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

The de-Broglie wavelength $\lambda$ for a charged particle accelerated through a potential difference $V$ is determined by the relation $\lambda = \frac{h}{\sqrt{2mqV}}$, where $h$ is Planck's constant, $m$ is the mass of the particle, and $q$ is its charge. Because both the alpha particle and the proton are accelerated by the same potential, the ratio of their wavelengths depends inversely on the square root of the product of their respective masses and charges, expressed as $\frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p q_p}{m_\alpha q_\alpha}}$.

Considering that an alpha particle has a mass $m_\alpha = 4m_p$ and a charge $q_\alpha = 2q_p$, substituting these into the ratio equation leads to $\sqrt{\frac{m_p q_p}{4m_p \cdot 2q_p}} = \sqrt{\frac{1}{8}}$. Given the required form of the answer as $\frac{1}{\sqrt{m}}$, it becomes clear that $m = 8$.

Einstein's photoelectric equation, $KE_{max} = \frac{hc}{\lambda} - \phi$, establishes that the maximum kinetic energy of a photoelectron is determined by the energy of the incident photon and the work function of the metal. Since the stopping potential is related to this energy via the expression $V_S = \frac{KE_{max}}{e}$, it depends exclusively on the frequency or wavelength of the radiation and remains unaffected by the total power or intensity of the light source. Therefore, both claims hold true because light intensity influences the quantity of ejected electrons rather than their individual energy levels.

The De-Broglie wavelength associated with a particle is defined by the relation $\lambda = \frac{h}{\sqrt{2mK}}$, where $h$ is Planck's constant, $m$ denotes mass, and $K$ represents kinetic energy. Since the kinetic energy is constant for all three particles, the wavelength is inversely proportional to the square root of their masses, expressed as $\lambda \propto \frac{1}{\sqrt{m}}$. Because the mass hierarchy follows $m_{\alpha} > m_{p} > m_{e}$, the resulting wavelengths necessarily follow the inverse order of $\lambda_{\alpha} < \lambda_{p} < \lambda_{e}$.

The de-Broglie wavelength of an electron accelerated through a potential difference $V$ follows the relation $\lambda = \frac{h}{\sqrt{2meV}}$. Rearranging this formula reveals that the potential is inversely proportional to the square of the wavelength, which can be expressed as $V \propto \frac{1}{\lambda^2}$.

When the wavelength increases by 50%, the final wavelength becomes $1.5\lambda$. Comparing the two states, the ratio of the initial potential to the final potential is $\frac{V_1}{V_2} = \left( \frac{1.5\lambda}{\lambda} \right)^2$. Squaring the value $1.5$ results in the fraction $\frac{9}{4}$.

The De Broglie wavelength of a particle is defined by the relationship $\lambda = \frac{h}{\sqrt{2mKE}}$, where $h$ is Planck's constant, $m$ is the mass, and $KE$ is the kinetic energy. Substituting the mass relationship $m_\alpha \approx 4m_p$ and the given energy values-4K for the electron, K for the proton, and 2K for the alpha particle-into the formula yields the specific expressions for each wavelength. These are $\lambda_e = \frac{h}{\sqrt{8m_eK}}$, $\lambda_p = \frac{h}{\sqrt{2m_pK}}$, and $\lambda_\alpha = \frac{h}{\sqrt{16m_pK}}$. Comparing these values, the alpha particle has the largest denominator, which corresponds to the shortest wavelength, while the extremely small mass of the electron creates the smallest denominator, resulting in the longest wavelength and the final order $\lambda_\alpha < \lambda_p < \lambda_e$.

The de-Broglie wavelength of a charged particle accelerated from rest is determined by the formula $\lambda = \frac{h}{\sqrt{2mqV}}$, where $h$ is Planck's constant, $m$ is the mass of the particle, $q$ is its charge, and $V$ is the accelerating potential. For a proton with mass $m$, charge $q$, and an accelerating potential of $2V$, the wavelength is $\lambda_p = \frac{h}{\sqrt{2(m)(q)(2V)}} = \frac{h}{\sqrt{4mqV}}$.

An alpha particle has a mass of $4m$ and a charge of $2q$, and it is accelerated through a potential of $4V$. Substituting these specific values into the same formula yields $\lambda_\alpha = \frac{h}{\sqrt{2(4m)(2q)(4V)}} = \frac{h}{\sqrt{64mqV}}$. Comparing the two wavelengths by taking their ratio, $\frac{\lambda_p}{\lambda_\alpha} = \frac{\sqrt{64mqV}}{\sqrt{4mqV}}$, the expression simplifies to $\sqrt{16}$, which results in a ratio of $4:1$.

Einstein's photoelectric equation dictates that the maximum kinetic energy of emitted photoelectrons equals the difference between the incident photon energy and the metal's work function, expressed as $\frac{1}{2}mv^2 = E - \phi$. For the first case, the incident photon energy is $5\phi$, leading to a maximum kinetic energy of $\frac{1}{2}mv_1^2 = 5\phi - \phi = 4\phi$. In the second case, with an incident energy of $10\phi$, the maximum kinetic energy is $\frac{1}{2}mv_2^2 = 10\phi - \phi = 9\phi$. By taking the ratio of the two kinetic energy expressions, we find $\frac{v_1^2}{v_2^2} = \frac{4}{9}$, and taking the square root of this result yields the ratio of the maximum velocities $v_1 : v_2$ as $2 : 3$.

When a particle of mass $m$ and charge $q$ is accelerated from rest through a potential difference $V$, it acquires kinetic energy $K = qV$. The linear momentum $p$ is related to this kinetic energy by the equation $p = \sqrt{2mK}$, which allows us to express the momentum as $p = \sqrt{2mqV}$. Since both the alpha particle and the proton are accelerated through the same potential, the ratio of their momenta simplifies to the square root of the ratio of their mass-charge products, specifically $\sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}}$.

Given that an alpha particle has four times the mass of a proton and carries twice the elementary charge, substituting $m_\alpha = 4m_p$ and $q_\alpha = 2q_p$ into this ratio yields $\sqrt{\frac{4m_p \cdot 2q_p}{m_p \cdot q_p}}$, which simplifies to $\sqrt{8}$. This results in a final ratio of $2\sqrt{2}:1$.

The relationship between the de Broglie wavelengths of an electron and a photon depends on how their shared kinetic energy $E$ determines their momentum. For a photon, the energy-momentum relation $E = pc$ yields $\lambda_p = \frac{hc}{E}$, while for an electron, $E = \frac{p^2}{2m}$ leads to $\lambda_e = \frac{h}{\sqrt{2mE}}$. Squaring the electron's wavelength expression reveals that $E$ is inversely proportional to $\lambda_e^2$. By equating both energy expressions, it follows that $\lambda_p \propto \lambda_e^2$ because the physical constants do not alter the proportionality.

The photoelectric effect is described by the principle that the energy of an incoming photon is split between overcoming the metal's work function and providing kinetic energy to the emitted photoelectron. Mathematically, this is expressed as:

$K_{max} = \frac{hc}{\lambda} - \phi$

where $K_{max}$ is the kinetic energy, $\frac{hc}{\lambda}$ is the photon energy, and $\phi$ is the work function. In this scenario, we are presented with two distinct conditions: one with a wavelength of 800 nm, giving $K = \frac{1230}{800} - \phi$, and another with a wavelength of 500 nm, resulting in a doubled kinetic energy of $2K = \frac{1230}{500} - \phi$.

To solve for $\phi$, we can manipulate the first equation to match the $2K$ term from the second equation by multiplying the entire expression by 2. This gives $2K = 2 \times \frac{1230}{800} - 2\phi$, which simplifies to $2K = 3.075 - 2\phi$. Now, equating this to the second expression results in $3.075 - 2\phi = 2.46 - \phi$. By rearranging the variables to isolate $\phi$, we arrive at the following equation:

$\phi = 3.075 - 2.46$

which yields a work function of 0.615 eV.

According to Einstein's photoelectric equation, the maximum kinetic energy of emitted electrons is determined by the most energetic photons striking the metal surface. The electric field equation provided reveals two distinct oscillation components with angular frequencies of $\omega_1 = 6 \times 10^{15} \text{ rad/s}$ and $\omega_2 = 9 \times 10^{15} \text{ rad/s}$. To find the energy of these photons, we use the relation $E = \frac{h \omega}{2\pi}$ with the given Planck's constant $h = 4.14 \times 10^{-15} \text{ eV s}$. Calculating for both frequencies yields photon energies of approximately $3.95 \text{ eV}$ and $5.93 \text{ eV}$, respectively.

Since the photoelectric effect dictates that the maximum kinetic energy is achieved by the most energetic incident photons, we utilize the $5.93 \text{ eV}$ value. Subtracting the metal's work function of $2.50 \text{ eV}$ from this maximum photon energy results in $5.93 \text{ eV} - 2.50 \text{ eV} = 3.43 \text{ eV}$, which corresponds to the value observed in the provided options.

According to Einstein's photoelectric equation, the kinetic energy of an emitted electron is determined by the energy of the incident photon minus the work function of the metal, expressed as $E = \frac{hc}{\lambda} - \phi$. When the incident wavelength changes to $\lambda'$ such that the kinetic energy doubles to $2E$, the new relationship becomes $2E = \frac{hc}{\lambda'} - \phi$. Subtracting the first equation from the second eliminates the constant work function, resulting in $E = \frac{hc}{\lambda'} - \frac{hc}{\lambda}$. Rearranging this expression to isolate the term containing the new wavelength yields $\frac{hc}{\lambda'} = E + \frac{hc}{\lambda}$, which simplifies to $\frac{hc}{\lambda'} = \frac{E\lambda + hc}{\lambda}$. Inverting this equation to solve for $\lambda'$ provides the final expression $\lambda' = \frac{hc\lambda}{E\lambda + hc}$.

The de-Broglie wavelength of a particle relates to its mass and kinetic energy according to the formula $\lambda = \frac{h}{\sqrt{2mK}}$ Since the kinetic energy $K$ is constant for both particles, the ratio of the wavelengths of the \alpha particle and the carbon-12 atom depends only on their masses, resulting \in the relation $\frac{\lambda_\alpha}{\lambda_{C12}} = \sqrt{\frac{m_{C12}}{m_\alpha}}$ Given that the mass of an \alpha particle is 4 atomic mass units and the mass of carbon-12 is 12 atomic mass units, substituting these values into the expression yields $\sqrt{3}$. Consequently, the ratio of the de-Broglie wavelengths is $\sqrt{3} : 1$.

The photoelectric effect dictates that the energy of an incident photon is partitioned between the work function of a metal and the kinetic energy of an emitted photoelectron. The energy supplied by the radiation is determined by $E = \frac{hc}{\lambda}$. When this electron enters a perpendicular magnetic field, it moves in a circular path because the magnetic Lorentz force provides the necessary centripetal force, leading to the relation $R = \frac{\sqrt{2m K E}}{qB}$.

Rearranging this expression to solve for kinetic energy yields $K E = \frac{q^2 B^2 R^2}{2m}$. Using the standard values for an electron where the charge is 1.6 x 10^-19 C, the field strength is 2 mT, the radius is 2 mm, and the mass is 9.1 x 10^-31 kg, the kinetic energy is approximately 1.40 eV. Since the incident energy from the 4500 A wavelength radiation is approximately 2.76 eV, subtracting the kinetic energy of the photoelectron from this total energy reveals the work function to be approximately 1.36 eV.

The intensity of light incident on a surface defines the energy arriving per unit area per unit time, while the particle nature of light dictates that this energy arrives in discrete packets known as photons. To determine the number of photons striking the target each second, we first calculate the total power delivered to the area by multiplying the intensity of 100 Wm^-2 by the area of 10^-4 m^2, which results in a power of 10^-2 W.

Every photon carries a specific amount of energy determined by its wavelength, defined as $E = \frac{hc}{\lambda}$. Because the total power is equivalent to the number of photons per second, denoted by $n$, multiplied by the energy of a single photon, we can express the relationship as $P = n \frac{hc}{\lambda}$. Rearranging this formula to isolate the photon rate gives:

$n = \frac{P \lambda}{hc}$

Using the standard values for Planck's constant of 6.63e-34 Js and the speed of light of 3e8 m/s, along with the wavelength of 900 nm, we substitute these into our equation:

$n = \frac{10^{-2} \times 900 \times 10^{-9}}{6.63 \times 10^{-34} \times 3 \times 10^8}$

Completing this calculation reveals that the number of photons crossing the defined area every second is approximately 4.5e16.

The power of a monochromatic light source is defined by the product of the rate of photon emission and the energy of a single photon, which is given by the relationship $P = \frac{n}{t} \frac{hc}{\lambda}$. Since the source operates at a constant power, the product of the number of photons emitted per second and the wavelength must remain invariant, meaning the emission rate is directly proportional to the wavelength. This allows the state of the source at different wavelengths to be related by the equation $\frac{n_1}{t_1} \lambda_1 = \frac{n_2}{t_2} \lambda_2$, where the new emission rate is $\frac{n_2}{t_2} = \frac{n_1}{t_1} \left( \frac{\lambda_2}{\lambda_1} \right)$. Substituting the values $n_1/t_1 = 9 \times 10^{20}$, $\lambda_1 = 600\text{ nm}$, and $\lambda_2 = 800\text{ nm}$ into this ratio yields $9 \times 10^{20} \times \frac{800}{600}$, resulting in $12 \times 10^{20}$ photons per second.

The photoelectric effect follows the energy conservation principle defined by $h\nu = W_0 + K_{max}$, where incident energy must overcome the work function $W_0$ to eject an electron, making the assertion logically sound as emission fails below this threshold. Similarly, the reason is correct because if the incident energy perfectly matches the work function, the electron is liberated with $K_{max} = 0$ since no excess energy remains for kinetic movement. Although both statements are individually true, the reason describes the threshold boundary condition rather than providing the physical justification for why the effect is absent at lower energy levels.

When an electron is accelerated through a potential difference $V$ in volts, its associated wavelength simplifies to a widely used empirical relationship that depends directly on the accelerating voltage. By substituting the physical constants into the general de-Broglie equation $\lambda = \frac{h}{\sqrt{2meV}}$, we obtain the result:

$\lambda = \frac{1.227}{\sqrt{V}}$ nm

Comparing this standardized expression to the given formula $\lambda = \frac{1.227}{x}$ nm, it is clear that the variable $x$ must be equal to $\sqrt{V}$.

Einstein's photoelectric equation describes the maximum kinetic energy of emitted photo-electrons as $K = \frac{hc}{\lambda} - \phi$, where $h$ is Planck's constant, $c$ is the speed of light, $\lambda$ is the incident wavelength, and $\phi$ represents the metal's work function. Given that the two incident beams have wavelengths related by $\lambda_1 = 3 \lambda_2$, the respective kinetic energies are defined as $K_1 = \frac{hc}{3 \lambda_2} - \phi$ and $K_2 = \frac{hc}{\lambda_2} - \phi$.

To determine the relationship between $K_1$ and $\frac{K_2}{3}$, we can manipulate the second equation to isolate the common wavelength term by dividing the entire expression by three, which yields $\frac{K_2}{3} = \frac{hc}{3 \lambda_2} - \frac{\phi}{3}$. Comparing this result to $K_1 = \frac{hc}{3 \lambda_2} - \phi$, we observe that the term involving the wavelength is identical in both expressions. Because subtracting the full work function $\phi$ results in a smaller total value than subtracting only a fraction of the work function, specifically $\frac{\phi}{3}$, it follows that $K_1 < \frac{K_2}{3}$.

Einstein's photoelectric equation governs this interaction by stating that the energy of an incident photon is distributed between the work function of the metal and the maximum kinetic energy of the ejected electron, represented by the relation $h \nu = \phi + \frac{1}{2} m v^2$. Since the work function $\phi$ corresponds to the energy of a photon at the threshold frequency, we can represent it as $\phi = h \nu_0$.

For incident light with a frequency of twice the threshold, the balance equation is $2 \phi = \phi + \frac{1}{2} m v_1^2$, which simplifies to $\phi = \frac{1}{2} m v_1^2$. When the frequency increases to five times the threshold, the relationship becomes $5 \phi = \phi + \frac{1}{2} m v_2^2$, resulting in $4 \phi = \frac{1}{2} m v_2^2$. Dividing the second kinetic energy equation by the first yields $\frac{v_2^2}{v_1^2} = 4$, which indicates that the ratio of the velocities is $2$, meaning $x$ is equal to 2.