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

The photoelectric effect is defined by the governing equation $(KE)_{max} = hv - \phi$, where the maximum kinetic energy of photoelectrons is determined solely by the frequency of the incident radiation rather than its intensity. Since intensity represents the quantity of incident photons, it directly determines the number of emitted photoelectrons and thus the magnitude of the photocurrent, but it has no influence on individual electron energy. Therefore, the maximum kinetic energy depends strictly on the incident frequency, making it independent of intensity and unrelated to any minimum threshold intensity requirement.

According to Einstein's photoelectric equation, the maximum kinetic energy is expressed as $K_{max} = hf - \phi$. Substituting the relationship between stopping potential and maximum kinetic energy, $eV_s = hf - \phi$, allows us to derive the linear equation for the stopping potential as $V_s = \frac{h}{e}f - \frac{\phi}{e}$. In this form, the slope of the graph is represented by the term $\frac{h}{e}$. Although Aluminium has a work function of 4.1 eV and Gold has a work function of 5.1 eV, the slope depends exclusively on the universal constants $h$ and $e$. Because this slope remains constant regardless of the material, the ratio of the slopes for Gold and Aluminium is 1.

The de-Broglie wavelength of a particle is fundamentally determined by the expression $\lambda = \frac{h}{p}$, where $h$ is Planck's constant and $p$ denotes the momentum. When two particles such as a proton and an electron are specified to have identical de-Broglie wavelengths, their respective expressions must satisfy the equality $\frac{h}{p_p} = \frac{h}{p_e}$. Simplifying this equation by dividing both sides by $h$ reveals that the momenta must be exactly equal, meaning the ratio $p_p:p_e$ is $1:1$, rendering the provided mass relationship irrelevant for this specific calculation.

The relationship between kinetic energy and the de Broglie wavelength is defined by the expression $K = \frac{p^2}{2m}$, where the particle's momentum $p$ is linked to the wavelength $\lambda$ through the de Broglie relation $p = \frac{h}{\lambda}$. Substituting this momentum expression into the kinetic energy formula yields $K = \frac{h^2}{2m\lambda^2}$. Because the problem specifies that both the proton and the alpha particle share the same de Broglie wavelength, the term $\frac{h^2}{2\lambda^2}$ remains constant, implying that the kinetic energy is inversely proportional to the mass of the particle, or $K \propto \frac{1}{m}$.

This inverse proportionality dictates that the ratio of the kinetic energies of the two particles is the inverse of the ratio of their masses, expressed as $\frac{K_p}{K_\alpha} = \frac{m_\alpha}{m_p}$. Since an alpha particle has a mass four times that of a proton, the resulting ratio of the kinetic energy of the proton to that of the alpha particle is $\frac{4}{1}$.

The de-Broglie wavelength of a particle is related to its kinetic energy $K$ and mass $m$ through the formula:
$\lambda = \frac{h}{\sqrt{2mK}}$
When comparing a proton and an electron that share the same kinetic energy, the terms involving the Planck constant and the kinetic energy cancel out in the ratio, leaving the comparison dependent solely on the inverse ratio of their square roots of mass:
$\frac{\lambda_p}{\lambda_e} = \sqrt{\frac{m_e}{m_p}}$
Substituting the given mass ratio $m_p = 1849 m_e$ simplifies the expression to $\sqrt{\frac{m_e}{1849 m_e}}$. Canceling the common electron mass term from the numerator and denominator results in $\sqrt{\frac{1}{1849}}$, which evaluates to $\frac{1}{43}$.

The photoelectric effect occurs only when the incident photon energy exceeds the work function of the metal surface, which represents the minimum energy required to liberate an electron. We determine the energy of the incoming radiation using the relation $E = \frac{hc}{\lambda}$, approximating the product $hc$ as $1240 \text{ eV}\cdot\text{nm}$. Plugging in the given wavelength of $350 \text{ nm}$, the energy of the incident photons is:

$E = \frac{1240}{350} \approx 3.54 \text{ eV}$

Comparing this incident energy to the specific work functions of the metals reveals that Metal B, with a work function of $2.2 \text{ eV}$, will successfully emit photo-electrons because the available $3.54 \text{ eV}$ exceeds this threshold. Conversely, Metal A has a work function of $4.8 \text{ eV}$, which is greater than the $3.54 \text{ eV}$ provided by the incident radiation. Since the incident energy is insufficient to overcome the energy barrier of Metal A, it will not emit photo-electrons.

When hydrogen atoms in the ground state absorb monochromatic light, they transition to a higher energy level and subsequently emit six distinct spectral lines as the electrons cascade back to the ground state. The total number of emission lines generated during these transitions is related to the principal quantum number $n$ of the excited state by the formula $\frac{n(n-1)}{2}$. By setting this expression equal to 6, we find that $n(n-1) = 12$, which reveals that the electrons were excited to the fourth energy level.

The energy required for an electron to jump from the ground state $n=1$ to the state $n=4$ is determined by the Bohr model energy difference $\Delta E = 13.6 \left( 1 - \frac{1}{n^2} \right)$ eV. Substituting $n=4$ into this equation gives an energy of $13.6 \left( 1 - \frac{1}{16} \right)$ eV, which simplifies to $13.6 \times \frac{15}{16}$ eV, yielding $12.75$ eV. This absorbed energy corresponds to the energy of the incident photon, governed by the equation $\Delta E = hf$. Given the constant $h = 4.25 \times 10^{-15}$ eVs, the frequency is calculated as $f = \frac{12.75}{4.25 \times 10^{-15}}$ Hz, which results in $3 \times 10^{15}$ Hz. Comparing this result to the given format $x \times 10^{15}$ Hz, the value of $x$ is 3.

The threshold wavelength for photoemission, which represents the maximum wavelength of light capable of ejecting an electron from a metal surface, is governed by the inverse relationship $\lambda = \frac{hc}{\phi}$. By utilizing the constant $hc = 1242 \text{ eV nm}$, the threshold wavelength for any metal can be determined by dividing this constant by the respective work function.

For metal A with a work function of 9 eV, the calculation is $\lambda_A = \frac{1242}{9} = 138$ nm. For metal B with a work function of 4.5 eV, the threshold wavelength is $\lambda_B = \frac{1242}{4.5} = 276$ nm. The difference between these two threshold wavelengths is calculated as $\lambda_B - \lambda_A = 276 - 138 = 138$ nm.

The photoelectric effect dictates that the maximum kinetic energy of an ejected electron is the difference between the energy of the incident photon and the work function of the metal, represented mathematically as $K_{max} = hf - hf_0$. Because the maximum kinetic energy is linked to the electron's velocity through the relationship $K_{max} = \frac{1}{2}mv^2$, changes in incident light frequency directly alter the velocity of the emitted photoelectrons.

In the first scenario with an incident frequency of $2f_0$, the energy equation gives $\frac{1}{2}mv_1^2 = 2hf_0 - hf_0$, simplifying to $\frac{1}{2}mv_1^2 = hf_0$. When the frequency is raised to $5f_0$, the energy equation yields $\frac{1}{2}mv_2^2 = 5hf_0 - hf_0$, which simplifies to $\frac{1}{2}mv_2^2 = 4hf_0$. Comparing these two results by dividing the first equation by the second gives $\frac{v_1^2}{v_2^2} = \frac{hf_0}{4hf_0} = \frac{1}{4}$. Taking the square root of this result confirms that the ratio of the velocities is $\frac{v_1}{v_2} = \frac{1}{2}$.

The photoelectric effect demonstrates the particle nature of light, where the maximum kinetic energy of emitted photoelectrons is governed by the Einstein relation $K_{max} = h\nu - \Phi$. Because increasing the intensity of incident light simply increases the photon flux, it leads to a higher rate of electron emission and a proportional increase in the saturation current. Since the stopping potential depends on both the frequency of the incident light and the metal's work function, and because the maximum kinetic energy is determined by photon frequency rather than light intensity, wave theory fails to explain these observations while particle theory successfully accounts for them.

Electromagnetic radiation carries linear momentum, and when an object fully absorbs a light pulse, it experiences a change in momentum equivalent to the total energy of that pulse divided by the speed of light. This relationship is described by the equation $p = E/c$. Since energy is calculated as power multiplied by duration, the formula expands to $p = (P \times t)/c$.

With a power of 20 mW, which is 20 x 10^-3 W, and a duration of 300 ns, which is 300 x 10^-9 s, the energy of the pulse is determined using these values. Dividing this energy by the given speed of light of 3 x 10^8 m/s yields:

$p = \frac{20 \times 10^{-3} \times 300 \times 10^{-9}}{3 \times 10^8}$

Performing the multiplication in the numerator produces 6 x 10^-9, which leads to a final momentum of $p = 2 \times 10^{-17}$ kg m/s.

The net energy change in an atom during the absorption and emission of photons is governed by the conservation of energy, where the total energy retained is the difference between the energy absorbed and the energy released as a photon. The governing expression for this energy difference is:
$E = hc \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right)$
Using the provided constants for Planck's constant and the speed of light along with the wavelengths of 500 nm and 600 nm, the net energy is calculated as 6.6 × 10^-20 J. Converting this value into electron volts gives 4.125 × 10^-1 eV. Comparing this to the target form $n \times 10^{-4} \text{ eV}$, we find that $n = 4125$.

The de Broglie wavelength of an electron is inversely proportional to the square root of its kinetic energy, as defined by the relationship $\lambda = \frac{h}{\sqrt{2mE}}$. Because Planck's constant and the mass of the electron remain unchanged, the wavelength scales according to the expression $\lambda \propto \frac{1}{\sqrt{E}}$.

When the kinetic energy is reduced to $\frac{E}{4}$, the new wavelength $\lambda'$ is related to the original wavelength by the ratio $\frac{\lambda'}{\lambda} = \sqrt{\frac{E}{E/4}}$. Simplifying the term under the square root leads to $\sqrt{4}$, which equals $2$. Consequently, the new de Broglie wavelength becomes $2\lambda$.

The photoelectric effect establishes that electron emission from a metal surface requires incident photons to have at least a minimum energy, known as the work function $\phi$. When graphing stopping potential as a function of incident frequency, the x-intercept represents the threshold frequency $\nu_{th}$ at which the energy of the incident light is exactly sufficient to overcome the binding energy of the surface electrons. Calculating the work function is then a direct application of the equation $\phi = h \nu_{th}$, where $h$ is Planck's constant. Using the values $h = 6.6 \times 10^{-34} \, \text{J}\cdot\text{s}$ and $\nu_{th} = 5 \times 10^{14} \, \text{Hz}$ identified from the graph, the energy is calculated as:

$\phi = (6.6 \times 10^{-34} \, \text{J}\cdot\text{s}) \times (5 \times 10^{14} \, \text{Hz}) = 3.3 \times 10^{-19} \, \text{J}$

To convert this quantity into the standard unit of electron-volts, divide the result by the elementary charge constant of $1.6 \times 10^{-19} \, \text{C}$:

$\phi = \frac{3.3 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{C}} = 2.07 \, \text{eV}$

Einstein's photoelectric equation dictates that the energy of an incident photon equals the sum of the metal's work function and the maximum kinetic energy of emitted electrons, expressed as the stopping potential energy $eV_0$. The initial condition is represented by the equation $\frac{hc}{\lambda} = eV_0 + \phi_0$. Upon increasing the wavelength to $2\lambda$, the stopping potential drops to $\frac{V_0}{4}$, leading to a second equation: $\frac{hc}{2\lambda} = \frac{eV_0}{4} + \phi_0$.

Eliminating the variable $eV_0$ by multiplying the second equation by four provides $\frac{2hc}{\lambda} = eV_0 + 4\phi_0$. Subtracting the initial equation from this derived expression yields $\frac{2hc}{\lambda} - \frac{hc}{\lambda} = (eV_0 + 4\phi_0) - (eV_0 + \phi_0)$, which simplifies to $\frac{hc}{\lambda} = 3\phi_0$. Replacing the work function with its equivalent definition, $\phi_0 = \frac{hc}{\lambda_0}$, results in $\frac{hc}{\lambda} = \frac{3hc}{\lambda_0}$, identifying the threshold wavelength as $3\lambda$.

Photoelectric emission occurs only if the wavelength of the incident radiation is less than the material's threshold wavelength, which is 5500 Å. This criterion means that for the effect to take place, the incident wavelength must satisfy the condition $\lambda < 5500 \text{ \AA}$.

Because the power of a light source influences only its intensity and not the individual photon energy, the wattage of the lamps does not determine whether emission occurs. Ultra-violet light naturally falls within the range of wavelengths shorter than 5500 Å, while infra-red light resides in a range with wavelengths much greater than this threshold. Consequently, both the 75 W and 10 W ultra-violet lamps satisfy the necessary condition for emission, whereas neither of the infra-red lamps can trigger the release of photoelectrons.

The de-Broglie wavelength of an electron accelerated through a potential difference $V$ is defined by the equation $\lambda = \frac{h}{\sqrt{2meV}}$, where $h$ is Planck's constant, $m$ is the mass of the electron, and $e$ is its charge. This relationship establishes that the wavelength is inversely proportional to the square root of the applied voltage, expressed as $\lambda \propto \frac{1}{\sqrt{V}}$. To determine the new wavelength when the voltage is increased from $20\,\text{kV}$ to $40\,\text{kV}$, we set up the ratio of the new wavelength to the initial wavelength $\lambda_0$ as $\frac{\lambda}{\lambda_0} = \sqrt{\frac{20\,\text{kV}}{40\,\text{kV}}}$. Simplifying this ratio results in $\sqrt{\frac{1}{2}}$, which is $\frac{1}{\sqrt{2}}$, leading to a final wavelength of $\frac{\lambda_0}{\sqrt{2}}$.

The frequency of electromagnetic radiation is fundamentally linked to the energy of its photons via the relation $E = h\nu$, which allows us to classify the radiation by calculating the frequency from the source's power output. Since power represents the total energy emitted per second, the energy of a single photon is obtained by dividing the power of 15 kW by the emission rate of $10^{16}$ photons per second, yielding $E = 15 \times 10^{-13} \text{ J}$. Applying the relation $\nu = \frac{E}{h}$ with the given Planck constant of $6 \times 10^{-34} \text{ Js}$ results in a frequency of $\nu = 2.5 \times 10^{21} \text{ Hz}$, which places this radiation firmly within the gamma ray portion of the electromagnetic spectrum.

The de Broglie wavelength of a gas molecule is governed by the relationship $\lambda = \frac{h}{\sqrt{3mkT}}$, which indicates that the wavelength is inversely proportional to the square root of the absolute temperature. To determine the effect of changing the temperature, we examine the ratio of the new wavelength to the initial wavelength using the proportionality $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_1}{T_2}}$. Substituting the given values of $300\text{ K}$ and $600\text{ K}$ into this expression results in $\sqrt{\frac{300}{600}}$, which simplifies to $\sqrt{\frac{1}{2}}$ or $\frac{1}{\sqrt{2}}$. Consequently, the new wavelength becomes $\frac{1}{\sqrt{2}} \lambda_1$.

The photoelectric effect relies on incident photons having sufficient energy to overcome a metal's work function, and since photon energy is directly proportional to frequency, ultraviolet rays are the most effective at triggering emission because they carry the highest frequency among microwaves, infrared, and ultraviolet radiation. Einstein's photoelectric equation, given by $KE_{max} = hf - hf_0$, describes the maximum kinetic energy of photoelectrons as a linear function of incident frequency $f$, where $f_0$ is the threshold frequency. Because this equation demonstrates that kinetic energy increases as frequency rises above the threshold, it contradicts the claim of inverse proportionality, identifying the second statement as false.