely proportional to the square root of the energy. Since the new wavelength is $\frac{\lambda}{2}$, the ratio of the initial state to the final state implies that $2 = \sqrt{\frac{E + \Delta E}{E}}$. Squaring both sides of this equation yields $4 = \frac{E + \Delta E}{E}$, which simplifies to $4 = 1 + \frac{\Delta E}{E}$. Solving for the added energy gives $\Delta E = 3E$.

The emission of photoelectrons is governed by the energy balance where each incident photon must provide enough energy to overcome the material's work function, with the final rate of electron emission determined by the photon flux and the surface's quantum efficiency. First, determine the energy of an individual incident photon using the relation $E = \frac{hc}{\lambda}$ Substituting the provided values gives $E = \frac{1240 \, \text{eVnm}}{310 \, \text{nm}} = 4 \, \text{eV}$. Since the energy of each photon is $4 \, \text{eV}$ and the work function is $2 \, \text{eV}$, photoelectric emission occurs because the incident energy exceeds the work function. To perform the subsequent rate calculation \in SI units, convert this energy to Joules: $E = 4 × 1.6 × 10^{-19} \, \text{J} = 6.4 × 10^{-19} \, \text{J}$.

The total incident power on the surface is the product of intensity and area, which amounts to $6.4 \times 10^{-5} \, \text{W/cm}^2 \times 1 \, \text{cm}^2 = 6.4 \times 10^{-5} \, \text{J/s}$. The number of photons striking the surface per second, $n$, is the ratio of this total power to the energy per photon: $n = \frac{6.4 \times 10^{-5} \, \text{J/s}}{6.4 \times 10^{-19} \, \text{J}} = 10^{14} \, \text{photons/s}$ With the efficiency specified as one electron ejected for every $10^3$ incident photons, multiply the total photon flux by the efficiency factor of $10^{-3}$ to find the rate of electron emission: $10^{14} \times 10^{-3} = 10^{11} \, \text{electrons/s}$ Matching this result to the given expression $10^x$ identifies the value as $x = 11$.

Einstein's photoelectric equation, $eV = \frac{hc}{\lambda} - \phi$, establishes that the stopping potential $V$ for a given wavelength $\lambda$ is determined entirely by Planck's constant $h$, the speed of light $c$, the electronic charge $e$, and the work function $\phi$. Because these parameters are fixed properties of the metal and universal constants, they are completely independent of the intensity of incident radiation. Since increasing light intensity only raises the number of incident photons and the resulting photocurrent without changing the energy of individual photoelectrons, the slope and intercept of the graph remain entirely unchanged.

Einstein's photoelectric equation relates the incident photon energy to the work function of a metal and the maximum kinetic energy of the emitted electrons, expressed as $h\nu = \phi + eV_s$. In this context, the work function $\phi$ represents the minimum energy required to eject an electron from the surface, which occurs at the threshold frequency where the stopping potential $V_s$ is zero.

Identifying the threshold frequency from the graph as 5.5 x 10^14 Hz, we use the relation $\phi = h\nu_0$ to determine the energy. To calculate this value, we multiply Planck's constant, 6.62 x 10^-34 J·s, by the threshold frequency, 5.5 x 10^14 Hz. Since the answer is required in electron-volts, we divide the resulting energy in Joules by the elementary charge of 1.6 x 10^-19 C.

$\phi = \frac{h \nu_0}{e}$

Substituting the known values into the equation, we perform the calculation: 6.62 x 10^-34 multiplied by 5.5 x 10^14, divided by 1.6 x 10^-19. This results in an energy value of 2.27 eV.

An electron subjected to a constant electric field $E$ experiences a constant force $F = |e|E$, which imparts an acceleration $a = \frac{|e|E}{m}$ according to Newton's second law. Since the electron starts from rest, its velocity at time $t$ increases linearly as $v = \frac{|e|Et}{m}$. By substituting this velocity into the de-Broglie wavelength formula $\lambda = \frac{h}{mv}$, we find that $\lambda = \frac{h}{|e|Et}$. Differentiating $\lambda$ with respect to time $t$ provides the rate of change of the wavelength:

$\frac{d\lambda}{dt} = -\frac{h}{|e|Et^2}$

A photon with energy $E$ is characterized by the wavelength $\lambda_{photon} = \frac{hc}{E}$. For an electron of mass $m$ and energy $E$, the de-Broglie wavelength is defined as $\lambda_{electron} = \frac{h}{\sqrt{2mE}}$, which stems from the fundamental relationship between kinetic energy and momentum where $E = \frac{p^2}{2m}$ and $\lambda = \frac{h}{p}$.

Dividing the electron's wavelength by the photon's wavelength eliminates the Planck constant $h$, resulting in the expression $\frac{\lambda_{electron}}{\lambda_{photon}} = \frac{h}{\sqrt{2mE}} \cdot \frac{E}{hc}$. Canceling the common factors simplifies the ratio to $\frac{E}{c\sqrt{2mE}}$, which further reduces to the final form of $\frac{1}{c} \left( \frac{E}{2m} \right)^{1/2}$.

The de Broglie wavelength is defined by the relationship $\lambda = \frac{h}{mv}$, where $h$ is Planck's constant, $m$ is the mass, and $v$ is the velocity. Expressing the ratio of the wavelength of a particle to that of an electron as $\frac{\lambda_p}{\lambda_e} = \frac{m_e v_e}{m_p v_p}$ allows us to incorporate the given condition that the particle moves $5$ times faster than the electron, resulting in $v_p = 5v_e$ and simplifying the ratio to $\frac{\lambda_p}{\lambda_e} = \frac{m_e}{5m_p}$. Given this ratio is $1.878 \times 10^{-4}$ and the electron mass is $9.1 \times 10^{-31}$ kg, the mass of the particle is determined as follows:

$m_p = \frac{9.1 \times 10^{-31}}{5 \times 1.878 \times 10^{-4}} = \frac{9.1 \times 10^{-31}}{9.39 \times 10^{-4}} \approx 9.7 \times 10^{-28} \text{ kg}$

Einstein's photoelectric equation establishes that the energy of an incident photon is shared between the work function of the metal and the maximum kinetic energy of the ejected electrons. Since the maximum kinetic energy corresponds to the product of the electron charge and the stopping potential, the stopping potential serves as a direct measure of the incident photon energy. Increasing the wavelength of the light reduces the energy per photon, which in turn leads to a measurable decrease in the stopping potential when comparing the 300 nm and 400 nm states.

The difference in stopping potential is derived by subtracting the photoelectric equations for the two wavelengths, which yields the expression $\Delta V = \frac{hc}{e} \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right)$. Substituting the values $\lambda_1 = 300\ \text{nm}$, $\lambda_2 = 400\ \text{nm}$, and $\frac{hc}{e} = 1240\ \text{nm-V}$ into the equation results in $1240 \times \left( \frac{1}{300} - \frac{1}{400} \right)$. Simplifying the expression leads to $1240 \times \frac{100}{120000}$, which gives a calculated value of approximately $1.033\ \text{V}$, confirming that the decrease in stopping potential is close to $1.0\ \text{V}$.

In the Frank-Hertz experiment, an incident electron transfers kinetic energy to mercury atoms through inelastic collisions, and the mercury atom subsequently releases this energy by emitting a photon as its excited electrons return to a lower energy state. The energy of this emitted photon exactly corresponds to the kinetic energy lost by the electron during the interaction. Calculating this change in energy gives $5.6 - 0.7 = 4.9$ eV. By utilizing the relationship $\lambda = 1240 / E$, where $\lambda$ is the wavelength in nm and $E$ is the photon energy in eV, we find $\lambda = 1240 / 4.9$ nm. This calculation yields approximately 253 nm, which is closest to 250 nm.