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

The de Broglie wavelength for a particle is defined by the relationship $\lambda = \frac{h}{\sqrt{2mK}}$, which establishes that for a constant kinetic energy, the wavelength is inversely proportional to the square root of the mass of the particle. Because the masses of these particles follow the order $m_\alpha > m_n > m_p > m_e$, it logically follows that the particle with the greatest mass will have the smallest wavelength. Consequently, this inverse relationship dictates that the wavelengths are ordered as $\lambda_\alpha < \lambda_n < \lambda_p < \lambda_e$.

The Davisson-Germer experiment provides the definitive experimental confirmation that electrons exhibit wave-like characteristics, consistent with the de Broglie relation $\lambda = \frac{h}{p}$. Because interference and diffraction are phenomena unique to wave motion, the presence of these behaviors in electrons is a direct consequence of their established wave nature. Therefore, both the experimental proof of the electron's wave character and the logical link between this nature and diffraction effects are true.

Einstein's photoelectric equation dictates that the maximum kinetic energy of an emitted electron is the difference between the incident photon's energy and the metal's work function, given by $K_{max} = E - \Phi$. Since kinetic energy relates to velocity through the relation $K = \frac{1}{2}mv^2$, the square of the speed is directly proportional to the kinetic energy.

For the first photon, the maximum kinetic energy is 3.8 eV - 0.6 eV, which equals 3.2 eV. For the second photon, the calculation results in a maximum kinetic energy of 1.4 eV - 0.6 eV, which equals 0.8 eV. By taking the ratio of these two kinetic energies, we find
$\frac{K_1}{K_2} = \frac{3.2}{0.8} = 4$
and since the speed is proportional to the square root of the kinetic energy, the ratio of the speeds is simply
$\frac{v_1}{v_2} = \sqrt{4} = 2$
resulting in a final ratio of 2:1.

Einstein's photoelectric equation states that $\frac{1}{2} m v_m^2 = h \nu - \phi$, where $\phi$ represents the metal's work function. Because $m$ and $\phi$ are constants, the square of the maximum velocity $v_m^2$ must vary linearly with the frequency of the incident radiation $\nu$, confirming the first statement. Regarding the wave model, it predicts that light energy would be distributed uniformly across the wavefront, implying that emission should occur at any frequency given enough time and intensity, which contradicts the observed existence of a threshold wavelength; thus, the wave nature of light cannot account for this phenomenon. Other statements fail because saturation current is linked to light intensity rather than source distance, kinetic energy remains independent of light power, and the particle nature of light specifically provides, rather than inhibits, an explanation for the immediate emission of photoelectrons.

The de-Broglie wavelength of a particle is inversely proportional to its momentum, expressed as $\lambda = \frac{h}{p}$. Since the electron and the photon are given as having the same de-Broglie wavelength, their momenta must be identical, allowing us to denote both as $p$.

The kinetic energy of the electron can be written in terms of its momentum as $E_e = \frac{1}{2}mv^2$, which simplifies to $E_e = \frac{pv}{2}$ when substituting $p = mv$. In contrast, the energy of a photon is defined by the relation $E_{ph} = pc$. Taking the ratio of the electron's kinetic energy to the photon's energy, the common momentum term cancels out, yielding $\frac{E_e}{E_{ph}} = \frac{pv/2}{pc} = \frac{v}{2c}$.

Einstein's photoelectric equation establishes the relationship between incident photon energy, the material's work function, and the maximum kinetic energy of emitted photoelectrons. This is expressed as $hf = hf_{th} + eV_0$, where $f$ is the frequency of incident light, $f_{th}$ is the threshold frequency, and $eV_0$ represents the maximum kinetic energy defined by the stopping potential. By rearranging this equation, the threshold frequency is found using the expression $f_{th} = \frac{c}{\lambda} - \frac{eV_0}{h}$.

Calculating the incident frequency yields $\frac{c}{\lambda} = \frac{3 \times 10^8}{6630 \times 10^{-10}} \approx 45.25 \times 10^{13} \text{ Hz}$. The frequency equivalent of the stopping potential is determined as $\frac{eV_0}{h} = \frac{1.6 \times 10^{-19} \times 0.42}{6.63 \times 10^{-34}} \approx 10.14 \times 10^{13} \text{ Hz}$. Subtracting this term from the incident frequency gives a threshold frequency of approximately $35.11 \times 10^{13} \text{ Hz}$, making the nearest integer value for $x$ equal to 35.

The de-Broglie wavelength $\lambda$ of a particle is tied to its kinetic energy $E$ by an inverse square root relationship, represented by the formula $\lambda = \frac{h}{\sqrt{2mE}}$. Because the Planck constant $h$ and the mass $m$ of the particle remain fixed throughout the process, any change in the wavelength is dictated entirely by the change in kinetic energy. Consequently, the ratio of the initial wavelength $\lambda_1$ to the final wavelength $\lambda_2$ corresponds to the square root of the ratio of the final kinetic energy $E_2$ to the initial kinetic energy $E_1$, expressed as $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{E_2}{E_1}}$.

When the wavelength is reduced to 75% of its initial value, the final wavelength becomes $\lambda_2 = 0.75 \lambda_1$. Substituting this into the ratio gives $\frac{\lambda_1}{0.75 \lambda_1} = \frac{1}{0.75} = \frac{4}{3}$, which is equal to $\sqrt{\frac{E_2}{E_1}}$. By squaring both sides of this equation, we obtain $\frac{16}{9} = \frac{E_2}{E_1}$. Since the final energy is the sum of the initial energy $E$ and the added energy $\Delta E$, we set the equation as $\frac{E + \Delta E}{E} = \frac{16}{9}$. Expanding this leads to $16E = 9E + 9\Delta E$, which simplifies to $9\Delta E = 7E$, indicating that the required additional energy is $\Delta E = \frac{7}{9}E$.

The cut-off wavelength in an X-ray tube marks the point where the maximum energy of the incident electron is fully converted into the energy of a photon. The kinetic energy $K$ of the electron, based on its momentum $P$, is represented as $K = \frac{P^2}{2m}$, and substituting the de-Broglie wavelength relation $P = \frac{h}{\lambda}$ transforms this into $K = \frac{h^2}{2m\lambda^2}$. Because the energy of the resulting photon with cut-off wavelength $\lambda_c$ is defined by $E = \frac{hc}{\lambda_c}$, the conservation of energy requires the equality $\frac{h^2}{2m\lambda^2} = \frac{hc}{\lambda_c}$. Isolating $\lambda_c$ from this expression yields $\frac{2mc\lambda^2}{h}$.

The de-Broglie wavelength of a particle is defined by the relation $\lambda = \frac{h}{p}$, where $h$ is Planck's constant and $p$ represents the particle's momentum. Since both the proton and electron have the same de-Broglie wavelength, their momenta must be identical.

Relating this to energy, we use the expression $K = \frac{p^2}{2m}$, where $m$ denotes mass and $K$ stands for kinetic energy. Because the momentum is constant for both, kinetic energy is inversely proportional to mass; consequently, since the proton has a greater mass than the electron, the kinetic energy of the proton must be lower, resulting in $K_p < K_e$.

The photoelectric effect dictates that the energy of an incident photon is divided between the metal's work function and the maximum kinetic energy of the ejected electron. A hydrogen atom transitioning from the $n=3$ to $n=2$ state emits a photon with an energy of 1.89 eV. Once these photoelectrons enter a magnetic field $B$ of 5 x 10^-4 T, they follow a circular path with a radius $r$ defined by the balance between magnetic and centripetal forces. The governing equation for the radius of the electron's path is:

$r = \frac{mv}{qB}$

From this relationship, we can derive the electron's kinetic energy $K$ using the expression for momentum $p = mv = qBr$ to obtain:

$K = \frac{(qBr)^2}{2m}$

By substituting the electron mass $m$ of 9.1 x 10^-31 kg, the charge $q$ of 1.6 x 10^-19 C, the magnetic field $B$ of 5 x 10^-4 T, and the radius $r$ of 7 x 10^-3 m into the expression, the maximum kinetic energy is calculated to be 1.077 eV. Subtracting this value from the initial photon energy of 1.89 eV reveals the work function $\phi$ of the metal to be 0.813 eV, which corresponds to 0.82 eV.

When a hydrogen atom transitions from an excited state to the ground state, it sheds energy by emitting a photon. Because the system is initially at rest, the law of conservation of linear momentum requires that the atom recoil with a momentum exactly equal in magnitude to the momentum carried away by the photon. This relationship is described by $mv = p$, where $m$ represents the mass of the hydrogen atom, $v$ is its recoil velocity, and $p$ denotes the momentum of the photon.

The photon's momentum is linked to its wavelength $\lambda$ through the de Broglie expression $p = h/\lambda$, where $h$ is Planck's constant. The wavelength of the emitted photon is derived from the Rydberg formula, $1/\lambda = R(1/n_2^2 - 1/n_1^2)$. By substituting the energy levels $n = 5$ and $n = 1$ with the Rydberg constant $R = 1.09 \times 10^7$, the expression $1/\lambda = 1.09 \times 10^7 \times (24/25)$ represents the inverse wavelength.

Combining these principles, the velocity of the atom is defined by $v = h/(m\lambda)$. Inserting the known values $h = 6.63 \times 10^{-34}$ and $m = 1.6 \times 10^{-27}$ into the equation, we arrive at:
$v = \frac{h \times R \times 24}{m \times 25} = \frac{6.63 \times 10^{-34} \times 1.09 \times 10^7 \times 24}{1.6 \times 10^{-27} \times 25}$
This calculation yields a recoil speed of 4.17 m/s.

The photoelectric effect is governed by the principle of energy conservation, where the energy of an incident photon is shared between the metal's work function and the maximum kinetic energy of the emitted electrons. This relationship is mathematically expressed as $eV = \frac{hc}{\lambda} - \phi$, where $V$ is the stopping potential, $\lambda$ is the wavelength, and $\phi$ is the work function. Using the constant approximation for $hc$ as 1240 eV nm, the expression becomes $V = \frac{1240}{\lambda} - \phi$ when the potential is measured in volts and the wavelength in nanometers.

For the initial conditions, the relationship is $0.710 = \frac{1240}{491} - \phi$. When the stopping potential shifts to 1.43 V with a new wavelength, the equation becomes $1.43 = \frac{1240}{\lambda_2} - \phi$. Subtracting the initial equation from the second eliminates the constant work function, leaving the calculation $1.43 - 0.710 = 1240 \left( \frac{1}{\lambda_2} - \frac{1}{491} \right)$. This reduces to $0.72 = 1240 \left( \frac{1}{\lambda_2} - \frac{1}{491} \right)$. Rearranging the terms results in $\frac{1}{\lambda_2} = \frac{0.72}{1240} + \frac{1}{491}$, which simplifies to approximately 0.00262. Inverting this value provides the final wavelength of 382 nm.

The minimum wavelength of an X-ray photon occurs when an accelerated electron releases its total kinetic energy in a single collision at the target. This energy conversion is defined by the relationship $eV = \frac{hc}{\lambda}$ where $e$ is the elementary charge, $V$ is the accelerating potential, $h$ represents Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the photon.

To determine the shortest wavelength, rearrange the equation to $\lambda = \frac{hc}{eV}$. Using the values $h = 6.67 \times 10^{-34}$ J s, $c = 3 \times 10^8$ m/s, $e = 1.6 \times 10^{-19}$ C, and $V = 1.24 \times 10^6$ V, we substitute these into the expression: $\lambda = \frac{6.67 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 1.24 \times 10^6}$ This calculation yields a result of $\lambda \approx 1.009 × 10^{-12}$ meters. Given that one nanometer equals $10^{-9}$ meters, the final wavelength is approximately $\lambda = 10^{-3}$ nm.

The de-Broglie wavelength of a particle is fundamentally linked to its mass and kinetic energy through the relationship $\lambda = \frac{h}{\sqrt{2mE}}$. When the kinetic energy $E$ remains constant for different particles, the wavelength is inversely proportional to the square root of the particle's mass, expressed as $\lambda \propto \frac{1}{\sqrt{m}}$. Because an alpha particle possesses a significantly greater mass than a proton, which in turn is much heavier than an electron ($m_\alpha > m_p > m_e$), this inverse relationship dictates that the particle with the smallest mass will yield the largest wavelength. Consequently, the electron, being the lightest of the three, results in the longest wavelength, followed by the proton, with the heaviest alpha particle resulting in the shortest wavelength, establishing the order $\lambda_e > \lambda_p > \lambda_\alpha$.

The behavior of a photon is governed by the fundamental de Broglie relation $\lambda = \frac{h}{p}$, where wavelength is inversely proportional to momentum, and the energy formula $E = \frac{hc}{\lambda}$. Because wavelength and momentum share this inverse dependency, two photons with equal linear momenta must mathematically possess identical wavelengths, confirming that the first statement is true. Conversely, decreasing a photon's wavelength leads to a corresponding increase in both its momentum and energy due to these inverse relationships, which demonstrates that the second statement is false.

The resolving power of any microscope is fundamentally limited by the wavelength of the radiation used, following the relationship $\text{Resolving Power} \propto \frac{1}{\lambda}$. Because the de-Broglie wavelength of electrons is significantly shorter than the wavelength of visible light, an electron microscope can distinguish much finer details than an optical instrument. Since the smaller wavelength is the direct physical cause enabling superior resolution, the stated reason provides the correct explanation for the assertion.

When an electron and a proton are accelerated through the same potential difference $\Delta V$, they both acquire the same kinetic energy $K = e\Delta V$. The de-Broglie wavelength for a particle is defined by the relationship $\lambda = \frac{h}{\sqrt{2mK}}$, where $m$ is the mass of the particle. By substituting the kinetic energy expression into this equation, the wavelength becomes $\lambda = \frac{h}{\sqrt{2m(e\Delta V)}}$. Since the constants $h$, $e$, and $\Delta V$ are identical for both particles, the wavelength is inversely proportional to the square root of the particle's mass, or $\lambda \propto \frac{1}{\sqrt{m}}$. Applying this proportionality to the ratio of the electron's wavelength $\lambda_e$ to the proton's wavelength $\lambda_P$ yields $\frac{\lambda_e}{\lambda_P} = \sqrt{\frac{m_p}{m_e}}$.

The resolving power of a microscope is governed by the de Broglie wavelength of the particles used, where a smaller wavelength enables higher resolution. This wavelength follows the relationship $\lambda = \frac{h}{mv}$, and since resolving power is inversely proportional to the wavelength, it scales as $RP \propto \frac{mv}{h}$. As the velocity is held constant, the resolving power is directly proportional to the mass of the particle, expressed as $RP \propto m$. Consequently, the ratio of the resolving power of protons to that of electrons is determined by the ratio of their masses, $\frac{RP_p}{RP_e} = \frac{m_p}{m_e}$. Since a proton is $1837$ times more massive than an electron, the resolving power of the scanning proton microscope increases by a factor of $1837$.

Every entity, whether a photon or a moving particle, possesses a characteristic wavelength inversely proportional to its momentum as defined by the de Broglie relationship. This allows us to define the wavelength of a photon as $\lambda_1 = h/p$ and a particle as $\lambda_2 = h/(mv)$, where $h$ is Planck's constant, $m$ is the particle mass, $v$ is the particle velocity, and $p$ is the photon momentum.

Determining the relationship between these wavelengths involves taking the ratio $\lambda_1/\lambda_2$, which simplifies to $mv/p$ as the Planck constant cancels out. Using the given particle mass of 9.1 x 10^-31 kg, velocity of 10^6 m/s, and photon momentum of 10^-27 kg m/s, the ratio calculation becomes:

$\frac{\lambda_1}{\lambda_2} = \frac{9.1 \times 10^{-31} \times 10^6}{10^{-27}} = 910$

The photon wavelength is 910 times that of the particle.

The De-Broglie wavelength establishes a fundamental relationship between a particle's wavelength and its momentum, expressed as $\lambda = \frac{h}{p}$. Since the electron and the photon have identical wavelengths, they must also share the same momentum, which we can represent as $p$. The kinetic energy of the electron is defined by the relation $K_e = \frac{p^2}{2m}$, whereas the energy of the photon is given by $E_{ph} = pc$.

To determine the ratio of their energies, we divide the kinetic energy of the electron by the energy of the photon, resulting in the expression $\frac{p^2 / 2m}{pc}$. Simplifying this fraction leads to $\frac{p}{2mc}$. Because the momentum of the electron is given by $p = mv$, substituting this into the expression yields $\frac{mv}{2mc}$, which simplifies to $\frac{v}{2c}$.