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

The behavior of photoelectrons is governed by Einstein's photoelectric equation, which dictates that the kinetic energy of an emitted electron equals the photon's energy minus the material's work function, expressed as $\frac{1}{2}mv^2 = hf - \phi$. Since both photocathodes are identical, they share the same work function $\phi$, allowing us to establish two separate equations for the respective incident frequencies $f_1$ and $f_2$ and resulting velocities $v_1$ and $v_2$. Subtracting the second equation from the first eliminates the work function term, resulting in $\frac{1}{2}mv_1^2 - \frac{1}{2}mv_2^2 = h(f_1 - f_2)$. Rearranging this expression by multiplying both sides by $\frac{2}{m}$ isolates the difference of the squares of the velocities, leading to the relationship $v_1^2 - v_2^2 = \frac{2h}{m}(f_1 - f_2)$.

The relationship between the wavelength and energy is distinct for massive particles and massless quanta. For an electron of mass $m$, the kinetic energy $E$ relates to momentum $p$ through $E = p^2 / (2m)$, which allows us to express its de Broglie wavelength as:

$\lambda_e = \frac{h}{\sqrt{2mE}}$

Conversely, the wavelength of a photon with energy $E$ is defined by the energy-momentum relation $E = hc / \lambda$, which yields:

$\lambda_p = \frac{hc}{E}$

Dividing the wavelength of the electron by the wavelength of the photon allows us to find their ratio:

$\frac{\lambda_e}{\lambda_p} = \frac{h / \sqrt{2mE}}{hc / E}$

Canceling the Planck constant $h$ from both the numerator and the denominator simplifies the expression to:

$\frac{\lambda_e}{\lambda_p} = \frac{E}{c\sqrt{2mE}}$

By combining the terms inside the square root, we arrive at the final ratio:

$\frac{\lambda_e}{\lambda_p} = \frac{1}{c} \left( \frac{E}{2m} \right)^{1/2}$

The photoelectric effect relies on the principle that the kinetic energy of an ejected electron corresponds to the incident photon energy minus the work function of the metal, expressed by the equation $V_s = \frac{hc}{\lambda} - \phi$. Applying this to the provided scenarios, we have $4.8 = \frac{hc}{\lambda} - \phi$ for the first case and $1.6 = \frac{hc}{2\lambda} - \phi$ for the second. Subtracting the second equation from the first allows us to eliminate the work function, resulting in $3.2 = \frac{hc}{\lambda} - \frac{hc}{2\lambda}$. This simplifies to $\frac{hc}{2\lambda} = 3.2$, which implies that $\frac{hc}{\lambda} = 6.4$. Substituting this value back into the first equation shows that the work function equals 1.6. Since the threshold wavelength $\lambda_{\text{th}}$ is defined by the condition $\phi = \frac{hc}{\lambda_{\text{th}}}$, we can use the relation $1.6 = \frac{6.4\lambda}{\lambda_{\text{th}}}$, which confirms that $\lambda_{\text{th}} = 4\lambda$.

Einstein's photoelectric equation describes the energy balance of the interaction, stating that the maximum kinetic energy of emitted photoelectrons equals the difference between the incident photon energy and the work function of the metal surface. Given that the maximum kinetic energy corresponds to the product of the electron charge and the stopping potential, the relationship for any wavelength $\lambda$ is:
$eV_s = \frac{hc}{\lambda} - \phi$

Applying this expression to the two provided scenarios, we formulate a system of equations. The first condition gives $3eV_0 = \frac{hc}{\lambda} - \phi$, while the second condition, using a radiation wavelength of $2\lambda$, leads to:
$eV_0 = \frac{hc}{2\lambda} - \phi$

Subtracting the second equation from the first eliminates the work function $\phi$ and yields $2eV_0 = \frac{hc}{\lambda} - \frac{hc}{2\lambda}$, which simplifies to $2eV_0 = \frac{hc}{2\lambda}$, or $eV_0 = \frac{hc}{4\lambda}$. Substituting this expression for $eV_0$ back into the second equation allows for the calculation of the work function as $\phi = \frac{hc}{2\lambda} - \frac{hc}{4\lambda} = \frac{hc}{4\lambda}$. Since the threshold wavelength $\lambda_t$ is defined by the relation $\phi = \frac{hc}{\lambda_t}$, we equate $\frac{hc}{\lambda_t} = \frac{hc}{4\lambda}$, which indicates that the threshold wavelength for the surface is $4\lambda$.

Energy conservation governs the transition of an electron captured by a proton, as the total energy released during the atom's formation appears as an emitted photon. The electron begins with a kinetic energy of 3 eV and settles into the hydrogen atom's second excited state, which corresponds to the principal quantum number $n = 3$. The energy level of this bound state is determined by the formula:
$E_n = -13.6 / n^2$
Substituting $n = 3$ yields an energy of -1.51 eV. Because the electron loses energy transitioning from a free state to this bound state, the emitted photon carries an energy equivalent to $3 - (-1.51) = 4.51$ eV.

When this photon strikes the photosensitive metal, it can eject a photoelectron if its energy exceeds the metal's work function. The work function $\phi$ is determined from the threshold wavelength 4000 Å using the equation:
$\phi = 12400 / 4000 = 3.1$
Applying Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectron is the difference between the incident photon energy and the work function:
$K_{\text{max}} = 4.51 - 3.1 = 1.41$
The resulting maximum kinetic energy is 1.41 eV.

The decay of a particle of mass 4M initially at rest relies on the law of conservation of linear momentum. Since the initial momentum of the stationary particle is zero, the two resulting fragments with masses M and 3M must recoil in opposite directions with momenta of equal magnitude, so $p_M = p_{3M}$. De-Broglie's relation states that $\lambda = \frac{h}{p}$, where $h$ represents Planck's constant. Because the magnitude of momentum is identical for both pieces, their wavelengths must be equivalent, making the ratio 1:1.

The de Broglie wavelength formula, expressed as $\lambda = \frac{h}{mv}$, dictates that if two particles share the same wavelength, they must necessarily share the same momentum since Planck's constant $h$ is universal. Because the momentum of the proton equals that of the alpha particle, we can set their mass-velocity products equal, specifically $m_p v_p = m_\alpha v_\alpha$. Recognizing that an alpha particle has four times the mass of a proton, we substitute $m_\alpha = 4m_p$ into the relationship to obtain $m_p v_p = 4m_p v_\alpha$. Dividing both sides by the mass of the proton leads directly to $v_p = 4v_\alpha$, resulting in a velocity ratio of 4:1.

The photoelectric effect is governed by the principle that the maximum kinetic energy of emitted electrons is determined by the energy of individual incident photons rather than the total brightness or intensity of the light. According to Einstein's photoelectric equation, $eV_0 = h\nu - \varphi$, where $V_0$ is the stopping potential, $h$ is Planck's constant, $\nu$ is the frequency, and $\varphi$ is the work function of the metal. Because the work function is a constant property of the material, the stopping potential depends exclusively on the frequency of the radiation, as this determines how much excess energy is available to impart kinetic energy to the electrons.

The de Broglie wavelength of a moving particle is inversely proportional to its mass when velocity remains constant, as defined by the relationship $\lambda = \frac{h}{mv}$. Because Planck's constant $h$ and the velocity $v$ are identical for both the electron and the proton, the ratio of their wavelengths is equivalent to the inverse ratio of their masses, or $\frac{\lambda_{\text{electron}}}{\lambda_{\text{proton}}} = \frac{m_{\text{proton}}}{m_{\text{electron}}}$. Given that the proton mass is 1836 times the mass of the electron, substituting $m_{\text{proton}} = 1836 m_e$ into this relationship allows the $m_e$ terms to cancel out, leaving a final value of 1836.

The intensity of incident light in the photoelectric effect corresponds to the density of photons striking the surface, meaning that an increase in intensity leads to a larger number of photons interacting with the material and subsequently a higher number of ejected electrons. The kinetic energy of these ejected electrons, however, is determined by the energy of individual photons rather than the total light intensity, a relationship defined by $\frac{1}{2} m_{e} v_{e}^{2} = e V_{0}$. Since the stopping potential $V_{0}$ remains constant regardless of variations in light intensity, the kinetic energy of the emitted electrons does not change when the intensity is increased.

The de-Broglie wavelength of a particle is defined by the relationship $\lambda = \frac{h}{mv}$, which indicates an inverse proportionality between wavelength and the product of mass and velocity. Because the particle's wavelength is twice that of the electron and its velocity is four times greater, the equation $\frac{\lambda_p}{\lambda_e} = \frac{m_e v_e}{m_p v_p}$ becomes $2 = \frac{m_e}{m_p} \left( \frac{v_e}{4v_e} \right)$. Solving this expression for the mass of the particle results in $m_p = \frac{1}{8} m_e$.

The de-Broglie wavelength of a charged particle accelerated through a potential difference $V$ is defined by the relationship $\lambda = \frac{h}{\sqrt{2mqV}}$, where $h$ is Planck's constant, $m$ is the particle's mass, and $q$ is its charge. Because both the electron and the proton are subjected to the same potential $100 \text{ V}$ and carry equal magnitudes of charge, the ratio of their wavelengths simplifies to $\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}}$.

Substituting the given mass values of $m_p = 1.00727 \text{ u}$ and $m_e = 0.00055 \text{ u}$ leads to the calculation $\sqrt{\frac{1.00727}{0.00055}}$, which is approximately $\sqrt{1831.4}$. Evaluating this square root yields a value of about $42.79$, demonstrating that the ratio of the wavelengths is nearly $43:1$.

In thermal equilibrium, the average kinetic energy of gas particles is determined strictly by the temperature, meaning that for an electron and a proton at the same temperature, their average kinetic energies must be identical. Using the definition of kinetic energy in terms of momentum, $KE = \frac{p^2}{2m}$, we can equate the expressions for both particles to find that $\frac{p_e^2}{2m_e} = \frac{p_p^2}{2m_p}$. Rearranging this equation demonstrates that the ratio of the squared momenta is equal to the ratio of their masses, specifically $\frac{p_p^2}{p_e^2} = \frac{m_p}{m_e}$, which implies that the momentum of each particle is proportional to the square root of its mass.

According to the Heisenberg uncertainty principle, the uncertainty in determining a particle's position is inversely proportional to its momentum, approximated as $\Delta x \approx \frac{h}{\Delta p}$. Since the uncertainty in momentum scales with the particle's momentum, the ratio of the position uncertainty of the electron to that of the proton corresponds to the inverse ratio of their momenta, $\frac{\Delta x_e}{\Delta x_p} = \frac{p_p}{p_e}$. Substituting the square root relationship derived from the kinetic energy comparison into this ratio leads to the result of $\sqrt{\frac{m_p}{m_e}}$.

In a three-dimensional ideal gas, the average kinetic energy of the particles is linked to temperature through the relation $E = \frac{3}{2} k_B T$. By incorporating this into the de-Broglie wavelength formula $\lambda = \frac{h}{\sqrt{2mE}}$, we derive an expression that connects the wavelength directly to the thermal energy:

$\lambda = \frac{h}{\sqrt{3 m k_B T}}$

Using the given values-Planck's constant as 6.6 × 10^-34 Js, mass of the electron as 9 × 10^-31 kg, Boltzmann constant as 1.38 × 10^-23 JK^-1, and temperature as 300 K-the calculation is:

$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{3 \times 9 \times 10^{-31} \times 1.38 \times 10^{-23} \times 300}}$

This calculation results in approximately 6.26 × 10^-9 meters, which is equivalent to 6.26 nm.

The relationship between incident photon energy, the metal's work function, and the stopping potential is governed by Einstein's photoelectric equation. This principle establishes that the energy of an incoming photon is split between the energy required to overcome the work function of the metal and the maximum kinetic energy of the emitted photoelectron, which is proportional to the stopping potential $V_0$. This is expressed by the formula $eV_0 = \frac{hc}{\lambda} - \phi$. By using the constant $hc = 1240\text{ eV}\cdot\text{nm}$, the calculation simplifies to $V_0 = \frac{1240}{\lambda} - \phi$, where $V_0$ is in volts and $\lambda$ is in nanometers.

For the initial condition with a wavelength of 280 nm and a work function of 2.5 eV, the stopping potential is calculated as $V_0 = \frac{1240}{280} - 2.5$, which yields 1.93 V. When the wavelength is changed to 400 nm, the new stopping potential becomes $V_0 = \frac{1240}{400} - 2.5$, resulting in 0.6 V. Subtracting the second potential from the first, 1.93 V - 0.6 V, gives a change of 1.33 V, which is approximately 1.3 V.

Einstein's photoelectric equation dictates that the maximum kinetic energy of an emitted photoelectron is the difference between the incident photon energy and the metal's work function, expressed as $\frac{1}{2}mv^2 = E - \phi$. In this expression, $m$ is the mass of the electron, $v$ is the maximum velocity, $E$ is the incident energy, and $\phi$ is the work function.

For the first case, the incident energy is $2\phi$, which yields a maximum kinetic energy of $\frac{1}{2}mv_1^2 = 2\phi - \phi = \phi$. In the second case, the incident energy is $10\phi$, resulting in a maximum kinetic energy of $\frac{1}{2}mv_2^2 = 10\phi - \phi = 9\phi$. Dividing the first kinetic energy equation by the second eliminates the common terms and gives $\frac{v_1^2}{v_2^2} = \frac{\phi}{9\phi} = \frac{1}{9}$. Taking the square root of both sides results in a velocity ratio of $\frac{v_1}{v_2} = \frac{1}{3}$. By comparing this result to the given ratio $x:y$, we find that the value of $x$ is 1.

The photoelectric effect dictates that the maximum kinetic energy of an ejected electron is the difference between the incident photon energy and the metal's work function. The incident photon energy is determined by the formula $E = \frac{hc}{\lambda}$, which, using the provided $hc = 20 \times 10^{-26} \text{ J-m}$ and a wavelength of 500 nm, equals $4 \times 10^{-19} \text{ J}$. Dividing this value by the elementary charge $1.6 \times 10^{-19} \text{ C}$ converts the photon energy to $2.5 \text{ eV}$. Subtracting the work function of $1.25 \text{ eV}$ leaves a maximum kinetic energy of $1.25 \text{ eV}$, which corresponds to $2 \times 10^{-19} \text{ J}$.

When the emitted electron enters the magnetic field, the Lorentz force acts as the centripetal force responsible for the circular path of radius 0.3 m. The relationship between the radius $r$ and kinetic energy $K$ is $r = \frac{\sqrt{2mK}}{eB}$, which is rearranged to $B = \frac{\sqrt{2mK}}{er}$. Substituting the mass $m = 9 \times 10^{-31} \text{ kg}$, the kinetic energy $K = 2 \times 10^{-19} \text{ J}$, and the radius $r = 0.3 \text{ m}$ into the equation gives $B = \frac{\sqrt{2 \times 9 \times 10^{-31} \times 2 \times 10^{-19}}}{1.6 \times 10^{-19} \times 0.3}$. This calculation simplifies to $12.5 \times 10^{-6} \text{ T}$, which is equal to $125 \times 10^{-7} \text{ T}$.

The energy liberated during the capture of a free electron into a bound state of a hydrogen atom is determined by the principle of conservation of energy. The photon emitted carries away the difference between the initial kinetic energy of the free electron and the final potential energy of the atom in its $n=2$ excited state. The energy of this state is calculated as $E_2 = -\frac{13.6}{2^2}\text{ eV} = -3.4\text{ eV}$. Since the electron starts with $2.6\text{ eV}$, the energy difference available for the photon is $\Delta E = 2.6\text{ eV} - (-3.4\text{ eV}) = 6.0\text{ eV}$.

Converting this energy into standard units using the conversion $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ provides $\Delta E = 6.0 \times 1.6 \times 10^{-19}\text{ J} = 9.6 \times 10^{-19}\text{ J}$. Applying the relationship between energy and frequency, $\nu = \frac{\Delta E}{h}$, where Planck's constant is $h = 6.6 \times 10^{-34}\text{ Js}$, results in $\nu = \frac{9.6 \times 10^{-19}}{6.6 \times 10^{-34}}\text{ Hz}$. This division yields a frequency of approximately $1.45 \times 10^{15}\text{ Hz}$. Finally, dividing by $10^6$ to express this value in megahertz gives a frequency of $1.45 \times 10^9\text{ MHz}$.

The de Broglie wavelength for a charged particle accelerated through a potential difference of 200 V is defined by the relationship $\lambda = \frac{h}{\sqrt{2mqV}}$. Since both the alpha particle and the proton are accelerated by this same voltage, the wavelength of each depends inversely on the square root of the product of its mass and charge, causing the voltage to cancel out in the final ratio. This means the ratio $\frac{\lambda_p}{\lambda_\alpha}$ simplifies to $\sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}}$.

An alpha particle has a mass of $4m_p$ and a charge of $2e$, whereas a proton has a mass of $m_p$ and a charge of $e$. Substituting these values into the ratio provides $\sqrt{\frac{4m_p \times 2e}{m_p \times e}}$, which simplifies to $\sqrt{8}$. Calculating this expression yields approximately 2.828, which results in a final ratio of 2.8.

Einstein's photoelectric equation defines the relationship between the incident photon energy, the material's work function, and the maximum kinetic energy of the emitted photoelectrons. Since the stopping potential $V$ relates to this kinetic energy via the expression $eV = \frac{hc}{\lambda} - \phi_0$, we recognize that the material's work function $\phi_0$ remains an invariant property regardless of the incident light source.

By comparing the energy balance for both the initial and the new wavelengths, we can isolate the difference in stopping potentials directly in terms of the incident photon energies. This leads to the derived expression $V_2 = V_1 + \frac{hc}{e} \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \right)$.

Substituting the physical constants $h = 6.63 \times 10^{-34}$ J s, $c = 3 \times 10^8$ m/s, and $e = 1.6 \times 10^{-19}$ C, alongside the provided wavelengths $\lambda_1 = 670.5$ nm and $\lambda_2 = 474.6$ nm, the shift in energy adds 0.77 V to the original 0.48 V. Consequently, the new stopping voltage is 1.25 V.