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

The photoelectric effect dictates that the maximum kinetic energy of an emitted electron is the difference between the incident photon energy and the metal's work function, while the number of emitted electrons is determined by the total incident power scaled by the efficiency. Total incident power is calculated by multiplying intensity by the plate area using the formula
$P = I \times A$
which results in 16 × 10^-7 W. Dividing this power by the incident photon energy of 16 × 10^-19 J indicates that 10^12 photons strike the surface per second. Since only 10% of these photons trigger electron emission, the flux of photoelectrons becomes 10^11 electrons per second. Finally, the maximum energy of the emitted electrons is obtained from the relation
$K_{max} = E_{photon} - \Phi$
which gives 10 eV - 5 eV, resulting in 5 eV.

Einstein's photoelectric effect equation dictates that the maximum kinetic energy of emitted photoelectrons depends on the energy of the incident photon and the work function of the metal surface, which is expressed as $K_{max} = E_{ph} - \Phi$. Since the light wave consists of two frequency components derived from the oscillating magnetic field, we identify the photon with the higher frequency to determine the maximum kinetic energy, as higher frequency correlates to higher photon energy. By analyzing the sine function arguments where $\omega = 2\pi\nu$, the higher frequency oscillation corresponds to $\nu = 10^7 c$.

The photon energy is calculated using the relation $E_{ph} = h\nu$. Substituting the provided constants, the energy is $6.6 × 10^{-34} × 10^7 × 3 × 10^8$ Joules, which equals $1.98 × 10^{-18}$ Joules. To convert this energy into electron-volts, we divide by the elementary charge of $1.6 × 10^{-19}$, yielding $12.375$ eV. Subtracting the work function of $4.7$ eV from this photon energy results \in a maximum kinetic energy of $7.675$ eV, which is the value that most closely aligns with the provided options.

Laser power represents the total energy released per second, which equals the number of photons emitted per second multiplied by the energy of a single photon, given by the expression $E = \frac{hc}{\lambda}$. Isolating the rate of emission, defined as $\frac{n}{t}$, results in the relationship:
$\frac{n}{t} = \frac{P \lambda}{hc}$
Given the power $P$ of 2 mW (or $2 \times 10^{-3}$ W), a wavelength $\lambda$ of 500 nm (or $500 \times 10^{-9}$ m), Planck's constant $h$ of $6.6 \times 10^{-34}$ Js, and the speed of light $c$ of $3.0 \times 10^{8}$ m/s, the formula becomes:
$\frac{n}{t} = \frac{2 \times 10^{-3} \times 500 \times 10^{-9}}{6.6 \times 10^{-34} \times 3.0 \times 10^{8}}$
Evaluating this division yields a total of $5 \times 10^{15}$ photons emitted each second.

The physical behavior of particles during a collision is governed by the conservation of linear momentum, meaning the vector sum of initial momenta equals the resulting final momentum. Because the particles approach at right angles, the magnitude of the final momentum vector is determined by the Pythagorean relationship $P^2 = P_1^2 + P_2^2$. By substituting the de Broglie relation $P = \frac{h}{\lambda}$ into this equation, the terms involving Planck's constant cancel out, leading to the relationship:

$\frac{1}{\lambda^2} = \frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2}$

The photoelectric effect dictates that the work function of a metal surface corresponds to the minimum photon energy required for electron emission, which occurs at the threshold frequency. This relationship is represented by the formula $W = hv_0$, where $v_0$ represents the x-intercept on the plot showing the relationship between stopping potential and frequency.

Identifying the threshold frequency as 4 × 10^14 Hz from the provided figure, the work function is determined by multiplying this value by Planck's constant and dividing by the elementary charge. Using the constants h = 6.63 × 10^-34 Js and e = 1.6 × 10^-19 C, the calculation is as follows:

$W = \frac{6.63 \times 10^{-34} \times 4 \times 10^{14}}{1.6 \times 10^{-19}}$

This calculation yields a final value of 1.66 eV.

The principle of conservation of linear momentum dictates that the initial momentum of the nucleus must equal the vector sum of the momenta of the two fission products. Because the fragments move in opposite directions, the magnitude of the parent nucleus's momentum corresponds to the difference between the momenta of the two fragments, expressed as $P_A = P_B - P_C$. Since both fragments have equal mass and the velocity of fragment C is half that of fragment B, the relationship between their momenta is $P_C = P_B / 2$.

Substituting this momentum relationship into the conservation equation results in $P_A = P_B - P_B / 2 = P_B / 2$. Utilizing the de Broglie relation $\lambda = h/P$, we can convert these momentum values into wavelengths. The wavelength of fragment B is $\lambda_B = h/P_B = h/(2P_A) = \lambda_A / 2$. Similarly, for fragment C, the wavelength is $\lambda_C = h/P_C = h/(P_B / 2) = 2(h/P_B) = 2\lambda_B$. Substituting the value of $\lambda_B$ leads to $\lambda_C = 2(\lambda_A / 2) = \lambda_A$.

The resolution of an electron microscope is fundamentally determined by the wave nature of electrons, where the achievable resolution is limited by the de Broglie wavelength. To effectively resolve a structure of a given width, the wavelength $\lambda$ of the electron must be approximately equal to that width. The relationship between momentum and wavelength is given by $\lambda = \frac{h}{p}$, while the kinetic energy of an electron relates to its momentum through $KE = \frac{p^2}{2m}$. By substituting the momentum from the wavelength equation into the kinetic energy expression, we derive the relation $KE = \frac{(\frac{h}{\lambda})^2}{2m}$.

With the Planck constant $h$ approximately 6.6 x 10^-34 J·s, the electron mass $m$ approximately 9.1 x 10^-31 kg, and the required resolution width of 7.5 x 10^-12 m, we can compute the minimum energy:

$KE = \frac{(\frac{6.6 \times 10^{-34}}{7.5 \times 10^{-12}})^2}{2 \times 9.1 \times 10^{-31}}$

Solving this expression yields the energy in Joules. Dividing this resulting value by the elementary charge of 1.6 x 10^-19 C converts the unit into electronvolts, which provides a kinetic energy value of approximately 25 keV.

The de Broglie wavelength of an electron is defined by the expression $\lambda = \frac{h}{mv}$, while the wavelength of a photon is given by $\lambda = \frac{c}{f}$. Equating the de Broglie wavelength to the specified fraction of the photon's wavelength leads to the relation $\frac{h}{mv} = 10^{-3} \left(\frac{c}{f}\right)$. Rearranging this equality to solve for the velocity of the electron yields the formula $v = \frac{hf}{10^{-3}mc}$. Inserting the values $h = 6.63 \times 10^{-34}$ J·s, $f = 6 \times 10^{14}$ Hz, $m = 9.1 \times 10^{-31}$ kg, and $c = 3 \times 10^8$ m/s into this equation results in a speed of $v = 1.45 \times 10^6$ m/s.

The behavior of particles after a completely inelastic collision is defined by the principle of conservation of linear momentum. Because particles $x$ and $y$ are moving in opposite directions, the magnitude of the total momentum of the resulting particle $P$ is equal to the difference between their individual momenta, where each momentum relates to its de-Broglie wavelength by the equation $p = h/\lambda$.

Mathematically, the resultant momentum of particle $P$ is expressed as $P_P = |h/\lambda_x - h/\lambda_y|$, which simplifies to $P_P = h |1/\lambda_x - 1/\lambda_y|$. Since the de-Broglie wavelength of the combined particle is defined by the relationship $P_P = h/\lambda_P$, we equate these expressions:
$\frac{h}{\lambda_P} = h \left| \frac{1}{\lambda_x} - \frac{1}{\lambda_y} \right|$
By canceling Planck's constant $h$ and finding a common denominator for the terms inside the absolute value, the expression resolves to:
$\lambda_P = \frac{\lambda_x \lambda_y}{|\lambda_x - \lambda_y|}$

The photoelectric effect is governed by the principle of conservation of energy, where the energy of an incident photon is utilized to overcome the surface binding energy, known as the work function, while any remaining energy is converted into the maximum kinetic energy of the ejected electron. To find this kinetic energy, we first determine the work function by using the threshold wavelength of $380\text{ nm}$ in the relation $E = \frac{1237}{\lambda}$, which results in a work function of approximately $3.25\text{ eV}$.

Next, we calculate the energy of the incident photon by applying the same constant to the incident wavelength of $260\text{ nm}$, which yields an incident energy of approximately $4.76\text{ eV}$. The maximum kinetic energy of the emitted photoelectron is found by subtracting the work function from the incident photon energy, leading to $4.76\text{ eV} - 3.25\text{ eV} = 1.5\text{ eV}$.

The de-Broglie wavelength of a charged particle accelerated through a potential difference is inversely proportional to the square root of the product of its mass, charge, and accelerating potential, expressed as $\lambda = \frac{h}{\sqrt{2mqV}}$. This relationship is derived from the fact that the kinetic energy gained by the particle, $K = qV$, is directly related to its momentum through the equation $p = \sqrt{2mK}$.

Comparing the two particles requires evaluating the ratio $\frac{\lambda_A}{\lambda_B}$ as:
$\frac{\lambda_A}{\lambda_B} = \sqrt{\frac{m_B q_B V_B}{m_A q_A V_A}}$
Substituting the specific parameters, where particle A has mass $m$, charge $q$, and potential 50 V, while particle B has mass $4m$, charge $q$, and potential 2500 V, leads to the expression:
$\frac{\lambda_A}{\lambda_B} = \sqrt{\frac{4m \cdot q \cdot 2500}{m \cdot q \cdot 50}}$
Canceling the mass and charge variables simplifies the expression to $\sqrt{200}$, which evaluates to approximately 14.14.

Einstein's photoelectric equation relates the maximum kinetic energy of emitted photoelectrons to the incident photon energy and the work function of the metal via the expression $K_{\text{max}} = \frac{1240}{\lambda} - \Phi$, where $\lambda$ is in nanometers and the resulting energy is in electron-volts. For the first case with $\lambda_1 = 350\text{ nm}$, the maximum kinetic energy is $K_1 = \frac{1240}{350} - \Phi$, while for the second case with $\lambda_2 = 540\text{ nm}$, it is $K_2 = \frac{1240}{540} - \Phi$. Since the maximum speed of the photoelectrons in the first case is twice that of the second, and kinetic energy is proportional to the square of the speed, the kinetic energy in the first case must be four times that of the second, leading to the relation $K_1 = 4K_2$.

Substituting the energy expressions into this relationship provides the equation $\frac{1240}{350} - \Phi = 4 \left( \frac{1240}{540} - \Phi \right)$. Distributing the factor of four on the right side results in $\frac{1240}{350} - \Phi = \frac{4960}{540} - 4\Phi$. By moving all terms involving the work function to one side, we get $3\Phi = \frac{4960}{540} - \frac{1240}{350}$. Calculating these values results in $3\Phi \approx 9.185 - 3.543$, which simplifies to $3\Phi \approx 5.642$. Dividing this result by three yields a work function of approximately $1.8\text{ eV}$.

The stopping potential of photo-electrons is determined by finding the energy of the incident light wave and applying the principle of energy conservation in the photoelectric effect. The wave equation provided identifies the wave number as $k = \frac{2\pi}{5 \times 10^{-7}}$, which corresponds to a wavelength $\lambda = 5 \times 10^{-7} \text{ m}$, or $5000 \text{ \AA}$. Based on the provided relationship $E = \frac{12375}{\lambda(\text{\in } \text{\AA})}$, the incident photon energy is $E = \frac{12375}{5000} \text{ eV}$, which equals $2.48 \text{ eV}$.

Einstein's photoelectric equation relates the photon energy to the work function and the maximum kinetic energy, which is equivalent to the stopping potential $V_s$ multiplied by the elementary charge. With a work function of $2 \text{ eV}$, the maximum kinetic energy of the electrons is $2.48 \text{ eV} - 2 \text{ eV} = 0.48 \text{ eV}$. Therefore, the stopping potential is $0.48 \text{ V}$.

According to Einstein's photoelectric equation, the maximum kinetic energy of photoelectrons is related to the incident light frequency $v$ and the threshold frequency $v_0$ by $eV_s = hv - hv_0$, where $V_s$ is the stopping potential. For the first case, with frequency $v$ and stopping potential $V_0/2$, the equation is $e(V_0/2) = hv - hv_0$, which simplifies to $eV_0 = 2hv - 2hv_0$. In the second case, with frequency $v/2$ and stopping potential $V_0$, the equation is $eV_0 = h(v/2) - hv_0$.

Equating these two expressions for $eV_0$ results in $2hv - 2hv_0 = hv/2 - hv_0$. Solving for the threshold frequency $v_0$ leads to $hv_0 = 2hv - hv/2 = 1.5hv$, which gives $v_0 = 1.5v = 3v/2$. This result indicates that the threshold frequency is $3v/2$, which is greater than the incident frequency $v$ used in the first case, implying that photoelectric emission should not occur under the given initial conditions. Since this contradicts the premise of the problem that photoemission does occur, the calculated value is physically inconsistent, and thus none of the listed options are correct.

The wavelength of an emitted photon is related to the energy change of the system by the equation $\lambda = \frac{hc}{\Delta E}$. Comparing the atomic and nuclear de-excitations, the ratio of the wavelengths is given by $\frac{\lambda_N}{\lambda_A} = \frac{\Delta E_A}{\Delta E_N}$. Electronic transitions in atoms typically involve energy differences on the order of electron volts, while transitions between nuclear energy levels involve energy differences on the order of mega electron volts, which is $10^6$ times larger. Substituting these orders of magnitude into the ratio yields $\frac{\lambda_N}{\lambda_A} \approx \frac{1}{10^6} = 10^{-6}$.

The de Broglie wavelength is related to momentum by $\lambda = h/p$. For two electrons of mass $m$ with perpendicular velocities $\vec{v}_1$ and $\vec{v}_2$, the individual momenta are $\vec{p}_1 = m\vec{v}_1$ and $\vec{p}_2 = m\vec{v}_2$, corresponding to magnitudes $p_1 = h/\lambda_1$ and $p_2 = h/\lambda_2$. The velocity of the centre of mass is $\vec{v}_{CM} = (\vec{v}_1 + \vec{v}_2)/2$, making the momentum of the first electron in the centre of mass frame $\vec{p}'_1 = m(\vec{v}_1 - \vec{v}_{CM}) = (\vec{p}_1 - \vec{p}_2)/2$.

Since the momenta $\vec{p}_1$ and $\vec{p}_2$ are perpendicular, the magnitude of the momentum in the centre of mass frame is
$p_{CM} = |\vec{p}'_1| = \frac{1}{2} \sqrt{p_1^2 + p_2^2}$
Substituting $p_1 = h/\lambda_1$ and $p_2 = h/\lambda_2$ yields
$p_{CM} = \frac{1}{2} \sqrt{\left(\frac{h}{\lambda_1}\right)^2 + \left(\frac{h}{\lambda_2}\right)^2} = \frac{h}{2} \frac{\sqrt{\lambda_1^2 + \lambda_2^2}}{\lambda_1 \lambda_2}$
Using the relation $\lambda_{CM} = h/p_{CM}$, the wavelength in the centre of mass frame is
$\lambda_{CM} = \frac{h}{\frac{h}{2} \frac{\sqrt{\lambda_1^2 + \lambda_2^2}}{\lambda_1 \lambda_2}} = \frac{2\lambda_1\lambda_2}{\sqrt{\lambda_1^2 + \lambda_2^2}}$

The de-Broglie wavelength of an electron in the $n$-th orbit of a hydrogen atom is given by $2\pi r_n = n\lambda_n$, where $r_n$ is the radius of the orbit and $n$ is the principal quantum number. Since the radius $r_n$ is proportional to $n^2$, the wavelength scales as $\lambda_n \propto \frac{n^2}{n} = n$. Given that $n = 1$ for the ground state and $n = 3$ for the second excited state, the wavelengths are related by $\lambda_B / \lambda_G = 3 / 1$, leading to $\lambda_B = 3\lambda_G$.

The de Broglie wavelength is defined by the relation $\lambda = \frac{h}{mv}$, where $h$ is Planck's constant, $m$ is the mass, and $v$ is the velocity. Equating the wavelengths for the proton ($p$) and the $\alpha$-particle ($\alpha$), the following relation holds:

$\frac{h}{m_p v_p} = \frac{h}{m_\alpha v_\alpha}$

This simplifies to the velocity ratio:

$\frac{v_p}{v_\alpha} = \frac{m_\alpha}{m_p}$

Given that the mass of an $\alpha$-particle is approximately four times the mass of a proton, $m_\alpha = 4m_p$, substituting this into the expression yields:

$\frac{v_p}{v_\alpha} = \frac{4m_p}{m_p} = 4$

The ratio of the velocity of the proton to that of the $\alpha$-particle is therefore 4:1.

When particles undergo a head-on elastic collision, we must satisfy the conservation of both linear momentum and kinetic energy. This interaction implies that the coefficient of restitution is one, so the relative velocity of separation is identical to the relative velocity of approach. These two conservation laws are sufficient to uniquely determine the final velocities of both bodies.

For a particle A of mass $m$ and velocity $v$ colliding with a particle B of mass $m/2$ at rest, conservation of linear momentum gives $m v = m v_1 + (m/2) v_2$, which simplifies to:
$2v = 2v_1 + v_2$
Using the elasticity condition $v_2 - v_1 = v$, we substitute $v_2 = v + v_1$ into the momentum equation to find that particle A moves with $v_1 = v/3$ and particle B moves with $v_2 = 4v/3$.

The de-Broglie wavelength of a particle is determined by the relation $\lambda = h/p$. Comparing the two wavelengths, the ratio is the inverse of the ratio of their momenta:
$\frac{\lambda_A}{\lambda_B} = \frac{p_B}{p_A} = \frac{(m/2)v_2}{m v_1}$
Substituting the specific velocities $v_1 = v/3$ and $v_2 = 4v/3$ into this expression yields $2/1$, or simply $2$.

The production of the shortest wavelength X-ray is a direct consequence of energy conservation, where the kinetic energy of an accelerating electron, $eV$, is entirely converted into the energy of a single photon, $\frac{hc}{\lambda_{\min}}$. Equating these energies, we obtain the fundamental relationship:
$\frac{hc}{\lambda_{\min}} = eV$
Rearranging this expression to solve for $\lambda_{\min}$ yields $\lambda_{\min} = \frac{hc}{eV}$, and applying logarithms to both sides results in $\log(\lambda_{\min}) = -\log(V) + \log\left(\frac{hc}{e}\right)$. This equation matches the linear form $y = mx + c$, confirming that plotting $\log(\lambda_{\min})$ against $\log(V)$ produces a straight line with a slope of $-1$. Consequently, the graph shows a constant downward linear trend, consistent with the inverse proportionality between the shortest wavelength and the accelerating voltage.