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

When a monochromatic light strikes the metallic plate, it ejects photoelectrons with a maximum kinetic energy defined by the relation $K = \frac{hc}{\lambda} - \phi$. As the electron enters a uniform magnetic field perpendicular to its velocity, the resulting Lorentz force acts as a centripetal force, forcing the particle into a circular path. The radius of this circle is given by $R = \frac{mv}{eB}$, and because the kinetic energy relates to the velocity through $v = \sqrt{\frac{2K}{m}}$, the radius can be expressed as $R = \frac{\sqrt{2mK}}{eB}$. The distance between the emission point $A$ and the impact point $B$ is equal to the diameter of this circular path, which is $2R$. Substituting the energy expression into the diameter formula yields $2R = \frac{2\sqrt{2m(\frac{hc}{\lambda} - \phi)}}{eB}$, which simplifies by bringing the coefficient into the square root to give $\frac{\sqrt{8m(\frac{hc}{\lambda} - \phi)}}{eB}$.

When light strikes a perfectly reflecting surface, it transfers double the momentum compared to an absorbing surface because the photons reverse direction upon impact. The radiation pressure is determined using the relation $p = \frac{2I}{c}$, where $I$ is the intensity of the light source at the given distance and $c$ is the speed of light.

Intensity at a distance of 2 m is calculated by dividing the 450 W source power by the surface area of a sphere, resulting in $I = \frac{450}{4\pi(2)^2}$. Substituting this into the pressure equation, the final calculation becomes
$p = \frac{2 \times 450}{16\pi \times 3 \times 10^8}$
which yields a result of approximately $6 \times 10^{-8}$ Pascals.

The photoelectric effect follows the linear relationship $V_0 = (\frac{h}{e})\nu - \frac{\phi_0}{e}$, which dictates that a plot of stopping potential $V_0$ versus frequency $\nu$ is a straight line. Because the slope of this line is $\frac{h}{e}$, Planck's constant $h$ is fundamentally tied to this gradient, though the dependence on the electron charge $e$ means $h$ cannot be determined from the curve without knowing $e$. Conversely, the y-intercept of the graph corresponds to $-\frac{\phi_0}{e}$, allowing the work function $\phi_0$ to be calculated directly from the intercept and the known charge of the electron, independent of any prior assumption about $h$.

The photoelectric effect dictates that the maximum kinetic energy of an ejected electron equals the incident photon energy minus the metal's work function, represented by the relationship $KE = \frac{hc}{\lambda} - \phi_0$. With an initial kinetic energy of 2 eV and a work function of 1 eV, the initial incident photon energy must be 3 eV. Since photon energy is inversely proportional to wavelength, reducing the wavelength to $\frac{\lambda}{2}$ doubles the incoming energy to 6 eV. Subtracting the unchanged work function of 1 eV from this new energy value results in a final maximum kinetic energy of 5 eV.

To initiate the photoelectric effect, incoming light must possess photons with energy at least equal to the metal's work function of $3 \text{ eV}$, which dictates a threshold wavelength calculated by the relationship $\lambda = \frac{12400}{\phi}$:

$\lambda = \frac{12400}{3} \text{ Å} = 4133 \text{ Å}$

Given that visible light spans from $3800 \text{ Å}$ to $7600 \text{ Å}$ with colors progressing from violet to red, a calculated value of $4133 \text{ Å}$ falls within the blue region of the spectrum, making blue light the correct requirement for emission.

The interaction relies on the principle of conservation of momentum, where the momentum of the incident light pulse is transferred to the mirror. Since a reflected photon pulse of energy $E$ reverses its direction, the total change in momentum, or impulse imparted to the mirror, is $\Delta p = \frac{E}{c} - (-\frac{E}{c}) = \frac{2E}{c}$. Consequently, the mirror acquires an initial velocity $v = \frac{\Delta p}{m} = \frac{2E}{mc}$.

Applying conservation of mechanical energy to the pendulum, the initial kinetic energy $\frac{1}{2}mv^2$ equals the potential energy $mgh$ at the maximum deflection angle $\theta$, where $h = l(1 - \cos \theta)$. For small angles, $1 - \cos \theta \approx \frac{\theta^2}{2}$, so the energy relation becomes $\frac{1}{2}mv^2 = mg \frac{l \theta^2}{2}$, which simplifies to $v = \theta \sqrt{gl}$. Equating the two expressions for the initial velocity yields $\frac{2E}{mc} = \theta \sqrt{gl}$, resulting in $\theta = \frac{2E}{mc\sqrt{gl}}$.

An infinite non-conducting sheet with uniform charge density generates a constant electric field given by $E = \frac{\sigma}{2E_0}$. When an electron is released from rest in this region, it experiences a constant force $F = qE$, which causes the electron's momentum to increase linearly over time following the relation $p = Ft$. Since the de-Broglie wavelength is defined by $\lambda = \frac{h}{p}$, calculating the rate of change of this wavelength involves differentiating with respect to time.

Applying the chain rule to the wavelength expression leads to $\frac{d\lambda}{dt} = -\frac{h}{p^2} \frac{dp}{dt}$. Because the force is constant, $\frac{dp}{dt}$ is equal to $F$, and substituting both the momentum and the rate of change of momentum into the equation results in $\frac{d\lambda}{dt} \propto \frac{F}{p^2} = \frac{F}{(Ft)^2}$. This simplifies to a relationship where the rate of change is proportional to $\frac{1}{t^2}$, which confirms that the value of $n$ is 2.

The photoelectric effect occurs if and only if the energy of each incident photon exceeds the material's work function, which is the minimum energy required to liberate an electron from the metal surface. We calculate the incident photon energy using the formula $E = \frac{1240}{\lambda}$ By substituting the given wavelength of 550 nm, we find that the incident photon energy is 2.25 eV. Because this energy is greater than the work function of cesium (1.9 eV), photoemission occurs for this metal. However, since 2.25 eV is less than the work function of lithium (2.5 eV), the incident light does not provide sufficient energy to eject electrons from the lithium surface, meaning the effect is observed only in cesium.

When an electron moves through a magnetic field, the magnetic force acts perpendicularly to its velocity at every instant, which means it performs no work on the particle. Because no work is done, the kinetic energy of the electron and its speed $v$ remain constant over time. Given the de Broglie relationship $\lambda = \frac{h}{mv}$, the wavelength remains unchanged from its initial value $\lambda_0$.

The photoelectric effect provides a direct link between incident photon energy and the kinetic energy of emitted electrons through the principle of conservation of energy. The incident radiation provides energy defined by $\frac{hc}{\lambda_i}$, which must first overcome the material's work function $\phi = \frac{hc}{\lambda_0}$ before transferring any remaining energy into kinetic energy, expressed as $KE = \frac{p^2}{2m} = \frac{h^2}{2m\lambda_e^2}$.

Balancing these components yields the energy equation $\frac{hc}{\lambda_i} = \frac{hc}{\lambda_0} + \frac{h^2}{2m\lambda_e^2}$. Isolating the kinetic energy term results in $\frac{h^2}{2m\lambda_e^2} = hc \left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)$, which simplifies to $\frac{1}{\lambda_e^2} = \frac{2mc}{h} \left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)$. Consequently, solving for the de-Broglie wavelength gives:

$\lambda_e = \sqrt{\frac{h}{2mc \left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)}}$

De Broglie's hypothesis establishes that a moving particle possesses an associated wavelength determined by its momentum, which allows us to categorize its behavior within the electromagnetic spectrum. The relationship between the wavelength $\lambda$, Planck's constant $h$, mass $m$, and velocity $v$ is expressed by the formula $\lambda = h / (mv)$.

Substituting the given values into this relationship provides the following calculation:
$\lambda = \frac{6.63 \times 10^{-34}}{10^{-30} \times 2.21 \times 10^6}$
This results in a wavelength of $3 \times 10^{-10}$ meters, a value equivalent to 3 Angstroms. Since electromagnetic radiation within this specific wavelength range corresponds to the X-ray region of the spectrum, the particle exhibits behavior characteristic of X-rays.

In the photoelectric effect, electrons are emitted from a material with varying kinetic energies, but this photocurrent can be reduced to zero by applying a sufficiently negative potential, known as the stopping potential, to the collector plate to repel even the fastest electrons. This stopping potential ($V_s$) is governed by Einstein's photoelectric equation, expressed as $eV_s = hf - \phi_0$, which confirms that the stopping potential indeed varies linearly with the frequency ($f$) of the incident radiation. While both statements describe valid physical principles, the linear mathematical relationship between frequency and potential does not provide the causal mechanism for why a retarding electric field is capable of halting electron emission; therefore, the reason does not explain the assertion.

The stopping potential in the photoelectric effect is determined solely by the frequency of the incident radiation and the work function of the metal according to the relation $eV_s = h\nu - \Phi$, making it entirely independent of the light's intensity. Increasing the intensity of incident light only increases the number of photons hitting the metal surface per unit time, which enhances the rate of photoelectron emission-and thus the photocurrent-provided the frequency exceeds the threshold frequency. Therefore, the assertion that stopping potential increases with intensity is incorrect, while the reason correctly describes the effect of intensity on the emission rate.

The de Broglie wavelength is fundamentally linked to an electron's momentum via the relation $\lambda = \frac{h}{mv}$. Initially, the electron moves with velocity $v_0$ in the x-direction, establishing an initial wavelength of $\lambda_0 = \frac{h}{mv_0}$. Upon entering the electric field $\vec{E} = -E_0 \hat{k}$, the electron experiences an electric force $\vec{F} = eE_0 \hat{k}$, which induces a constant acceleration in the z-direction while the x-component of the velocity remains constant.

After time $t$, the z-component of the velocity is $v_z = \frac{e E_0 t}{m}$. Consequently, the magnitude of the resulting velocity vector is given by
$v = \sqrt{v_0^2 + \left(\frac{e E_0 t}{m}\right)^2}$
Substituting this expression into the definition of wavelength, we find
$\lambda = \frac{h}{m \sqrt{v_0^2 + \left(\frac{e E_0 t}{m}\right)^2}}$
Factoring the initial velocity out of the denominator allows us to rewrite this in terms of the initial wavelength, yielding
$\lambda = \frac{\lambda_0}{\sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}$

According to the de Broglie hypothesis, the circumference of an electron's orbit corresponds to an integer multiple of its wavelength, given by the relationship $2\pi r_n = n \lambda_n$. By rearranging this formula to solve for the wavelength, we find that $\lambda_n = \frac{2\pi r_n}{n}$. Since the orbital radius of an electron in a hydrogen atom is proportional to the square of the principal quantum number, or $r_n \propto n^2$, we can substitute this into our wavelength expression to see that $\lambda_n \propto \frac{n^2}{n}$, which simplifies to a linear proportionality where $\lambda_n \propto n$.

For the ground state, where $n = 1$, the orbital radius is $r_1 = 5.3 \times 10^{-11} \text{ m}$. For the third excited state, where $n = 4$, the radius is $r_4 = 8.48 \times 10^{-10} \text{ m}$. Using the simplified proportionality $\lambda_n \propto n$, the ratio of the wavelength in the third excited state to the ground state is calculated as $\frac{\lambda_4}{\lambda_1} = \frac{4}{1} = 4$. Verifying this with the given radii, $\lambda_1 = \frac{2\pi(5.3 \times 10^{-11})}{1}$ and $\lambda_4 = \frac{2\pi(8.48 \times 10^{-10})}{4}$. Substituting $8.48 \times 10^{-10} \text{ m}$ as $16 \times 5.3 \times 10^{-11} \text{ m}$ into the second equation gives $\lambda_4 = \frac{2\pi(16 \times 5.3 \times 10^{-11})}{4} = 8\pi(5.3 \times 10^{-11}) \text{ m}$, which confirms that the wavelength in the excited state is 4 times larger than in the ground state.

The wave model of light successfully explains phenomena like reflection, refraction, and diffraction by treating electromagnetic radiation as continuous, oscillating waves. However, the Compton effect involves a collision between a photon and a free electron, where the photon shifts in wavelength upon impact. This specific interaction cannot be described by wave propagation and instead necessitates the particle nature of light, where energy and momentum are transferred as discrete packets during the collision.

The de Broglie wavelength is fundamentally linked to a particle's mass and velocity through the relation $\lambda = \frac{h}{mv}$, where $h$ is Planck's constant. Isolating velocity in this equation reveals that $v = \frac{h}{m\lambda}$, indicating that the velocity depends inversely on both mass and wavelength. By comparing the proton and the alpha particle, we can express the ratio of their velocities as
$\frac{v_p}{v_\alpha} = \frac{m_\alpha}{m_p} \times \frac{\lambda_\alpha}{\lambda_p}$
Since an alpha particle possesses four times the mass of a proton and we are given $\lambda_p = \lambda$ and $\lambda_\alpha = 2\lambda$, substituting these values into the expression yields $4 \times 2 = 8$, resulting in a final ratio of 8:1.

The rate at which a light source emits photons is determined by its total power output and the energy characteristic of each individual photon. Since the energy of a photon is inversely proportional to its wavelength, described by the relation $E = \frac{hc}{\lambda}$, light with a longer wavelength carries less energy per photon. Because the total power is the product of the number of photons emitted per second and the energy of each photon, a constant power output across different sources implies that the frequency of photon emission must balance the difference in energy per photon.

Setting the power output for both sources as $P = n \frac{hc}{\lambda}$ reveals that the rate of photon emission is directly proportional to the wavelength, given by $n \propto \lambda$. Consequently, the ratio of the number of photons emitted by the first source to the number emitted by the second source is equivalent to the ratio of their wavelengths. Substituting the provided values of 300 nm and 500 nm into this relationship yields the calculation $\frac{n_1}{n_2} = \frac{300}{500} = \frac{3}{5}$.

The de-Broglie wavelength of a particle is fundamentally determined by the ratio of Planck's constant to its momentum. For a photon, energy and momentum are related by the expression $E_p = p_p c$, which allows its wavelength to be represented as $\lambda_p = \frac{hc}{E_p}$. In the case of a non-relativistic electron, the de-Broglie wavelength is defined by $\lambda_e = \frac{h}{p_e}$. Using the kinetic energy relation $K_e = \frac{1}{2}m_e v_e^2$, we can express the electron's momentum as $p_e = m_e v_e = \frac{2 K_e}{v_e}$, which transforms the wavelength formula into $\lambda_e = \frac{h v_e}{2 K_e}$.

Equating the two wavelengths as required by the problem conditions yields $\frac{hc}{E_p} = \frac{h v_e}{2 K_e}$. By canceling Planck's constant from both sides and substituting the given velocity $v_e = 0.25c$ into the equation, we obtain $\frac{c}{E_p} = \frac{0.25c}{2 K_e}$. This simplifies to $\frac{1}{E_p} = \frac{0.25}{2 K_e}$, which further reduces to $\frac{1}{E_p} = \frac{1}{8 K_e}$. Rearranging this result leads to the final ratio of the kinetic energy of the electron to the energy of the photon as $\frac{K_e}{E_p} = \frac{1}{8}$.

The relationship between the energy of a photon and its wavelength is fundamentally defined by the equation $E = \frac{hc}{\lambda}$, where the energy of the photon corresponds to the difference in energy levels of the atom. When a hydrogen atom is excited to its first level, it absorbs 10.2 eV of energy, and the subsequent return to the ground state results in the emission of a photon with that same energy value.

Applying these values leads to the following calculation:
$\lambda = \frac{1245 \text{ eV nm}}{10.2 \text{ eV}} = 122.06 \text{ nm}$
Rounding this result provides the wavelength of 122 nm.