Dual Nature of Radiation

Step 1: Formula for de-Broglie Wavelength

The de-Broglie wavelength is given by: $ \lambda = \frac{h}{\sqrt{2mk}} $

Step 2: Rearranging to Find Kinetic Energy

We can rearrange the formula to find kinetic energy $k$: $ \lambda^2 = \frac{h^2}{2mk} $ So, $ k = \frac{h^2}{2m\lambda^2} $

Step 3: Substitute the Values

Here,

• Planck's constant, $ h = 6.6 \times 10^{-34} $ Js
• Mass of electron, $ m = 9 \times 10^{-31} $ kg
• Wavelength, $ \lambda = 2 $ nm $ = 2 \times 10^{-9} $ m

Step 4: Calculation

Calculate the value: $ k = \frac{(6.6 \times 10^{-34})^2}{2 \times 9 \times 10^{-31} \times (2 \times 10^{-9})^2} $ This gives: $ k = 0.608 \times 10^{-19} \text{ J} $

Step 5: Convert Joules to Electron Volts

1 electron volt (eV) $ = 1.6 \times 10^{-19} $ J.
So, $ k = \frac{0.608 \times 10^{-19}}{1.6 \times 10^{-19}} \text{ eV} = 0.38 \text{ eV} $

The kinetic energy of the electron is about $ 0.38 $ eV.

$ \begin{aligned} &\text { de-Broglie wavelength }\\ &\begin{aligned} \lambda & =\frac{h}{\sqrt{2 m q V}} \\ \Rightarrow \quad \lambda & \propto \frac{1}{\sqrt{m q}} \\ \frac{\lambda_p}{\lambda_\alpha} & =\sqrt{\frac{m_\alpha}{m_p}} \sqrt{\frac{q_\alpha}{q_p}}=\sqrt{\frac{4}{1} \cdot \frac{2}{1}}=\frac{2 \sqrt{2}}{1} \\ \therefore \lambda_p: \lambda_\alpha & =2 \sqrt{2}: 1 \end{aligned} \end{aligned} $

Given:

Threshold wavelength,

$\lambda_0 = 6250 \, \text{Å} = 6250 \times 10^{-10} \, \text{m} = 6.25 \times 10^{-7} \, \text{m}$

Planck’s constant,

$h = 6.6 \times 10^{-34} \, \text{J·s}$

Speed of light,

$c = 3 \times 10^8 \, \text{m/s}$

Formula:

Work function $ \phi = h c / \lambda_0 $

Calculation:

$ \phi = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{6.25 \times 10^{-7}} $

$ \phi = \frac{19.8 \times 10^{-26}}{6.25 \times 10^{-7}} = 3.168 \times 10^{-19} \, \text{J} $

Converting to eV:

$1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$

$ \phi = \frac{3.168 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.98 \, \text{eV} $

✅ Answer: Option B — 1.98 eV

$ \begin{aligned} & \text { Since, } E=\frac{h c}{\lambda} \\ & \begin{aligned} \Rightarrow \lambda & =\frac{h c}{E}=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{3 \times 1.6 \times 10^{-19}} \\ & =4.13 \times 10^{-7}=413 \times 10^{-9} \mathrm{~m} \\ & =413 \mathrm{~nm} \end{aligned} \end{aligned} $

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

when $\lambda^{\prime}=\lambda+0.25 \%$ of $\lambda=1.0025 \lambda$

So, $\quad 1.0025 \lambda=\frac{h}{p-p_0}$

$\Rightarrow \quad p-p_0=\frac{h}{1.0025}=\frac{p \lambda}{1.0025 \lambda}$

$\Rightarrow \quad \frac{p-p_0}{p}=\frac{1}{1.0025}$

$\Rightarrow \quad p=400 p_0$

$ \text { } \begin{aligned} E & =\frac{h c}{\lambda} \\ \Rightarrow \quad \lambda & =\frac{h c}{E}=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{5 \times 1.6 \times 10^{-19}} \\ = & 248 \times 10^{-9} \mathrm{~m}=248 \mathrm{~nm} \end{aligned} $

To find the maximum kinetic energy of the ejected photoelectrons, we'll use the photoelectric effect equation:

where $KE_{\text{max}}$ is the maximum kinetic energy of the ejected electrons, $h$ is Planck's constant, $\nu$ is the frequency of the incident light, and $\phi$ is the work function of the metal.

However, in this problem, we are given the energy of the UV light in electronvolts (eV) directly, which simplifies the problem. The energy of the UV light in electronvolts also represents the energy of the photons ($h\nu$) hitting the metal surface. Thus, we can calculate the maximum kinetic energy of the ejected photoelectrons using the given energies directly:

Substituting the given values:

Therefore, the maximum kinetic energy of the ejected photoelectrons will be $1 \, \text{eV}$. The correct answer is Option B: 1 eV.

To determine the relationship between the de-Broglie wavelengths of a proton, an electron, and an alpha particle with the same energies, we need to use the de-Broglie wavelength formula:

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

where $ h $ is Planck's constant and $ p $ is the momentum of the particle.

For particles with the same kinetic energy $ E $, we have:

$E = \frac{p^2}{2m}$

Solving for $ p $:

$p = \sqrt{2mE}$

Substituting this into the de-Broglie equation, we get:

$\lambda = \frac{h}{\sqrt{2mE}}$

Since all three particles have the same energy $ E $, the de-Broglie wavelength is inversely proportional to the square root of the mass $ m $:

$\lambda \propto \frac{1}{\sqrt{m}}$

The masses of the particles are as follows:

• Mass of proton $ m_p $ is approximately $ 1.67 \times 10^{-27} $ kg
• Mass of electron $ m_e $ is approximately $ 9.11 \times 10^{-31} $ kg
• Mass of alpha particle $ m_{\alpha} $ is approximately $ 4 \times 1.67 \times 10^{-27} $ kg (since it has 2 protons and 2 neutrons)

Comparatively:

• Mass of alpha particle $ m_{\alpha} $ is the largest.
• Mass of proton $ m_p $ is intermediate.
• Mass of electron $ m_e $ is the smallest.

Therefore, the de-Broglie wavelength will be:

• Largest for the electron ($ \lambda_{\mathrm{e}} $)
• Intermediate for the proton ($ \lambda_{\mathrm{p}} $)
• Smallest for the alpha particle ($ \lambda_{\alpha} $)

Hence, the correct order is:

Option B:

$\lambda_\alpha<\lambda_{\mathrm{p}}<\lambda_{\mathrm{e}}$

To determine the correct relation between the kinetic energies of a proton ($\mathrm{K}_{\mathrm{p}}$) and an electron ($\mathrm{K}_{\mathrm{e}}$) when they have the same de Broglie wavelength, we need to use the de Broglie wavelength formula:

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

where $\lambda$ is the de Broglie wavelength, $h$ is Planck's constant, and $p$ is the momentum of the particle.

The momentum $p$ of a particle is given by:

$p = \sqrt{2mK}$

where $m$ is the mass of the particle and $K$ is its kinetic energy. For a proton and an electron with the same de Broglie wavelength:

$\lambda_{\mathrm{p}} = \lambda_{\mathrm{e}}$

This implies the momenta should be the same:

$p_{\mathrm{p}} = p_{\mathrm{e}}$

Thus, we can write:

$\sqrt{2 m_{\mathrm{p}} K_{\mathrm{p}}} = \sqrt{2 m_{\mathrm{e}} K_{\mathrm{e}}}$

Squaring both sides to eliminate the square roots:

$2 m_{\mathrm{p}} K_{\mathrm{p}} = 2 m_{\mathrm{e}} K_{\mathrm{e}}$

$m_{\mathrm{p}} K_{\mathrm{p}} = m_{\mathrm{e}} K_{\mathrm{e}}$

Rearranging to solve for $K_{\mathrm{p}}$ in terms of $K_{\mathrm{e}}$:

$K_{\mathrm{p}} = \frac{m_{\mathrm{e}}}{m_{\mathrm{p}}} K_{\mathrm{e}}$

Since the mass of a proton $m_{\mathrm{p}}$ is much greater than the mass of an electron $m_{\mathrm{e}}$:

$m_{\mathrm{p}} \gg m_{\mathrm{e}}$

This means:

$\frac{m_{\mathrm{e}}}{m_{\mathrm{p}}} \ll 1$

Therefore, it implies:

$K_{\mathrm{p}} < K_{\mathrm{e}}$

So, the correct option is:

Option C: $\mathrm{K}_{\mathrm{p}}<\mathrm{K}_{\mathrm{e}}$

To solve for the ratio of the kinetic energies of a proton and an electron with the same de-Broglie wavelength, let us first recall the relationship between kinetic energy, momentum, and the de-Broglie wavelength.

The de-Broglie wavelength $\lambda$ is given by:

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

where $h$ is Planck’s constant and $p$ is the momentum.

Since the proton and the electron have the same de-Broglie wavelength, their momenta must be equal:

$\lambda_{\text{electron}} = \lambda_{\text{proton}} \implies \frac{h}{p_{\text{e}}} = \frac{h}{p_{\text{p}}} \implies p_{\text{e}} = p_{\text{p}}$

Next, the kinetic energy $K$ of a particle is related to its momentum $p$ and mass $m$ by the following formula:

$K = \frac{p^2}{2m}$

Given that the momentum $p$ is the same for both the proton and the electron, we can write the kinetic energies as:

$K_{\text{e}} = \frac{p^2}{2m_{\text{e}}}$

$K_{\text{p}} = \frac{p^2}{2m_{\text{p}}}$

The ratio of the kinetic energies is therefore:

$\frac{K_{\text{e}}}{K_{\text{p}}} = \frac{\frac{p^2}{2m_{\text{e}}}}{\frac{p^2}{2m_{\text{p}}}} = \frac{m_{\text{p}}}{m_{\text{e}}}$

Given that the mass of the proton $m_{\text{p}}$ is 1836 times the mass of the electron $m_{\text{e}}$, we have:

$\frac{m_{\text{p}}}{m_{\text{e}}} = 1836$

Thus, the ratio of the kinetic energies is:

$\frac{K_{\text{e}}}{K_{\text{p}}} = 1836$

Therefore, the correct answer is:

Option C

$1: 1836$

To find the stopping potential ($V_s$) when UV light of wavelength $300 $ nm is incident on a metal surface with a work function of $2.13$ eV, we can use the photoelectric equation which relates the energy of the incident photons, the work function of the metal, and the kinetic energy of the emitted electrons.

The energy (E) of the photons can be calculated using the equation:

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

where $h$ is the Planck constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the incident light. Given that $hc = 1240 $ eV nm, we can calculate the energy of the UV light photons directly.

Substituting $hc = 1240 $ eV nm and $\lambda = 300 $ nm into the equation gives:

$E = \frac{1240 \, \text{eV nm}}{300 \, \text{nm}} = 4.13 \, \text{eV}$

Next, we can calculate the maximum kinetic energy of the emitted electrons using the photoelectric effect equation:

$K_{max} = E - \phi$

where $K_{max}$ is the maximum kinetic energy of the emitted electrons, $E$ is the energy of the incident photons, and $\phi$ is the work function of the metal.

Given $E = 4.13$ eV and the work function $\phi = 2.13$ eV, we have:

$K_{max} = 4.13 \, \text{eV} - 2.13 \, \text{eV} = 2 \, \text{eV}$

The stopping potential ($V_s$) is related to the maximum kinetic energy of the emitted electrons by the equation:

$K_{max} = eV_s$

where $e$ is the elementary charge (the charge of an electron), and $V_s$ is the stopping potential. Since $e = 1$ when using energy in eV and potential in volts, the stopping potential $V_s$ can be directly equated to the kinetic energy in eV:

$V_s = K_{max} = 2 \, \text{V}$

Therefore, the stopping potential is $2$ V, which corresponds to Option B.

The correct answer is Option A: C only.

Reflection, diffraction, interference, and polarization are all phenomena that can be explained by the wave nature of light. These phenomena are evidence that light behaves as a wave, evident through various experimental observations:

• Reflection is the change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated. The laws of reflection can be explained by treating light as waves.
• Diffraction is the bending of light around the corners of an obstacle or aperture into the region of geometrical shadow of the obstacle. This phenomenon is observed when a wave encounters an obstacle or a slit that is comparable in size to its wavelength, which is best explained by wave theory.
• Interference occurs when two or more waves superimpose to form a resultant wave of greater, lower, or the same amplitude. This phenomenon illustrates the wave nature of light through constructive and destructive interference patterns, such as those observed in the double-slit experiment.
• Polarization is a property applying to transverse waves that specifies the geometrical orientation of the oscillations. For light waves, this can be seen as the direction in which the electric field oscillates. Polarization can only occur with waves that have transverse components, further supporting the wave nature of light.

On the other hand, the photoelectric effect cannot be explained solely by the wave nature of light. It is the emission of electrons or other free carriers when light shines on a material. Electrons emitted in this manner can be called photoelectrons. The phenomenon is best explained by Albert Einstein's quantum theory of light, where light is considered as quanta of energy called photons. This effect demonstrates the particle aspect of light, wherein each photon has a discrete packet of energy equal to $hf$, where $h$ is Planck's constant and $f$ is the frequency of the light.

Thus, the photoelectric effect is the correct answer because it specifically requires the particle theory of light for its explanation, unlike reflection, diffraction, interference, and polarization, which are well-explained by the wave nature of light.

To find the work function of the photo sensitive material, we can use the photoelectric equation which relates the kinetic energy of the ejected electrons to the photon energy and the work function ($\phi$) of the material:

$KE_{\text{max}} = h\nu - \phi$

Where $KE_{\text{max}}$ is the maximum kinetic energy of the ejected electrons, $h\nu$ is the energy of the incoming photon, and $\phi$ is the work function of the material.

However, the kinetic energy of the ejected electrons can also be related to the stopping potential ($V_s$) by the equation:

$KE_{\text{max}} = e \cdot V_s$

Substituting this into the first equation gives:

$e \cdot V_s = h\nu - \phi$

Here, $e$ is the charge of an electron ($1.6 \times 10^{-19} \, \text{C}$), but since we are dealing with energies in electronvolts (eV), and $1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$, we can directly use the values given without converting the units:

$\phi = h\nu - e \cdot V_s$

Where $h\nu = 2.48 \, \text{eV}$ is the energy of the irradiating photons, and $V_s = 0.5 \, \text{V}$.

Substituting these values in, we get:

$\phi = 2.48 \, \text{eV} - 0.5 \, \text{eV} = 1.98 \, \text{eV}$

Therefore, the work function ($\phi$) of the photo sensitive material is $1.98 \, \text{eV}$, which corresponds to Option A.

Stopping potential is independent of intensity of light. It depends on frequency of light.

$\begin{aligned} & V_0=\frac{h}{e} f-\frac{h}{e} f_0 \\ & \Rightarrow \text { Slope }=\frac{h}{e} \text { (S-I is correct) } \\ & \because\left(f_0\right)_1<\left(f_0\right)_2 \\ & \therefore \quad(\mathrm{S}-I I \text { is incorrect) } \\ & \therefore \quad \text { Option (3) is correct. } \end{aligned}$

In order to determine the most appropriate answer to the question, let's analyze the given Assertion A and Reason R in detail.

Assertion A: Number of photons increases with increase in frequency of light.

This statement is not correct. The number of photons is determined by the intensity (or power) of the light and is given by the formula:

$N = \frac{P}{hf}$

where $N$ is the number of photons per second, $P$ is the power (intensity) of the light, $h$ is Planck's constant, and $f$ is the frequency of the light. As the frequency increases, the energy per photon increases, but it does not necessarily mean that the number of photons increases unless the power of the light also increases proportionally.

Reason R: Maximum kinetic energy of emitted electrons increases with the frequency of incident radiation.

This statement is correct. According to the photoelectric effect, the maximum kinetic energy of emitted electrons is given by:

$K.E_{\text{max}} = hf - \phi$

where $K.E_{\text{max}}$ is the maximum kinetic energy of the emitted electrons, $h$ is Planck's constant, $f$ is the frequency of the incident radiation, and $\phi$ is the work function of the material. As the frequency of the incident light increases, the kinetic energy of the emitted electrons increases.

Now, let's match the statements with the options given:

Option A: $\mathbf{A}$ is not correct but $\mathbf{R}$ is correct.

This option is correct because Assertion A is incorrect while Reason R is correct.

Option B: $\mathbf{A}$ is correct but $\mathbf{R}$ is not correct.

This option is incorrect because Assertion A is not correct.

Option C: Both $\mathbf{A}$ and $\mathbf{R}$ are correct and $\mathbf{R}$ is the correct explanation of $\mathbf{A}$.

This option is incorrect because Assertion A is not correct.

Option D: Both $\mathbf{A}$ and $\mathbf{R}$ are correct and $\mathbf{R}$ is NOT the correct explanation of $\mathbf{A}$.

This option is incorrect because Assertion A is not correct.

Therefore, the most appropriate answer is:

Option A: $\mathbf{A}$ is not correct but $\mathbf{R}$ is correct.

Stopping potential is independent on intensity but photocurrent increases non-linearly on increasing intensity.

To find out the number of photons emitted per second by the laser, we can use the relationship between the energy of a single photon, the total energy emitted per second (power), and the number of photons emitted per second. The energy $E$ of a single photon is given by Planck's equation:

$ E = hf $

where:

• $ h $ is Planck's constant ($ 6.63 \times 10^{-34} \mathrm{Js} $), and
• $ f $ is the frequency of the light ($ 6 \times 10^{14} \mathrm{Hz} $).

Let's first calculate the energy of one photon:

$ E = (6.63 \times 10^{-34} \mathrm{Js}) \times (6 \times 10^{14} \mathrm{Hz}) $

$ E = 3.978 \times 10^{-19} \mathrm{J} $

The power ($ P $) emitted by the laser is the total energy emitted per second,

$ P = E_{\text{total per second}} = 2 \times 10^{-3} \mathrm{W} = 2 \times 10^{-3} \mathrm{J/s} $

The number of photons ($ N $) emitted per second can be found by dividing the total energy emitted per second by the energy of one photon:

$ N = \frac{P}{E} $

Substitute the values we have:

$ N = \frac{2 \times 10^{-3} \mathrm{J/s}}{3.978 \times 10^{-19} \mathrm{J}} $

$ N = \frac{2 \times 10^{-3}}{3.978 \times 10^{-19}} $

$ N = 5.03 \times 10^{15} \text{ photons per second} $

The number of photons emitted per second is approximately $5 \times 10^{15}$. Therefore, the correct answer, rounded to one significant figure, is:

Option A: $5 \times 10^{15}$

To find the value of $ X $ in the band gap $ \left(\frac{\mathrm{X}}{8}\right) \mathrm{eV} $, we need to understand the relationship between the wavelength of light that can result in changes in the conductivity of a photodiode and the photodiode's band gap energy.

The band gap energy ($ E_{g} $) of a material is the minimum energy required for an electron to transition from the valence band to the conduction band, thus creating a hole-electron pair and allowing conductivity to occur. When light with a certain wavelength ($ \lambda $) is incident upon the photodiode, if the energy of the photons is greater than or equal to the band gap energy of the photodiode, then the photons can excite electrons and change the conductivity of the photodiode.

The energy of a photon ($ E_{photon} $) is given by the equation:

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

where:

• $ h $ is Planck's constant, which is given as $ 6.6 \times 10^{-34} $ J·s.
• $ c $ is the speed of light in a vacuum ($ \approx 3 \times 10^{8} $ m/s).
• $ \lambda $ is the wavelength of the incident light.

Then the band gap energy in terms of electron volts is found by converting the energy from joules to electron volts (eV) using the charge of an electron $ e $ ($ 1.6 \times 10^{-19} $ C):

$ E_{g} = \frac{hc}{\lambda e} $

Given that the photodiode starts conducting when the wavelength of light is less than $ 660 \mathrm{~nm} $, we should use this wavelength as the threshold wavelength $ \lambda_{threshold} $:

$ E_{g} = \frac{hc}{\lambda_{threshold}\times e} $

$ E_{g} = \frac{6.6 \times 10^{-34} \text{ J·s} \cdot 3 \times 10^{8} \text{ m/s}}{660 \times 10^{-9} \text{ m} \times 1.6 \times 10^{-19} \text{ C}} $

$ E_{g} = \frac{6.6 \times 3 \times 10^{-26}}{660 \times 1.6} \mathrm{eV} $

$ E_{g} = \frac{19.8}{660 \times 1.6} \mathrm{eV} $

$ E_{g} = \frac{19.8}{1056} \mathrm{eV} $

$ E_{g} = \frac{18.75}{1000} \mathrm{eV} $

$ E_{g} = 1.875 \mathrm{eV} $

Now, we have to compare this value to $ \left(\frac{\mathrm{X}}{8}\right) \mathrm{eV} $ to find the value of $ X $:

$ 1.875 = \frac{X}{8} $

$ X = 1.875 \times 8 $

$ X = 15 $

Therefore, the value of $ X $ is 15. The correct answer is Option C.

To find the ratio of velocities of two particles based on their de Broglie wavelengths, we can use the de Broglie wavelength formula, which relates the momentum of a particle to its wavelength. The de Broglie's wavelength formula is given by:

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

where:

$\lambda$ is the de Broglie wavelength,

$h$ is the Planck constant, and

$p$ is the momentum of the particle.

The momentum $p$ of a particle can also be expressed as the product of its mass $m$ and velocity $v$:

$ p = mv $

So, we can rewrite the de Broglie wavelength equation in terms of mass and velocity:

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

For the proton (let's use subscript $p$ for proton), the wavelength is $\lambda$:

$ \lambda_p = \frac{h}{m_p v_p}$

For the $\alpha$ particle (let's use subscript $\alpha$ for alpha), the wavelength is $2\lambda$:

$ 2\lambda = \frac{h}{m_\alpha v_\alpha}$

We are interested in finding the ratio of the velocities $\frac{v_p}{v_\alpha}$. Using the given data about wavelengths:

$ \lambda_p = \lambda $

$ \lambda_\alpha = 2\lambda $

Using the de Broglie equation for both particles:

$ \frac{h}{m_p v_p} = \lambda $

$ \frac{h}{m_\alpha v_\alpha} = 2\lambda $

Dividing the second equation by the first equation gives us:

$ \frac{\frac{h}{m_\alpha v_\alpha}}{\frac{h}{m_p v_p}} = \frac{2\lambda}{\lambda} $

$ \frac{m_p v_p}{m_\alpha v_\alpha} = \frac{2}{1} $

$ \frac{v_p}{v_\alpha} = \frac{2m_\alpha}{m_p} $

Since we know an $\alpha$ particle consists of 2 protons and 2 neutrons (essentially four nucleons), the mass of an $\alpha$ particle is roughly four times the mass of a proton ($m_\alpha \approx 4m_p$).

Substituting $m_\alpha$ with $4m_p$ in the equation:

$ \frac{v_p}{v_\alpha} = \frac{2(4m_p)}{m_p} $

$ \frac{v_p}{v_\alpha} = 2 \cdot 4 $

$ \frac{v_p}{v_\alpha} = 8 $

Therefore, the ratio of the velocities of proton to $\alpha$ particle is $8:1$, which corresponds to Option A.

Since $\frac{\mathrm{f}}{2}<\mathrm{f}_0$

i.e. the incident frequency is less than threshold frequency. Hence there will be no emission of photoelectrons.

$\Rightarrow \text { current }=0$

$\begin{aligned} & \mathrm{E}=\phi+\mathrm{K}_{\max } \\ & \phi=\frac{\mathrm{hc}}{\lambda_0} \\ & \mathrm{~K}_{\max }=\mathrm{eV}_0 \\ & 8 \mathrm{e}=\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_0} \ldots . . \text { (i) } \\ & 2 \mathrm{e}=\frac{\mathrm{hc}}{3 \lambda}-\frac{\mathrm{hc}}{\lambda_0} \ldots . . . \text { (ii) } \\ & \text { on solving (i) & (ii) } \\ & \lambda_0=9 \lambda \end{aligned}$

$\begin{aligned} & \text { K.E. }=\mathrm{hf}-\phi \\ & \tan \theta=\mathrm{h} \end{aligned}$

$\mathrm{p=\frac{E}{C}=\frac{6.48 \times 10^5}{3 \times 10^8}=2.16 \times 10^{-3}}$

$\begin{aligned} & \text { For P.E.E. : } \lambda \leq \frac{h c}{W_e} \\ & \lambda \leq \frac{1240 \mathrm{~nm}-\mathrm{eV}}{3 \mathrm{eV}} \\ & \lambda \leq 413.33 \mathrm{~nm} \\ & \lambda_{\max } \approx 414 \mathrm{~nm} \text { for P.E.E. } \end{aligned}$

$\begin{aligned} & \mathrm{n}_1 \times \frac{\mathrm{hc}}{\lambda_1}=200 \\ & \mathrm{n}_2 \times \frac{\mathrm{hc}}{\lambda_2}=200 \\ & \frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{\lambda_1}{\lambda_2}=\frac{300}{500} \\ & \frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{3}{5} \end{aligned}$

We know that the de-Broglie wavelength $\lambda$ of a particle is given by:

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

where:

• $h$ is Planck's constant,
• $p$ is the momentum of the particle.

For a photon (which has zero rest mass), its energy $E$ and momentum $p$ are related by the equation:

$E = cp$

and its de-Broglie wavelength $\lambda$ is given by:

$\lambda = \frac{h}{E} \times \frac{1}{c}$

Now, since the photon and electron are said to have the same de-Broglie wavelength:

$\lambda_{electron} = \lambda_{photon}$

$\frac{h}{p_{electron}} = \frac{h}{E_{photon}} \times \frac{1}{c}$

For the electron, its momentum $p_{electron}$ is given by:

$p_{electron} = m_{e}v_{electron}$

where:

• $m_{e}$ is the electron's rest mass,
• $v_{electron}$ is the electron's velocity.

The kinetic energy $K.E.$ of the electron is:

$K.E._{electron} = \frac{1}{2}m_{e}v_{electron}^2$

The question states that $v_{electron}$ is $25\%$ ($0.25c$) of the speed of light $c$. So we write:

$v_{electron} = 0.25c$

Plugging this into the kinetic energy formula, we get:

$K.E._{electron} = \frac{1}{2}m_{e}(0.25c)^2 = \frac{1}{2}m_{e} \times \frac{1}{16}c^2$

$K.E._{electron} = \frac{1}{32}m_{e}c^2$

For a photon, $p_{photon} = \frac{E_{photon}}{c}$ and hence its kinetic energy (which, for a photon, is simply its energy) is:

$K.E._{photon} = E_{photon} = cp_{photon}$

Now, comparing the kinetic energies:

$\frac{K.E._{electron}}{K.E._{photon}} = \frac{\frac{1}{32}m_{e}c^2}{cp_{photon}}$

Since $p_{electron} = p_{photon}$ (from the de-Broglie relation), we can replace $p_{photon}$ with $p_{electron}$ which is $m_{e}v_{electron}$:

$\frac{K.E._{electron}}{K.E._{photon}} = \frac{\frac{1}{32}m_{e}c^2}{m_{e}v_{electron}c}$

$\frac{K.E._{electron}}{K.E._{photon}} = \frac{\frac{1}{32}c}{0.25c}$

$\frac{K.E._{electron}}{K.E._{photon}} = \frac{1}{32} \times \frac{1}{0.25}$

$\frac{K.E._{electron}}{K.E._{photon}} = \frac{1}{32} \times 4$

$\frac{K.E._{electron}}{K.E._{photon}} = \frac{1}{8}$

So the correct answer is Option C $\frac{1}{8}$.

The threshold frequency, $ \nu_0 $, corresponds to the minimum frequency of light required to eject electrons from the surface of a metal, a phenomenon known as the photoelectric effect. The work function, represented by $ \phi $, is the minimum energy needed to remove an electron from the surface of the metal.

The energy of a photon is given by the equation $ E = h\nu $, where:

• $ E $ is the energy of the photon,
• $ h $ is Planck's constant ($ h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} $), and
• $ \nu $ is the frequency of the photon.

To find the threshold frequency for a metal with a work function of $ 6.63 \, \text{eV} $, we must first express the work function in joules (since Planck's constant is in joules per second). To convert electron volts to joules, use the conversion factor $ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} $:

The energy of the photon at the threshold frequency is equal to the work function:

Solve for $ \nu_0 $:

Thus, the correct answer is:

To solve this problem, we need to understand the relationship between the atomic number $ Z $, the characteristic X-ray wavelengths, and the cut-off wavelength in X-ray spectra.

The cut-off wavelength, also known as the minimum wavelength ($ \lambda_{\text{min}} $), corresponds to the maximum energy of the X-rays produced and can be related to the accelerating voltage used in the X-ray tube. The characteristic X-rays, such as the $ K_\alpha $-line, correspond to specific electron transitions in the atom and are dependent on the atomic number $ Z $ of the target material.

The wavelength of the $ K_\alpha $-line is given by Moseley's Law, which can be approximated as:

$ \lambda_{K_\alpha} \propto \frac{1}{(Z - 1)^2} $

The cut-off wavelength is given by the relationship between the energy of the incident electrons and the X-ray produced:

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

Now, given the ratio $ r $ for the first metal target with $ Z = 46 $ :

$ r = \frac{\lambda_{K_\alpha}}{\lambda_{\text{min}}} = 2 $

For the second metal target with $ Z = 41 $, let's denote the new ratio as $ r_2 $. We assume the same cutoff wavelength since the same electron beam is used.

The ratio for $ Z = 41 $ can be calculated using the relationship between the $ K_\alpha $-line wavelenghts:

$ \frac{\lambda_{K_\alpha (1)}}{\lambda_{K_\alpha (2)}} = \left( \frac{Z_2 - 1}{Z_1 - 1} \right)^2 $

Substituting $ Z_1 = 46 $ and $ Z_2 = 41 $:

$ \frac{\lambda_{K_\alpha (1)}}{\lambda_{K_\alpha (2)}} = \left( \frac{41 - 1}{46 - 1} \right)^2 = \left( \frac{40}{45} \right)^2 = \left( \frac{8}{9} \right)^2 = \frac{64}{81} $

Let $ \lambda_{K_\alpha (1)} = 2 \lambda_{\text{min}} $ because $ r = 2 $. Therefore:

$ \lambda_{K_\alpha (2)} = \lambda_{K_\alpha (1)} \cdot \frac{81}{64} = 2 \lambda_{\text{min}} \cdot \frac{81}{64} $

Thus, the new ratio $ r_2 $ is:

$ r_2 = \frac{\lambda_{K_\alpha (2)}}{\lambda_{\text{min}}} = 2 \cdot \frac{81}{64} = \frac{162}{64} = 2.53 $

So, the value of $ r $ for the second metal target with $ Z = 41 $ is:

Option A: 2.53

Given,

$ \phi_{A}: \phi_{B}=2: 3 \Rightarrow \phi_{A}=2 k, \phi_{B}=3 k $

Using Einstein's photoelectric equation

$ e V_{0}=h v-h v_{0} $

where, $ h= $ Planck's constant, $ v= $ frequency of incident light

$ v_{0}= $ threshold frequency, $ V_{0}= $ stopping potential

$ V_{0}=\frac{h}{e} v-\frac{h}{e} v_{0} $

Now comparing Eq. (i) with $ y=m x+C $

Here, slope $ m=\frac{h}{e} $.

The slopes $ x $ and $ y $ of the graphs drawn between the stopping potential and the frequency of incident light for surfaces

$A$ and $B$ depend only on the constant $h$ and $e$ not on the work function of $A$ and $B$. So, $x$ and $y$ are the same for both metals.

$ x=y=\frac{h}{e} $

Hence, $x: y=1: 1$

The wave picture of light fails to explain the photoelectric effect.

The photoelectric effect is the phenomenon where electrons are ejected from a material's surface when it is exposed to light of certain frequencies. According to the classical wave theory of light, increasing the intensity of light (regardless of frequency) should increase the energy of emitted electrons. However, experiments show that there is a minimum threshold frequency below which no electrons are emitted, regardless of light intensity. Moreover, the maximum kinetic energy of the emitted electrons depends on the frequency of the incident light, not its intensity.

Albert Einstein explained the photoelectric effect by proposing that light consists of discrete packets of energy called photons. Each photon has an energy given by the equation:

$ E = hf $

where:

$ E $ is the energy of the photon,

$ h $ is Planck's constant ($6.626 \times 10^{-34} \, \text{Js}$),

$ f $ is the frequency of the light.

Thus, only photons with enough energy (i.e., with frequency above the threshold frequency) can eject electrons, thereby providing a particle-like explanation for light, which is not accounted for by its wave-like nature.

In contrast, the wave picture of light successfully explains phenomena like interference, diffraction, and polarization of light. Interference and diffraction involve the superposition of light waves to produce patterns, and polarization relates to the orientation of light waves, all of which align with the wave nature of light.

Given:

Power of the bulb = $60 \, \text{W}$

Efficiency = $16\%$

Calculation of Actual Power:

The actual power is determined by multiplying the efficiency with the total power:

$ P = \frac{16}{100} \times 60 = 9.6 \, \text{W} $

Intensity of Electromagnetic Radiation:

The intensity, $I$, of the electromagnetic radiation is given by the formula:

$ I = \frac{1}{2} \varepsilon_{0} c E_{0}^{2} $

where:

$c$ is the speed of light.

$E_{0}$ is the peak electric field.

Relation of Intensity with Power and Area:

The intensity is also expressed as power per unit area:

$ I = \frac{\text{Power}}{\text{Area}} = \frac{P}{4 \pi r^{2}} $

Equating the two expressions for intensity:

$ \frac{P}{4 \pi r^{2}} = \frac{1}{2} \varepsilon_{0} c E_{0}^{2} $

Solving for $E_{0}^{2}$:

$ E_{0}^{2} = \frac{2P}{4 \pi \varepsilon_{0} c r^{2}} $

Substituting the Known Values:

$r = 2 \, \text{m}$

$c = 3 \times 10^8 \, \text{m/s}$

$\varepsilon_{0} = \frac{1}{4 \pi \times 9 \times 10^9}$

Substituting these values into the equation for $E_{0}^{2}$:

$ E_{0}^{2} = \frac{2 \times 9.6 \times 9 \times 10^9}{4 \times \pi \times \left(\frac{1}{4 \pi \times 9 \times 10^9}\right) \times (3 \times 10^8) \times 2^2} $

Simplifying further:

$ E_{0}^{2} = \frac{2 \times 9.6 \times 9 \times 10^9}{4 \times 3 \times 10^8} $

$ E_{0}^{2} = 144 $

Hence, the peak value of the electric field is:

$ E_{0} = \sqrt{144} = 12 \, \text{V/m} $

Given,

Work function, $ \phi=1.1 \mathrm{eV} $

Using photoelectric equation for each incident radiation.

$ E=\frac{1}{2} m v^{2}+\phi $

where $ E $ is energy of incident radiation $ \frac{1}{2} m v^{2}= $ kinetic energy of photoelectron

$ \begin{array}{l} E_{1}=\frac{1}{2} m v_{1}^{2}+\phi \\\\ E_{2}=\frac{1}{2} m v_{2}^{2}+\phi \end{array} $

From Eqs. (ii) and (iii)

$ \begin{aligned} \frac{v_{1}^{2}}{v_{2}^{2}} & =\frac{E_{1}-\phi}{E_{2}-\phi} \\\\ & =\frac{1.5-1.1}{2-1.1} \\\\ \frac{v_{1}^{2}}{v_{2}^{2}} & =\frac{0.4}{0.9} \Rightarrow \frac{v_{1}}{v_{2}}=\frac{2}{3} \\\\ v_{1} & : v_{2}=2: 3 \end{aligned} $

Given, de-Broglie wavelength of proton $ =2 \times $ de-Broglie wavelength of $ \alpha $-particle

de-Broglie wavelength in term of kinetic energy $ E $ and mass $ m $ is,

$ \lambda=\frac{h}{\sqrt{2 m E}} $

Now, $ \quad \lambda_{p}=2 \lambda_{\alpha} $

$ \Rightarrow \quad \frac{h}{\sqrt{2 m E_{1}}}=\frac{2 h}{\sqrt{2(4 m) E_{2}}} $

where $ E_{1} $ and $ E_{2} $ are KE respectively

$ \frac{1}{\sqrt{E_{1}}}=\frac{1}{\sqrt{E_{2}}} $

$ \therefore \quad \frac{E_{1}}{E_{2}}=\frac{1}{1} $

Ratio is $ 1: 1 $

Energy of neutron at temperature,

$ T=\frac{3}{2} k_B T $

de-Broglie wavelength

$ =\frac{h}{\sqrt{2 m E}} \propto \frac{h}{\sqrt{2 \times m_n \times T}} $

Here, $T_1=(273+77) \mathrm{K}=350 \mathrm{~K}$

$ \begin{aligned} T_2 & =(273+1127) \mathrm{K}=1400 \mathrm{~K} \\\\ \therefore \frac{\lambda_2}{\lambda_1} & =\frac{h}{\sqrt{2 m_n \times 1400}} \times \frac{\sqrt{2 m_n \times 350}}{h} \\\\ \lambda_2 & =\lambda \times \frac{1}{2} \end{aligned} $

The relationship between the cut-off voltage $ V_0 $ and the frequency $ f $ of incident light in a photoelectric experiment is given by the equation:

$ eV_0 = hf - \phi $

where:

$ e $ is the elementary charge ($1.6 \times 10^{-19} \, \mathrm{C}$),

$ h $ is Planck's constant ($6.63 \times 10^{-34} \, \mathrm{Js}$),

$ \phi $ is the work function of the material.

Rearranging the equation provides:

$ V_0 = \frac{h}{e} f - \frac{\phi}{e} $

The slope $ m $ of the graph of $ V_0 $ versus $ f $ is given by the factor of $ f $, which is $ \frac{h}{e} $.

Substituting the given values:

$ m = \frac{h}{e} = \frac{6.63 \times 10^{-34} \, \mathrm{Js}}{1.6 \times 10^{-19} \, \mathrm{C}} $

Calculating this gives:

$ m = \frac{6.63}{1.6} \times 10^{-15} \approx 4.14375 \times 10^{-15} \, \mathrm{Vs} $

Thus, the correct option is:

Option A: $ 414 \times 10^{-15} \, \mathrm{Vs} $

First, the photon energy is the sum of the work function and the maximum kinetic energy of the ejected electron:

• ϕ = 2 eV

• K_max = 2 eV

• E_photon = ϕ + K_max = 4 eV

Next, relate energy to wavelength via

$\lambda = \frac{hc}{E_{\rm photon}}$

with $hc = 1240\;\mathrm{eV\cdot nm}$. Therefore,

$\lambda = \frac{1240\;\mathrm{eV\cdot nm}}{4\;\mathrm{eV}} = 310\;\mathrm{nm} = 3100\;\text{\AA}.$

So the correct choice is Option B: 3100 Å.

Given:

Wavelength of incident light, $\lambda = 4000\, \text{Å} = 4 \times 10^{-7} \, \text{m}$

Threshold wavelength, $\lambda_0 = 5420\, \text{Å} = 5420 \times 10^{-10} \, \text{m}$

The relationship between the threshold wavelength $\lambda_0$ and the work function $\phi$ is:

$ \phi = \frac{hc}{\lambda_0} \, \text{J} \quad \Rightarrow \quad \phi = \frac{hc}{e \cdot \lambda_0} \, \text{eV} $

Calculating the work function:

$ \phi = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1.602 \times 10^{-19} \times 5420 \times 10^{-10}} $

$ \phi = \frac{19.87 \times 10^3}{8682.84} = 2.288 \times 10^{-3} \times 10^3 $

$ \phi = 2.29 \, \text{eV} $

According to Einstein's photoelectric equation:

$ \frac{1}{2} mv^2 = \frac{hc}{\lambda} - \phi $

Where:

$ v $ is the velocity of the photoelectrons.

$ h $ is Planck's constant.

$ c $ is the speed of light.

$ \lambda $ is the wavelength of the incident light.

$ \phi $ is the work function of the metal.

Given the ratio of maximum velocities of photoelectrons $ v_1 = 3v_2 $, we analyze two cases:

Case 1:

$ \frac{1}{2} m (3v_2)^2 = \frac{hc}{\lambda_1} - \phi $

where $\lambda_1 = 300 \, \text{nm}$.

Case 2:

$ \frac{1}{2} m (v_2)^2 = \frac{hc}{\lambda_2} - \phi $

where $\lambda_2 = 500 \, \text{nm}$.

Dividing Equations for Case 1 and Case 2:

$ \frac{hc}{\lambda_1} - \phi = 9\left(\frac{hc}{\lambda_2} - \phi\right) $

Solving for $\phi$:

$ 8\phi = \frac{9hc}{\lambda_2} - \frac{hc}{\lambda_1} $

Substitute $ \frac{hc}{\lambda} $ with $ \frac{1240 \, \text{eV nm}}{\lambda} $:

$ 8\phi = \left(9 \times \frac{1240}{500} - \frac{1240}{300}\right) \, \text{eV} $

$ 8\phi = 18.18 \, \text{eV} $

Thus, solving for $\phi$:

$ \phi = \frac{18.18}{8} \approx 2.28 \, \text{eV} $

Therefore, the work function $\phi$ of the metal is approximately $2.28 \, \text{eV}$.

The maximum speed ratio of emitted photoelectrons due to different light wavelengths incident on a photosensitive metal surface is determined using:

$ \frac{v_1}{v_2} = \sqrt{\frac{\left(\frac{1}{\lambda_1} - \frac{1}{\lambda_0}\right)}{\left(\frac{1}{\lambda_2} - \frac{1}{\lambda_0}\right)}} $

Here,

$ v_1 $ is the speed of photoelectrons emitted due to light with wavelength $ \lambda_1 $.

$ v_2 $ is the speed of photoelectrons emitted due to light with wavelength $ \lambda_2 $.

Given:

$ \lambda_1 = \frac{\lambda_0}{3} $

$ \lambda_2 = \frac{\lambda_0}{9} $

The formula becomes:

$ \frac{v_1}{v_2} = \sqrt{\frac{\left(\frac{3}{\lambda_0} - \frac{1}{\lambda_0}\right)}{\left(\frac{9}{\lambda_0} - \frac{1}{\lambda_0}\right)}} $

Simplifying, we find:

$ \frac{v_1}{v_2} = \sqrt{\frac{2}{8}} = \sqrt{\frac{1}{4}} = \frac{1}{2} $

Thus, the ratio of the maximum velocities of the photoelectrons is $ 1:2 $.

To find the number of photons emitted by a lamp, we can use Planck's quantum theory to calculate the energy of a photon. Given:

Mean wavelength ($\lambda$) = 4500 Å = $4500 \times 10^{-10}$ m

Planck's constant ($h$) = $6.626 \times 10^{-34}$ Js

Speed of light ($c$) = $3 \times 10^8$ m/s

Lamp power = 150 W

Efficiency = 8%

Energy of a Photon:

$ E = \frac{h c}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4500 \times 10^{-10}} $

$ E = 4.417 \times 10^{-19} \text{ J} $

Total Energy Emitted per Second:

Only 8% of the 150 W is emitted as light energy:

$ \text{Energy emitted as light} = \frac{8 \times 150}{100} = 12 \text{ J/s} $

Number of Photons Emitted:

Let $n$ be the number of photons emitted per second. The equation relating the total energy and the energy of photons is:

$ n \times \text{energy of a photon} = \text{total energy emitted as light} $

$ n \times 4.417 \times 10^{-19} = 12 $

Solve for $n$:

$ n = \frac{12}{4.417 \times 10^{-19}} $

$ n = 2.7167 \times 10^{19} $

$ n = 27167 \times 10^{18} $

Therefore, the number of photons emitted by the lamp per second is approximately $27.17 \times 10^{18}$.

The de Broglie wavelength is given by:

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

where $ p = \sqrt{2mK} $ represents the momentum, and $ K $ is the kinetic energy of the particle.

When the kinetic energy ($ K $) is decreased by 36%, the new energy ($ K' $) becomes:

$ K' = K - 0.36K = 0.64K $

The new momentum ($ p' $) can be calculated as follows:

$ p' = \sqrt{2mK'} = \sqrt{2m \cdot 0.64K} = 0.8p $

Given this new momentum, the new de Broglie wavelength ($ \lambda' $) is:

$ \lambda' = \frac{h}{p'} = \frac{h}{0.8p} = 1.25\lambda $

The change in wavelength ($ \Delta \lambda $) is:

$ \Delta \lambda = \lambda' - \lambda = 1.25\lambda - \lambda = 0.25\lambda $

The percentage increase in wavelength is:

$ \text{Percentage increase} = \frac{\Delta \lambda}{\lambda} \times 100 = 0.25 \times 100 = 25\% $

The de-Broglie wavelength is given by,

$ \lambda=\frac{h}{p}=\frac{h}{m v} $

Initial de-Broglie wavelength is

$ \lambda=\frac{h}{m v_0} $

Equating the forces, we get

$ \Rightarrow \quad F=m a=e E_0 $

Acceleration of electron

$ \Rightarrow \quad a=\frac{e E_0}{m} $

Velocity of electron after time $t$

$ \Rightarrow \quad v=v_0+\frac{e E_0}{m} t $

So, de-Broglie wavelength after time $t$.

$ \begin{aligned} & \lambda^{\prime}=\frac{h}{m v}=\frac{h}{m\left(v_0+\frac{e E_0}{m} t\right)} \\ \Rightarrow & \frac{h}{m v_0\left(1+\frac{e E_0}{m v_0} t\right)}=\frac{\lambda}{\left(1+\frac{e E_0}{m v_0} t\right)} \end{aligned} $

To determine the longest wavelength of light that can cause the photoelectric effect in a metal with a work function of 9 eV, we use the formula:

$ \phi_0 = \frac{hc}{\lambda_0} $

Where:

$\phi_0$ is the work function (9 eV in this case).

$h$ is Planck's constant ($6.62 \times 10^{-34} \, \text{J s}$).

$c$ is the speed of light ($3 \times 10^8 \, \text{m/s}$).

$\lambda_0$ is the wavelength.

Rearranging the formula to solve for $\lambda_0$, we get:

$ \lambda_0 = \frac{hc}{\phi_0} $

Substituting the given values:

$ \lambda_0 = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{9 \times 1.6 \times 10^{-19}} $

Calculating this yields:

$ \lambda_0 = 1.37 \times 10^{-7} \, \text{m} $

Hence, the longest wavelength capable of initiating the photoelectric effect is $1.37 \times 10^{-7} \, \text{m}$.

The energy required to remove an electron from the aluminium surface is $ \phi_0 = 4.2 \, \text{eV} $.

The wavelength of the light is given as $ \lambda = 2000 \, \text{Å} = 2 \times 10^{-7} \, \text{m} $.

Using Einstein's photoelectric equation:

$ \frac{1}{2} m v_{\text{max}}^2 = \frac{h c}{\lambda} - \phi_0 $

Given the constants:

Planck's constant $ h = 6.6 \times 10^{-34} \, \text{J s} $

Speed of light $ c = 3 \times 10^8 \, \text{m/s} $

Conversion factor $ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} $

Mass of electron $ m = 9.1 \times 10^{-31} \, \text{kg} $

The calculation becomes:

$ \frac{1}{2} m v_{\text{max}}^2 = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{2 \times 10^{-7} \times 1.6 \times 10^{-19}} \, \text{eV} $

Further simplifying:

$ \frac{1}{2} m v_{\text{max}}^2 = \frac{6.2 \, \text{eV} - 4.2 \, \text{eV}}{9.1 \times 10^{-31}} $

$ v_{\text{max}}^2 = \frac{2 \times 2 \times 1.6 \times 10^{-19} \, \text{J}}{9.1 \times 10^{-31}} $

$ v_{\text{max}}^2 = \frac{6.4}{91} \times 10^{12} \quad \Rightarrow \quad v_{\text{max}} = \sqrt{\frac{6.4}{91} \times 10^{12}} $

Thus, the velocity of the fastest ejected electron is:

$ v_{\text{max}} = 0.838 \times 10^6 \quad \Rightarrow \quad v_{\text{max}} = 8.4 \times 10^5 \, \text{m/s} $

Threshold wavelength ($\lambda_{threshold}$) is the maximum wavelength of light required to remove an electron from a metal surface, i.e., to overcome the work function ($\phi$). The relationship between work function and threshold wavelength is given by:

Where:

• $\phi$ is the work function in electron-volts (eV)
• $h$ is the Planck's constant
• $c$ is the speed of light
• $\lambda_{threshold}$ is the threshold wavelength

In this problem, we are given the product $hc = 1242 ~\mathrm{eV}\,\mathrm{nm}$, and the work functions $\phi_{A} = 9 ~\mathrm{eV}$ and $\phi_{B} = 4.5 ~\mathrm{eV}$. Let's calculate the threshold wavelengths for metal surfaces $\mathrm{A}$ and $\mathrm{B}$:

For metal surface $\mathrm{A}$:

For metal surface $\mathrm{B}$:

Now let's calculate the difference between the threshold wavelengths ($\Delta\lambda$):

By calculating the difference, we get:

So, the difference between the threshold wavelengths for the two metal surfaces is $\boxed{138}\,\mathrm{nm}$.