Step 1: Identifying Given Parameters and Constants

Based on the problem statement, we have the following physical quantities:

• Frequency of incident radiation ($ν$): $7.21 \times 10^{14} \, \text{Hz}$
• Maximum speed of emitted electrons ($v_{max}$): $6.0 \times 10^5 \, \text{m/s}$
• Planck’s constant ($h$): $6.626 \times 10^{-34} \, \text{J}\cdot\text{s}$
• Mass of an electron ($m$): $9.11 \times 10^{-31} \, \text{kg}$

Step 2: Theoretical Framework

According to Einstein’s Photoelectric Equation [Per the Law of Conservation of Energy applied to quantum interactions]:

$hν = \phi_0 + K_{max}$

Where $\phi_0$ is the work function defined as $\phi_0 = hν_0$ (where $ν_0$ is the threshold frequency) and $K_{max}$ is the maximum kinetic energy of the emitted photoelectrons.

Step 3: Calculating Maximum Kinetic Energy ($K_{max}$)

The maximum kinetic energy is calculated using the classical kinetic energy formula as the velocity is non-relativistic ($v \ll c$):

$K_{max} = \frac{1}{2}mv_{max}^2$

$K_{max} = 0.5 \times (9.11 \times 10^{-31} \, \text{kg}) \times (6.0 \times 10^5 \, \text{m/s})^2$

$K_{max} = 0.5 \times 9.11 \times 10^{-31} \times 36 \times 10^{10} \, \text{J}$

$K_{max} = 1.6398 \times 10^{-19} \, \text{J}$

Step 4: Deriving the Threshold Frequency ($ν_0$)

Rearranging the photoelectric equation to solve for the threshold frequency ($ν_0$):

$hν = hν_0 + K_{max}$

$hν_0 = hν - K_{max}$

$ν_0 = ν - \frac{K_{max}}{h}$

Substituting the numerical values:

$ν_0 = (7.21 \times 10^{14}) - \frac{1.6398 \times 10^{-19}}{6.626 \times 10^{-34}}$

Calculating the second term:

$\frac{1.6398 \times 10^{-19}}{6.626 \times 10^{-34}} \approx 0.24748 \times 10^{15} \, \text{Hz} = 2.475 \times 10^{14} \, \text{Hz}$

Final subtraction:

$ν_0 = 7.21 \times 10^{14} - 2.475 \times 10^{14}$

$ν_0 = 4.735 \times 10^{14} \, \text{Hz}$

Visual Representation of Photoelectric Effect

Final Solution: The threshold frequency for the metal surface is $4.74 \times 10^{14} \, \text{Hz}$.

Given Variables & Initial Setup

The photoelectric effect is governed by Einstein's Photoelectric Equation. We identify the provided physical constants and parameters:

• Wavelength of incident light ($\lambda$): $488 \text{ nm} = 488 \times 10^{-9} \text{ m}$
• Stopping potential ($V_0$): $0.38 \text{ V}$
• Planck’s constant ($h$): $6.626 \times 10^{-34} \text{ J}\cdot\text{s}$
• Speed of light ($c$): $3 \times 10^8 \text{ m/s}$
• Elementary charge ($e$): $1.602 \times 10^{-19} \text{ C}$

Step 1: Calculate Energy of Incident Photon

The energy of an incident photon ($E$) is determined by the relation [Per the Planck-Einstein Relation]:

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

$E = \frac{(6.626 \times 10^{-34} \text{ J}\cdot\text{s}) \times (3 \times 10^8 \text{ m/s})}{488 \times 10^{-9} \text{ m}}$

$E \approx 4.073 \times 10^{-19} \text{ J}$

To convert Joules to electron-volts (eV) [Conversion factor: $1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}$]:

$E (\text{in eV}) = \frac{4.073 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 2.54 \text{ eV}$

Step 2: Calculate Maximum Kinetic Energy

The maximum kinetic energy ($K_{max}$) of the photoelectrons is related to the stopping potential ($V_0$) by the relation [Per the definition of Stopping Potential]:

$K_{max} = eV_0$

$K_{max} = e \times 0.38 \text{ V} = 0.38 \text{ eV}$

Step 3: Determining the Work Function

Einstein’s Photoelectric Equation states that the energy of the incident photon is the sum of the work function ($\Phi_0$) and the maximum kinetic energy [Per the Conservation of Energy]:

$E = \Phi_0 + K_{max}$

Rearranging for the work function:

$\Phi_0 = E - K_{max}$

$\Phi_0 = 2.54 \text{ eV} - 0.38 \text{ eV}$

$\Phi_0 = 2.16 \text{ eV}$

Visual Representation of Photoelectric Effect

Final Solution: The work function of the material is 2.16 eV.

Step 1: Identifying Given Variables and Theoretical Foundation

To find the minimum wavelength ($\lambda_{min}$) of X-rays produced by an electron accelerated through a potential difference ($V$), we utilize the principle of conservation of energy. The kinetic energy gained by the electron is converted into a photon of maximum energy (corresponding to minimum wavelength) upon hitting the target.

• Potential Difference ($V$) = $30\text{ kV} = 30 \times 10^3\text{ V}$
• Elementary Charge ($e$) $\approx 1.602 \times 10^{-19}\text{ C}$
• Planck’s Constant ($h$) $\approx 6.626 \times 10^{-34}\text{ J}\cdot\text{s}$
• Speed of Light ($c$) $\approx 3.0 \times 10^8\text{ m/s}$

Step 2: Formulating the Governing Equation

The energy of the accelerated electron is given by $E = eV$ [Per the definition of electric potential energy]. The maximum energy of the emitted photon is given by $E = \frac{hc}{\lambda_{min}}$ [Per the Planck-Einstein relation]. Equating these:

$eV = \frac{hc}{\lambda_{min}}$

Rearranging to solve for $\lambda_{min}$:

$\lambda_{min} = \frac{hc}{eV}$

Step 3: Calculating the Numerical Value

Substitute the constants into the equation:

$\lambda_{min} = \frac{(6.626 \times 10^{-34}\text{ J}\cdot\text{s}) \times (3.0 \times 10^8\text{ m/s})}{(1.602 \times 10^{-19}\text{ C}) \times (30 \times 10^3\text{ V})}$

Calculating the numerator:

$hc = 19.878 \times 10^{-26}\text{ J}\cdot\text{m}$

Calculating the denominator:

$eV = 4.806 \times 10^{-15}\text{ J}$

Performing the division:

$\lambda_{min} = \frac{19.878 \times 10^{-26}}{4.806 \times 10^{-15}} \approx 4.136 \times 10^{-11}\text{ m}$

Step 4: Visualization of Photon Emission

Step 5: Final Unit Conversion

Converting meters to angstroms ($\text{\AA}$) where $1\text{ \AA} = 10^{-10}\text{ m}$:

$\lambda_{min} = 0.4136 \times 10^{-10}\text{ m} \approx 0.414\text{ \AA}$

Final Solution: The minimum wavelength of the X-rays produced is approximately $4.14 \times 10^{-11}\text{ m}$ or $0.414\text{ \AA}$.

Step 1: Given Parameters & Constants

To determine the de Broglie wavelength ($\lambda$) of a macroscopic object, we utilize the de Broglie relation. The provided parameters are as follows:

• Mass of the dust particle ($m$): $1.0 \times 10^{-9} \text{ kg}$
• Velocity of the dust particle ($v$): $2.2 \text{ m/s}$
• Planck’s constant ($h$): $6.626 \times 10^{-34} \text{ J}\cdot\text{s}$

Step 2: Theoretical Formulation

The de Broglie wavelength is related to the momentum ($p$) of the particle, where momentum is the product of mass and velocity. The governing equation is defined by the de Broglie hypothesis [Per the de Broglie relation]:

$\lambda = \frac{h}{p} = \frac{h}{mv}$

Step 3: Algebraic Calculation

Substitute the given numerical values into the equation:

$\lambda = \frac{6.626 \times 10^{-34} \text{ J}\cdot\text{s}}{(1.0 \times 10^{-9} \text{ kg}) \times (2.2 \text{ m/s})}$

Perform the multiplication in the denominator:

$m \times v = 2.2 \times 10^{-9} \text{ kg}\cdot\text{m/s}$

Perform the division:

$\lambda = \frac{6.626}{2.2} \times \frac{10^{-34}}{10^{-9}}$

$\lambda \approx 3.0118 \times 10^{-25} \text{ meters}$

Step 4: Visualization of the Wave-Particle Duality Context

Step 5: Analytical Interpretation

The calculated value of $\lambda \approx 3.0 \times 10^{-25} \text{ m}$ is infinitesimally small, far below the measurable limits of physical instrumentation (e.g., atomic dimensions are typically on the order of $10^{-10} \text{ m}$). This confirms that the wave nature of macroscopic objects is negligible [By the correspondence principle of quantum mechanics].

Final Solution: The de Broglie wavelength of the dust particle is approximately $3.0 \times 10^{-25} \text{ m}$.

Step 1: Given Variables & Constants

• Wavelength of light ($\lambda$): $632.8 \text{ nm} = 632.8 \times 10^{-9} \text{ m}$
• Power emitted ($P$): $9.42 \text{ mW} = 9.42 \times 10^{-3} \text{ W}$
• Planck's constant ($h$): $6.626 \times 10^{-34} \text{ J}\cdot\text{s}$
• Speed of light in vacuum ($c$): $3 \times 10^8 \text{ m/s}$

Step 2: Calculating the Energy of a Single Photon ($E$)

The energy of a photon is defined by the Planck-Einstein relation [Per the Quantum Theory of Radiation]:

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

Substituting the values:

$E = \frac{(6.626 \times 10^{-34} \text{ J}\cdot\text{s}) \times (3 \times 10^8 \text{ m/s})}{632.8 \times 10^{-9} \text{ m}}$

$E = \frac{19.878 \times 10^{-26}}{632.8 \times 10^{-9}} \text{ J}$

$E \approx 3.141 \times 10^{-19} \text{ J}$

Step 3: Calculating the Momentum of a Single Photon ($p$)

The momentum of a photon is derived from the De Broglie relation, which links wave-particle duality [Per the De Broglie Hypothesis]:

$p = \frac{h}{\lambda} = \frac{E}{c}$

Substituting the values:

$p = \frac{3.141 \times 10^{-19} \text{ J}}{3 \times 10^8 \text{ m/s}}$

$p \approx 1.047 \times 10^{-27} \text{ kg}\cdot\text{m/s}$

Visual Representation of Photon Properties

Summary Table of Photon Properties

Final Solution: The energy of each photon is $3.141 \times 10^{-19} \text{ J}$ and the momentum of each photon is $1.047 \times 10^{-27} \text{ kg}\cdot\text{m/s}$.

Step 1: Identifying Given Variables

• Wavelength of light, $\lambda = 632.8 \, \text{nm} = 632.8 \times 10^{-9} \, \text{m}$
• Power emitted, $P = 9.42 \, \text{mW} = 9.42 \times 10^{-3} \, \text{W}$
• Planck’s constant, $h = 6.626 \times 10^{-34} \, \text{J}\cdot\text{s}$
• Speed of light, $c = 3.00 \times 10^8 \, \text{m/s}$

Step 2: Theoretical Formulation of Photon Energy

The energy of a single photon ($E_{ph}$) is governed by the Planck-Einstein relation [Per the Photon Energy Postulate]:

$E_{ph} = \frac{hc}{\lambda}$

Step 3: Calculating Energy per Photon

Substituting the values into the equation:

$E_{ph} = \frac{(6.626 \times 10^{-34} \, \text{J}\cdot\text{s}) \times (3.00 \times 10^8 \, \text{m/s})}{632.8 \times 10^{-9} \, \text{m}}$

$E_{ph} \approx \frac{1.9878 \times 10^{-25}}{632.8 \times 10^{-9}} \, \text{J}$

$E_{ph} \approx 3.141 \times 10^{-19} \, \text{J}$

Step 4: Determining Photons per Second

Let $n$ be the number of photons per second. Since Power ($P$) is defined as energy per unit time, and the total energy emitted per second is $n \times E_{ph}$, we have [By the Principle of Energy Conservation]:

$P = n \times E_{ph}$

$n = \frac{P}{E_{ph}}$

Step 5: Final Computational Derivation

$n = \frac{9.42 \times 10^{-3} \, \text{J/s}}{3.141 \times 10^{-19} \, \text{J/photon}}$

$n \approx 3.0 \times 10^{16} \, \text{photons/s}$

Final Solution: The number of photons arriving at the target per second is $3.0 \times 10^{16} \, \text{photons/s}$.

Step 1: Theoretical Foundation and Variable Identification

In an X-ray tube, the kinetic energy of an accelerated electron is converted into photon energy upon collision with the target anode. The maximum frequency of the emitted X-ray radiation corresponds to the case where the entire kinetic energy of the incident electron is converted into a single photon [Per the Conservation of Energy Principle].

• Accelerating potential ($V$) = $30\text{ kV} = 30 \times 10^3\text{ V}$
• Charge of an electron ($e$) = $1.6 \times 10^{-19}\text{ C}$
• Planck's constant ($h$) = $6.626 \times 10^{-34}\text{ J}\cdot\text{s}$
• Speed of light ($c$) = $3 \times 10^8\text{ m/s}$

Step 2: Establishing the Governing Equation

The maximum energy ($E_{max}$) of the X-ray photon is equal to the work done on the electron by the accelerating electric field:

$E_{max} = eV = h\nu_{max}$

Rearranging the equation to isolate the maximum frequency ($\nu_{max}$):

$\nu_{max} = \frac{eV}{h}$

Step 3: Calculation of Maximum Frequency

Substitute the identified variables into the derived equation:

$\nu_{max} = \frac{(1.6 \times 10^{-19}\text{ C}) \times (30 \times 10^3\text{ V})}{6.626 \times 10^{-34}\text{ J}\cdot\text{s}}$

$\nu_{max} = \frac{4.8 \times 10^{-15}}{6.626 \times 10^{-34}}\text{ Hz}$

$\nu_{max} \approx 0.7244 \times 10^{19}\text{ Hz}$

$\nu_{max} \approx 7.24 \times 10^{18}\text{ Hz}$

Step 4: Visual Representation of Energy Conversion

Final Solution: The maximum frequency of the X-rays produced is $7.24 \times 10^{18}\text{ Hz}$.

Step 1: Given Variables & Physical Constants

To determine the stopping potential ($V_0$), we utilize the Einstein Photoelectric Equation. The following constants and given values are established:

• Work Function of Caesium ($\Phi_0$): $2.14 \, \text{eV}$
• Frequency of incident radiation ($\nu$): $6 \times 10^{14} \, \text{Hz}$
• Planck’s constant ($h$): $6.626 \times 10^{-34} \, \text{J}\cdot\text{s}$
• Elementary charge ($e$): $1.602 \times 10^{-19} \, \text{C}$
• Conversion factor: $1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}$

Step 2: Energy of Incident Photon

The energy of the incident photon ($E$) is calculated using the relation $E = h\nu$ [Per Planck’s Quantum Theory].

$E = (6.626 \times 10^{-34} \, \text{J}\cdot\text{s}) \times (6 \times 10^{14} \, \text{Hz})$

$E = 3.9756 \times 10^{-19} \, \text{J}$

Converting this value into electron-volts (eV) for consistency with the work function:

$E (\text{in eV}) = \frac{3.9756 \times 10^{-19} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} \approx 2.48 \, \text{eV}$

Step 3: Maximum Kinetic Energy of Photoelectrons

According to Einstein’s Photoelectric Equation [Per Conservation of Energy]:

$K_{\text{max}} = E - \Phi_0$

$K_{\text{max}} = 2.48 \, \text{eV} - 2.14 \, \text{eV}$

$K_{\text{max}} = 0.34 \, \text{eV}$

Step 4: Calculation of Stopping Potential

The stopping potential ($V_0$) is defined by the relation $K_{\text{max}} = eV_0$ [Per the definition of stopping potential in photoelectric effect studies].

$eV_0 = 0.34 \, \text{eV}$

$V_0 = 0.34 \, \text{V}$

Visual Representation of Photoelectric Emission

Final Solution: The stopping potential for the given frequency is $0.34 \, \text{V}$.