Correct answer is option B

Now let's analyze both statements:

Statement I is correct. According to the photoelectric effect, the ability to emit electrons from a metallic surface depends on the energy of the incident light. The energy of a photon is given by $E=h\nu$, where $h$ is Planck's constant and $\nu$ is the frequency of the light. Ultraviolet rays have higher frequencies and thus higher energies compared to microwaves and infrared rays, making them more effective for the emission of electrons from a metallic surface.

Statement II is incorrect. The maximum kinetic energy of photoelectrons is given by $K_{max}=h\nu -h{\nu }_0$, where ${\nu }_0$ is the threshold frequency. This equation shows that the maximum kinetic energy of photoelectrons is directly proportional to the frequency of the incident light (above the threshold frequency), not inversely proportional.

Based on the analysis, the correct answer is option:

Statement I is true, and Statement II is false.

Correct answer is option B

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 ($\varphi$). The relationship between work function and threshold wavelength is given by:

$\varphi =\frac{hc}{{\lambda }_{threshold}}$

Where:

• $\varphi$ 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 ${\varphi }_A=9\mathrm{eV}$ and ${\varphi }_B=4.5\mathrm{eV}$. Let's calculate the threshold wavelengths for metal surfaces $\mathrm{A}$ and $\mathrm{B}$:

For metal surface $\mathrm{A}$:

${\lambda }_A=\frac{1242}{{\varphi }_A}=\frac{1242}{9}$

For metal surface $\mathrm{B}$:

${\lambda }_B=\frac{1242}{{\varphi }_B}=\frac{1242}{4.5}$

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

$\mathrm{\Delta }\lambda ={\lambda }_B-{\lambda }_A=\frac{1242}{4.5}-\frac{1242}{9}$

By calculating the difference, we get:

$\mathrm{\Delta }\lambda =276-138=138\mathrm{nm}$

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

Correct answer is option A

The photoelectric effect occurs when light (or more generally, electromagnetic radiation) incident on a metallic surface causes the ejection of electrons from the surface. The energy of the incident photons must be greater than the work function of the metal (denoted by ${\varphi }_0$) for the electrons to be ejected.

The energy of the ejected electrons can be expressed as the difference between the energy of the incident photons and the work function of the metal. The maximum kinetic energy of the ejected electrons is given by:

$K_{max}=h\nu -{\varphi }_0$

where $h$ is the Planck's constant, $\nu$ is the frequency of the incident light, and ${\varphi }_0$ is the work function of the metal.

We can also write the maximum kinetic energy in terms of the stopping potential ($V_0$) and the elementary charge ($e$) of an electron:

$K_{max}=e{V}_0$

Equating these two expressions for the maximum kinetic energy, we get:

$e{V}_0=h\nu -{\varphi }_0$

We can express the frequency $\nu$ in terms of the speed of light $c$ and the wavelength $\lambda$:

$\nu =\frac{c}{\lambda }$

Substituting this expression for frequency in the equation, we get:

$e{V}_0=h\frac{c}{\lambda }-{\varphi }_0$

Now, we have two cases:

1. The stopping potential is $V_0$, and the wavelength is $\lambda$.
2. The stopping potential is $\frac{V_0}{4}$, and the wavelength is $2\lambda$.

For the first case, we use the photoelectric effect equation as is:

$e{V}_0=\frac{hc}{\lambda }-{\varphi }_0$

For the second case, we replace $V_0$ with $\frac{V_0}{4}$ and $\lambda$ with $2\lambda$:

$\frac{e{V}_0}{4}=\frac{hc}{2\lambda }-{\varphi }_0$

Now we have two equations:

(1) $e{V}_0=\frac{hc}{\lambda }-{\varphi }_0$

(2) $\frac{e{V}_0}{4}=\frac{hc}{2\lambda }-{\varphi }_0$

We can rewrite equation (1) as:

$\frac{e{V}_0}{4}=\frac{hc}{4\lambda }-\frac{{\varphi }_0}{4}$

Now we can equate the two expressions for $\frac{e{V}_0}{4}$:

$\frac{hc}{4\lambda }-\frac{{\varphi }_0}{4}=\frac{hc}{2\lambda }-{\varphi }_0$

Now we isolate the terms containing ${\varphi }_0$:

$\frac{3{\varphi }_0}{4}=\frac{hc}{4\lambda }-\frac{hc}{2\lambda }$

$\frac{3{\varphi }_0}{4}=\frac{hc}{4\lambda }$

Now we can write the work function ${\varphi }_0$ in terms of the threshold wavelength ${\lambda }_0$:

${\varphi }_0=\frac{hc}{{\lambda }_0}$

Substituting the expression for the work function ${\varphi }_0$ in terms of the threshold wavelength ${\lambda }_0$ into the previous equation, we get:

$\frac{3}{4}\frac{hc}{{\lambda }_0}=\frac{hc}{4\lambda }$

Now we can cancel out the common terms $hc$ from both sides:

$\frac{3}{4{\lambda }_0}=\frac{1}{4\lambda }$

Next, we can cross-multiply to solve for ${\lambda }_0$:

$3\lambda ={\lambda }_0$

Thus, the threshold wavelength for this metallic surface is $3\lambda$.

Correct answer is option D

$\begin{array}{rl}& e{V}_0=\frac{hc}{\lambda }-\varphi \\ \\ & V_0=\frac{hc}{e\lambda }-\left(\frac{\varphi }{e}\right)=\frac{hv}{e}-\left(\frac{\varphi }{e}\right)\end{array}$

When $V_0=0$,

$\begin{array}{rl}\frac{\varphi }{e}& =\left(\frac{hv}{e}\right)\\ \\ \varphi & =hv=\frac{6.626\times {10}^{-34}\times 5\times {10}^{14}}{1.6\times {10}^{-19}}\\ \\ & =\frac{6.626\times 5}{1.6}\times {10}^{-1}\\ \\ & =\frac{6.626\times 5}{16}\approx 2.07\mathrm{eV}\end{array}$

Correct answer is option D

The photoelectric effect is the phenomenon of emission of electrons (or photoelectrons) from the surface of a metal when it is illuminated by light of sufficient energy. The observations from the photoelectric effect led to the development of quantum theory. According to the principles of the photoelectric effect: A. The photocurrent (number of photoelectrons ejected per unit time) is indeed proportional to the intensity of the incident radiation. More intense light means more photons hitting the surface and thus more electrons being ejected. B. The maximum kinetic energy of the photoelectrons does not depend on the intensity of the incident light but rather on its frequency. Increasing the intensity of light increases the number of photoelectrons (current) but does not increase their maximum kinetic energy. C. The maximum kinetic energy of the photoelectrons does indeed depend on the frequency of the incident light. If the frequency of the incident light is below a certain threshold frequency specific to the metal, no photoelectrons are emitted regardless of the intensity of the light. Above this threshold, the maximum kinetic energy of the photoelectrons increases linearly with the frequency of the light. D. The emission of photoelectrons does not require a minimum threshold intensity of incident radiation, but rather a minimum threshold frequency. E. The maximum kinetic energy of the photoelectrons is not independent of the frequency of the incident light, but rather depends on it. Therefore, only statements A and C are correct.