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

The threshold wavelength and work function are related through the formula $\lambda = \frac{hc}{\phi}$, where h is Planck's constant, c is the speed of light, and $\phi$ is the work function. Substituting the constant product value of 1242 eV nm and the given work function of 4.125 eV into the equation yields the wavelength in nanometers. Converting this resulting nanometer value into Angstroms gives the final cut-off wavelength of 3000 Å.

Wave-particle duality dictates that any moving material particle exhibits wave characteristics with a wavelength inversely proportional to its momentum. Since momentum is the product of mass and velocity, substituting this definition into the de Broglie relation yields $\lambda = \frac{h}{mv}$. This expression successfully bridges quantum mechanics and classical physics by quantifying the wave nature of matter.

The linear momentum of an electron with kinetic energy $E$ is expressed as $p = \sqrt{2mE}$, while the momentum of a photon with energy $E$ is given by $p = E / c$. Setting up the ratio of the electron's momentum to that of the photon gives $\frac{p_e}{p_{\text{ph}}} = c\sqrt{\frac{2m}{E}}$. Substituting the given values where the energy is 20 eV converted to joules using $1.6 \times 10^{-19}$ J/eV, the electron mass is $9 \times 10^{-31}$ kg, and the speed of light is $3 \times 10^8$ m/s, the expression evaluates to $3 \times 10^8 \times \sqrt{\frac{2 \times 9 \times 10^{-31}}{20 \times 1.6 \times 10^{-19}}}$. Simplifying the fraction inside the square root results in $0.5625 \times 10^{-12}$, whose square root is $0.75 \times 10^{-6}$. Multiplying this outcome by the speed of light leads to 225.

The work function of a photoelectric emitter is inversely proportional to the threshold wavelength, expressed as $W_0 = \frac{hc}{\lambda_0}$. Because the work function depends inversely on this wavelength, the ratio of the work functions for the two emitters equals the inverse ratio of their wavelengths. Dividing 600 nm by 300 nm yields a ratio of 2.

According to Einstein's photoelectric equation, the energy of the incident photon is distributed between the work function and the maximum kinetic energy of the emitted electron, represented as $\frac{hc}{\lambda} = W + \frac{1}{2}mv^2$. Substituting the given wavelength of 3000 Å and work function of 1 eV into the equation allows us to determine the surplus kinetic energy. Solving this energy balance for velocity results in a maximum speed of approximately $10^6$ meters per second.

The photocurrent in a photoelectric setup is directly proportional to the number of photons striking the surface per second, whereas the magnitude of the stopping potential $|V|$ depends exclusively on the frequency of the incident radiation according to the relation $eV_0 = hf - \phi$. As the power of the light source decreases linearly over time $t$ while keeping the frequency constant, the photon emission rate drops linearly, which directly results in a linear decrease in the photocurrent $I$. Because the frequency of the light source does not change, the maximum kinetic energy of the ejected electrons remains unaffected, meaning the magnitude of the stopping potential stays completely constant throughout.

The internal appearance and regional breakdown of a gas discharge tube are directly determined by the gas pressure maintained inside. When this pressure is lowered to approximately $0.02 \text{ mm}$, Crooke's dark space expands to occupy the entire length of the tube as the cathode rays travel unobstructed.

Photoelectric emission and the resulting stopping potential depend strictly on whether the incident wavelength is shorter than the threshold wavelength of the target metal. For the first cell, the condition $\lambda_1 < \lambda$ means the incident wavelength exceeds the threshold, so no electrons are emitted and the stopping potential $V_1$ is zero. For the second cell, the condition $\lambda_2 > \lambda$ allows photoemission, generating a stopping potential $V_2 = \frac{hc}{e}\left(\frac{1}{\lambda} - \frac{1}{\lambda_2}\right)$. For the third cell, the relation $\lambda_3 \gg \lambda$ makes the threshold wavelength extremely large, which drastically reduces the work function and yields a significantly higher stopping potential $V_3 = \frac{hc}{e}\left(\frac{1}{\lambda} - \frac{1}{\lambda_3}\right)$ that exceeds $V_2$ because the $\frac{1}{\lambda_3}$ term becomes negligible.

The threshold wavelength for a given metal is determined by the formula $\lambda_0 = \frac{hc}{W}$, representing the maximum wavelength capable of ejecting photoelectrons. Converting the work function from electron-volts to joules by multiplying 6.6 eV by $1.6 \times 10^{-19}$ J/eV gives $1.056 \times 10^{-18}$ J. Multiplying Planck's constant $6.6 \times 10^{-34}$ J s by the speed of light $3 \times 10^8$ m/s yields a numerator $hc = 1.98 \times 10^{-25}$ J m. Substituting these values into the threshold formula yields $\lambda_0 = \frac{1.98 \times 10^{-25}}{1.056 \times 10^{-18}}$, which calculates to approximately 187.5 nm. Since radiation will only trigger photoemission if its wavelength is less than this threshold, the 200 nm wavelength exceeds 187.5 nm and thus fails to produce the photoelectric effect.

The intensity of light determines the total number of photons striking a metal surface per unit time rather than the energy of individual photons. When the light intensity increases, a greater number of photons interact with surface electrons, leading to a higher rate of photoelectron emission. Consequently, this surge in the number of emitted electrons directly results in an increase in the photoelectric current.

The minimum energy required to eject an electron from a metal surface is known as the work function, which is inversely proportional to the threshold wavelength through the equation $W_0 = \frac{hc}{\lambda_0}$. Substituting Planck's constant, the speed of light, and the threshold wavelength into this relation sets up the calculation with the product of 6.63 x 10^{-34} and 3 x 10^8 in the numerator and 5000 x 10^{-10} in the denominator. Completing this arithmetic yields a total energy value of 4 x 10^{-19} joules.

The energy carried by a single photon is directly proportional to its frequency and governed by Planck's constant. Expressing frequency as the ratio of the speed of light to the wavelength transforms the relation into $E = \frac{hc}{\lambda}$. This fundamental equation demonstrates that photon energy increases as the corresponding wavelength decreases.

The momentum of a photon is connected to its frequency through the combination of Planck's relation and relativistic energy-momentum equivalence, expressed as $p = \frac{h\nu}{c}$. Rearranging this formula to solve for frequency yields $\nu = \frac{cp}{h}$. Substituting the speed of light, the given photon momentum, and Planck's constant into this expression gives $\nu = \frac{3 \times 10^8 \times 3.3 \times 10^{-29}}{6.6 \times 10^{-34}}$. Evaluating these values directly results in the required frequency of the associated waves.

The total power emitted by a source represents the product of a single photon's energy and the total rate at which photons are emitted, expressed as $n = \frac{P}{h\nu}$. Given the transmitter power and operating frequency, the emission rate is found by dividing power by the photon energy term $h\nu$. Substituting the numerical values for power, Planck's constant, and frequency into the equation gives $n = \frac{10000}{6.6 \times 10^{-34} \times 880 \times 10^3}$. Carrying out this calculation produces the total count of emitted photons per second.

The de Broglie wavelength of any particle or quantum is fundamentally determined by the ratio of Planck's constant to its momentum through the relation $\lambda = \frac{h}{p}$. Because this inverse proportionality applies equally to both matter and radiation, identical wavelengths directly dictate identical linear momentum values regardless of the differing physical natures of electrons and photons.

Applying thermal energy to a metal surface imparts sufficient kinetic energy to free loosely bound outer shell charge carriers from the atomic lattice. These thermally liberated particles are specifically referred to as thermions. Because metals contain a sea of mobile negative charge carriers, the particles emitted during thermionic processes are electrons.

The kinetic energy acquired by a charged particle accelerated through an electric potential difference is determined by the equation $K = qV$. Substituting the elementary charge of an electron along with the potential difference of 100 volts into this relationship gives the total energy imparted. Multiplying these values results in a kinetic energy of $1.602 \times 10^{-17}$ joules.

The kinetic energy gained by a charged particle is calculated using the relation $\text{K.E.} = qV$, where q is the charge and V is the potential difference. Substituting the charge of a proton and an accelerating potential of 1 V yields an energy of 1.6 x 10^-19 Joules. Because one electron-volt represents this exact amount of energy, the resulting kinetic energy simplifies to 1 eV.

The energy of a photon is directly proportional to its frequency, defined by the Planck relation $E = h\nu$. Rearranging this fundamental equation shows that the ratio of energy to frequency equals Planck's constant, which structurally connects the expression $\frac{h\nu}{E}$ to unity and describes the fundamental energetic properties of radiation.

The magnitude of the photoelectric current depends entirely on the intensity of liquid light and not on the magnitude of the work function, provided the photon energy exceeds the threshold requirement. Because the light intensity remains constant while satisfying the emission condition, the rate at which electrons are liberated does not change. Consequently, the photoelectric currents before and after the change in work function remain equal to each other.