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

The de Broglie wavelength of a moving particle is fundamentally expressed by the equation $\lambda = \frac{h}{p}$, where $h$ represents Planck's constant and $p$ denotes the particle's momentum. Because Planck's constant remains fixed, the wavelength varies inversely with momentum, showing an inverse proportionality written as $\lambda \propto \frac{1}{p}$. When visualizing this mathematical dependence on a Cartesian plane, the resulting curve takes the shape of a rectangular hyperbola that asymptotically approaches both axes.

The power delivered by a light source represents the total energy of the emitted photons per unit time, expressed through the formula $P = \frac{n}{t}\left(\frac{hc}{\lambda}\right)$. Rearranging this equation allows us to calculate the rate of photon emission as $\frac{n}{t} = \frac{P\lambda}{hc}$. Substituting the given parameters into this derived expression yields a photon emission rate of $10^{16}$ photons per second.

Photoelectric emission only occurs when the frequency of incident radiation exceeds the threshold frequency of the photosensitive material. Halving the initial frequency of $1.5$ times the threshold frequency results in a new frequency of $0.75$ times the threshold frequency. Because this reduced frequency falls below the threshold limit, electron emission ceases completely and the photoelectric current becomes zero.

The de Broglie wavelength of an accelerated electron is given by the formula $\lambda = \frac{12.27}{\sqrt{V}}$ angstroms. Converting the given wavelength into angstroms and substituting it into the expression yields $0.1227 = \frac{12.27}{\sqrt{V}}$. Solving for the square root of the potential gives $\sqrt{V} = 100$, which upon squaring results in a potential difference of $10^4$ volts.

The de Broglie wavelength of an electron accelerated through a given potential is calculated with the formula $\lambda = \frac{12.27}{\sqrt{V}}$ angstroms. Inserting the potential value of $10000$ volts into this expression yields $\lambda = \frac{12.27}{100}$ angstroms. Converting this length into meters gives $12.27 \times 10^{-12}$ meters, which matches the required magnitude.

The threshold wavelength for photoelectric emission is related to the work function through the inverse proportionality equation $\lambda = \frac{hc}{\phi}$. Using the standard energy-wavelength constant $hc = 1240$ electron-volt nanometers simplifies the computation significantly. Substituting the work function of $4.0$ electron-volts into the formula yields $\lambda = \frac{1240}{4.0}$, which calculates directly to $310$ nanometers.

The de Broglie wavelength of a particle is determined by the momentum relation $\lambda = \frac{h}{mv}$. As the electron moves against the opposing electric field, its acceleration is given by $a = \frac{eE_0}{m}$, leading to a time-dependent velocity of $v = V_0 + \frac{eE_0}{m}t$. Substituting this velocity into the wavelength expression and incorporating the initial wavelength $\lambda_0 = \frac{h}{mV_0}$ simplifies the equation into the desired fractional form.

Einstein's photoelectric equation relates the maximum kinetic energy of emitted photoelectrons to the frequency of incident radiation through the relation $h\nu - h\nu_0 = \frac{1}{2}mv^2$. Applying this principle to the first scenario with incident frequency $2v_0$ gives $h(2v_0) - hv_0 = \frac{1}{2}mv_1^2$, which simplifies to $hv_0 = \frac{1}{2}mv_1^2$. For the second scenario with an increased frequency of $5v_0$, the equation becomes $h(5v_0) - hv_0 = \frac{1}{2}mv_2^2$, resulting in $4hv_0 = \frac{1}{2}mv_2^2$. Dividing the first expression by the second eliminates the common terms to yield $\frac{v_1^2}{v_2^2} = \frac{1}{4}$, which directly gives a velocity ratio of 1 to 2 after taking the square root.

The maximum kinetic energy of an emitted electron can be evaluated using the wavelength form of Einstein's photoelectric equation, expressed as $\frac{1}{2}mv^2 = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)$. Rearranging this expression for velocity gives $v = \sqrt{\frac{2hc}{m} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)}$. Substituting the given values for Planck's constant, the speed of light, electron mass, and the respective wavelengths into the formula evaluates the numerical magnitude. Computing this value yields a velocity on the order of 6 × 10^5 m s^{-1}.

The kinetic energy of a neutron in thermal equilibrium at temperature T is determined by the equipartition of energy as $\frac{3}{2}KT$, where K is the Boltzmann constant. Combining this thermal energy with the general de-Broglie wavelength formula $\lambda = \frac{h}{\sqrt{2mE}}$ yields the particle's wavelength expression. Substituting the thermal kinetic energy into the wavelength equation gives $\lambda = \frac{h}{\sqrt{2m\left(\frac{3}{2}KT\right)}}$. Simplifying the terms inside the square root results in the final expression $\frac{h}{\sqrt{3mKT}}$.

Einstein's photoelectric equation, given by $K_{\max} = E - \phi$, relates the maximum kinetic energy of photoelectrons to the incident photon energy and the work function. Subtracting the maximum kinetic energy of 2 eV from the incident photon energy of 5 eV determines the work function to be 3 eV. When the incident photon energy is raised to 6 eV, the new maximum kinetic energy becomes 3 eV, requiring a stopping potential of -3 V to completely halt the photocurrent.

The de-Broglie wavelength associated with an electron of mass m and energy E is expressed through the momentum-energy relation as $\lambda_e = \frac{h}{\sqrt{2mE}}$. For a photon carrying the same energy E, the wavelength is obtained from the Planck-Einstein relation combined with the wave speed equation, yielding $\lambda_p = \frac{hc}{E}$. Taking the ratio of the electron's wavelength to the photon's wavelength involves dividing the first expression by the second. Simplifying this ratio yields $\frac{1}{c} \left( \frac{E}{2m} \right)^{1/2}$.

The kinetic energy of an electron is connected to its de-Broglie wavelength through the relation $K = \frac{h^2}{2 m \lambda^2}$. Equating this kinetic energy to the maximum energy of an emitted X-ray photon gives the relation $\frac{hc}{\lambda_0} = \frac{h^2}{2 m \lambda^2}$. Rearranging this expression directly yields the cutoff wavelength as $\lambda_0 = \frac{2 m c \lambda^2}{h}$.

Einstein's photoelectric relation connects the stopping potential of a surface to the incident wavelength and the threshold wavelength via the equation $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_{\text{th}}}$. Applying this principle to the first condition with wavelength $\lambda$ and stopping potential V establishes the initial equation. For the second condition where the wavelength is $2\lambda$ and the stopping potential reduces to a quarter of its original value, a second parallel equation is formed. Solving these two simultaneous equations eliminates the stopping potential and work function parameters, isolating the threshold wavelength as $3\lambda$.

The stopping potential is linked to the incident wavelength and threshold wavelength via the photoelectric equation $e V_s = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$. Applying this principle to both illumination cases generates two simultaneous equations containing the unknown threshold wavelength. Eliminating the work function term between these equations allows the threshold wavelength to be evaluated directly as a multiple of the initial wavelength.

The de Broglie wavelength of an electron is determined by its kinetic energy through the relation $\lambda = \frac{h}{\sqrt{2 m K}}$. Subtracting the work function of 2.28 eV from the photon energy corresponding to the 500 nm wavelength yields a maximum kinetic energy of 0.2 eV. Substituting this maximum kinetic energy into the de Broglie formula computes the threshold value, establishing that the resulting electron wavelength must be greater than or equal to 2.8 × 10^-9 m.

Einstein's photoelectric equation expresses maximum kinetic energy as the difference between incident photon energy and the material work function, represented as $K_{\max} = \frac{hc}{\lambda} - \phi_0$. Setting up this equation for both wavelengths and applying the given condition that the second kinetic energy is triple the first yields $K_{\max 2} = 3 K_{\max 1}$. Solving this proportional relationship isolates the work function explicitly in terms of Planck's constant, speed of light, and wavelength.

The de-Broglie wavelength of a particle is inversely proportional to its momentum, as described by the relation $P = \frac{h}{\lambda}$. Because one variable decreases as the other increases while maintaining a constant product, their graphical representation forms a rectangular hyperbola.

Einstein's photoelectric equation relates the maximum kinetic energy of emitted photoelectrons to the energy of incident radiation and the work function of the metal through the expression $K = E - \phi$. Applying this principle to the initial scenario gives the relation $0.5 = E - \phi$, and increasing the radiation energy by 20 percent yields the second relation $0.8 = 1.2E - \phi$. Multiplying the first equation by 1.2 results in $0.6 = 1.2E - 1.2\phi$, which can then be subtracted from the second equation to eliminate the energy term and yield $0.2 = 0.2\phi$, leading directly to a work function value of 1.0 eV.

The de Broglie wavelength of a particle depends inversely on the square root of its kinetic energy according to the formula $\lambda = \frac{h}{\sqrt{2mK}}$. When the kinetic energy is increased to 16 times its original value, the new wavelength becomes $\lambda' = \frac{h}{\sqrt{2m(16K)}}$, which simplifies to one-fourth of the initial wavelength. The percentage change is then calculated as the ratio of the difference between the initial and final wavelengths to the initial wavelength multiplied by 100, resulting in a 75 percent reduction.