Correct option is (B)

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

$K{E}_{\text{max}}=h\nu -\varphi$

where $K{E}_{\text{max}}$ is the maximum kinetic energy of the ejected electrons, $h$ is Planck's constant, $\nu$ is the frequency of the incident light, and $\varphi$ 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:

$K{E}_{\text{max}}=E_{\text{photon}}-\varphi$

Substituting the given values:

$K{E}_{\text{max}}=4.13\text{eV}-3.13\text{eV}=1\text{eV}$

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

Correct option is (B)

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:

$K_{max}=E-\varphi$

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

Given $E=4.13$ eV and the work function $\varphi =2.13$ 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:

$K_{max}=e{V}_s$

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.

Correct option is (A)

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.

Correct option is (A)

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 ($\varphi$) of the material:

$K{E}_{\text{max}}=h\nu -\varphi$

Where $K{E}_{\text{max}}$ is the maximum kinetic energy of the ejected electrons, $h\nu$ is the energy of the incoming photon, and $\varphi$ 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:

$K{E}_{\text{max}}=e\cdot V_s$

Substituting this into the first equation gives:

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

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:

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

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:

$\varphi =2.48\text{eV}-0.5\text{eV}=1.98\text{eV}$

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

Correct option is (C)

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