Dual Nature Of Matter And Radiation – NEET UG PhysicsPractice Questions & PYQs

The number of photoelectrons emitted per second from a metal surface is directly proportional to the intensity of the incident light, because higher intensity delivers a greater flux of photons to the surface. Each photon has an independent probability of ejecting an electron, meaning that increasing the light intensity boosts the overall quantity of liberated electrons without changing their individual energies. Frequency and wavelength instead govern the kinetic energy of the electrons rather than their total count.

Einstein's photoelectric equation establishes that the maximum kinetic energy of an emitted photoelectron depends on the incident photon energy minus the work function, expressed as $\frac{1}{2}mv^2 = \frac{hc}{\lambda} - W_0$. When the excitation wavelength is shortened to $\frac{3\lambda}{4}$, the available photon energy increases, which directly raises the kinetic energy of the emitted electron. Substituting this into the energy relation yields $\frac{1}{2}mv_1^2 = \frac{4}{3}\left(\frac{1}{2}mv^2 + W_0\right) - W_0$, revealing an extra positive contribution from the work function that drives the final speed strictly past the scaled threshold of $v\left(\frac{4}{3}\right)^{1/2}$.

The rate of photoelectron emission from a photosensitive surface is directly governed by the number of incident photons striking it per unit time. Because light intensity correlates directly with this photon flux, supplying more intense light ejects a proportionally greater number of electrons to increase the photocurrent.

The energy of a photon is fundamentally tied to its frequency through the relation $E = h\nu$ and to its wavelength via $E = \frac{hc}{\lambda}$. Since photon energy is directly proportional to frequency, a $1 \text{ MeV}$ photon possesses one thousand times the energy of a $1 \text{ keV}$ photon. By establishing the ratio between their respective energy equations, the unknown frequency can be isolated as $\nu = \frac{10^3 c}{\lambda}$. Substituting the speed of light and the given wavelength into this expression yields $\nu = \frac{10^3 \times 3 \times 10^8}{1.24 \times 10^{-9}}$, which simplifies to the final frequency value.

The momentum of a photon is derived by combining the Planck-Einstein energy relation with the relativistic energy-momentum relation for massless particles. The energy of a photon is given by $E = \frac{hc}{\lambda}$, and its momentum relates to energy through the equation $E = pc$. Equating and simplifying these expressions yields the momentum formula $p = \frac{h}{\lambda}$.

The maximum kinetic energy of a photoelectron emitted from a metal surface is determined by Einstein's photoelectric equation, which subtracts the work function from the incident photon energy as $K = E - W_0$. Subtracting the work function of $4.2 \text{ eV}$ from the incident ultraviolet energy of $6.2 \text{ eV}$ yields a kinetic energy of $2.0 \text{ eV}$. Multiplying this value by the elementary charge converts the energy into Joules to obtain the final result.

Quantum mechanics bridges particle and wave perspectives through fundamental relations like $E = h v$ representing the energy of a photon and $\lambda = \frac{h}{p}$ giving the de Broglie wavelength. Phenomena like diffraction and interference directly showcase the wave nature of light, whereas the Compton effect provides concrete evidence of its particle nature. Pairing these principles with their correct physical interpretations yields the matching sequence where A corresponds to IV, B to III, C to I, and D to II.

The radius of an electron orbit in a hydrogen atom scales with the principal quantum number according to the relation $r = r_0 n^2$. For the second energy level where $n = 2$ and the Bohr radius is 0.052 nm, the orbital radius evaluates to $0.052 \times 4$, which equals 0.208 nm. The stationary orbit condition requires the circumference of the orbit to be an integer multiple of the de Broglie wavelength, expressed as $2\pi r = n\lambda$. Rearranging this relationship for the wavelength in the $n = 2$ state simplifies to $\lambda = \pi r$. Substituting the calculated radius of 0.208 nm into this expression gives $\pi \times 0.208$, which results in approximately 0.653 nm, closely rounding to 0.67 nm.

The de Broglie wavelength of a photon is derived from its energy relation as $\lambda_{\text{ph}} = \frac{hc}{E}$, while the de Broglie wavelength of a massive particle like an electron is expressed through its momentum as $\lambda_e = \frac{h}{\sqrt{2mE}}$. To find the required ratio, divide the photon's wavelength expression by the electron's wavelength expression. This substitution yields $\frac{\lambda_{\text{ph}}}{\lambda_e} = \frac{hc / E}{h / \sqrt{2mE}}$. Canceling Planck's constant from both numerator and denominator and simplifying the complex fraction results in the final expression $c \sqrt{\frac{2m}{E}}$.

The magnitude of the photoelectric current generated at a photosensitive surface depends directly upon the intensity of the incident light. When the light intensity increases, a larger flux of photons strikes the material, ejecting more electrons and producing a higher current in a linear fashion. Graph A is the only choice that depicts this straightforward, direct proportionality, accurately matching the expected physical behavior.

Einstein's photoelectric equation relates the maximum kinetic energy of an emitted photoelectron to the incident photon energy and the work function through the relation $K_{\max} = \frac{hc}{\lambda} - \phi$. Substituting the given wavelength value of $\lambda = \frac{hc}{e}$ into this expression results in the photon energy term simplifying directly to $e$. Subtracting the work function $\phi$ from this simplified energy leaves the maximum kinetic energy equal to $e - \phi$.

The de-Broglie wavelength of a charged particle accelerated through a potential difference $V$ is expressed by the formula $\lambda = \frac{h}{\sqrt{2mqV}}$. Since both the electron and the alpha particle experience identical accelerating potentials, their wavelengths vary inversely with the square root of the product of their mass and charge, represented as $\lambda \propto \frac{1}{\sqrt{mq}}$. Because an alpha particle possesses a substantially greater mass and charge than an electron, the resulting denominator for the alpha particle is much larger, making the de-Broglie wavelength of the electron greater than that of the alpha particle.

A photon is a fundamental quantum of electromagnetic radiation possessing specific relativistic energy and momentum characteristics. Its energy is directly proportional to frequency, and it travels through free space at the universal speed of light, while its momentum is given by the relation $p = \frac{h\nu}{c}$. Furthermore, during any collision with material particles such as electrons, fundamental conservation principles ensure that both total energy and total momentum remain conserved, and photons themselves carry zero electrical charge rather than a positive one.

The behavior of a particle's matter wave is described by the de Broglie wavelength formula, which connects wavelength $\lambda$ to momentum through Planck's constant $h$. Expressing momentum using kinetic energy $E$, the relationship transforms into $\lambda = \frac{h}{\sqrt{2mE}}$. Squaring both sides and rearranging to find the inverse square of the wavelength produces the equation $\frac{1}{\lambda^2} = \left(\frac{2m}{h^2}\right) E$. Because the mass and Planck's constant remain constant for a given particle, this equation takes the linear form of a straight line passing through the origin with a positive slope.

According to the laws of photoelectric emission, the maximum kinetic energy of emitted photoelectrons depends exclusively on the frequency of the incident radiation and the work function of the target metal surface. Increasing the intensity of the radiation simply raises the number of incident photons per second, which increases the total photocurrent without altering the individual energy packets delivered to electrons. Consequently, the maximum kinetic energy remains entirely independent of radiation intensity.

The de-Broglie wavelength of an electron accelerated by a potential difference $V$ can be directly computed using the relation $\lambda = \frac{12.27}{\sqrt{V}}$. Substituting the given potential value of 81 into the formula yields a denominator of 9, which evaluates to 1.36 angstroms. Converting this length from angstroms to nanometers by dividing by ten gives a final wavelength of 0.136 nm.

Photoelectric emission takes place only when the incident photon energy is greater than or equal to the work function of the metal surface, which is represented by $E \ge \phi$. Comparing the incident energy of 2.20 eV against the given work functions, only Caesium has a work function of 2.14 eV, which is less than the incoming energy. The work functions of Potassium at 2.30 eV and Sodium at 2.75 eV both exceed the incident energy, preventing any emission from those surfaces.

Einstein's photoelectric equation links the incident photon energy, work function, and stopping potential via the formula $h\nu = \phi + e V_s$. Substituting the given values yields an energy difference of $2.0 \text{ eV}$ for the product $e V_s$. Dividing this energy difference by the elementary charge directly gives a stopping potential value of 2 V.

The maximum kinetic energy of photoelectrons is determined by subtracting the work function from the energy of the incident photons using the equation $K_{\max} = h\nu - h\nu_0$. Substituting the incident frequency of $4\nu_0$ into this relation yields $K_{\max} = h(4\nu_0) - h\nu_0$. Simplifying this expression directly results in a maximum kinetic energy of $3h\nu_0$.

Applying Einstein's photoelectric relation $eV = h\nu - h\nu_0$ to both experimental scenarios provides a system of equations connecting the frequencies and stopping potentials. Substituting the second condition into the first yields the relation $\frac{1}{2}\left(h\frac{\nu}{2} - h\nu_0\right) = h\nu - h\nu_0$. Rearranging this algebraic expression isolates the threshold frequency as $\frac{3}{2}\nu$.