Einstein's Photoelectric Equation and Energy Quantum of Radiation Questions in English

(C) Einstein's photoelectric equation is given by ${E_k = h\nu - \phi}$.
In this equation,${h\nu}$ is the energy of the incident photon and ${\phi}$ is the work function of the metal surface.
When a photon strikes the metal surface,it transfers its energy to an electron.
Some energy is used to overcome the work function $({\phi})$,and the remaining energy is converted into the kinetic energy of the electron.
Since electrons can be emitted from different depths within the metal,they lose varying amounts of energy due to collisions before escaping.
Therefore,the electrons escaping from the surface with the least energy loss possess the maximum kinetic energy.
Thus,${E_k}$ represents the maximum kinetic energy of the emitted photoelectrons.

(A) According to Einstein's photoelectric equation,$K_{max} = h\nu - \phi$,where $K_{max}$ is the maximum kinetic energy of the emitted photoelectrons,$h$ is Planck's constant,$\nu$ is the frequency of incident radiation,and $\phi$ is the work function of the metal.
From this equation,it is clear that the kinetic energy of photoelectrons depends only on the frequency of the incident radiation $(\nu)$ and the work function of the metal $(\phi)$.
It is completely independent of the intensity of the incident light,which only affects the number of photoelectrons emitted per unit time.

(D) For photoelectric emission to occur, the wavelength of the incident radiation $(\lambda)$ must be less than or equal to the threshold wavelength $(\lambda_0)$ of the material.
Given: $\lambda_0 = 5200 \, \mathring{A}$.
Photoelectric emission occurs if $\lambda \le 5200 \, \mathring{A}$.
Ultraviolet $(UV)$ radiation typically has a wavelength range of $100 \, \mathring{A}$ to $4000 \, \mathring{A}$, which is less than $5200 \, \mathring{A}$.
Infrared $(IR)$ radiation has a wavelength greater than $7000 \, \mathring{A}$, which is greater than $5200 \, \mathring{A}$.
Therefore, both the $1 \, \text{W}$ ultraviolet lamp and the $50 \, \text{W}$ ultraviolet lamp will cause photoelectric emission, regardless of their power, as long as the wavelength is in the $UV$ range.

(C) Albert Einstein was awarded the Nobel Prize in $1921$ for his discovery of the law of the photoelectric effect. While he is famous for his work on the theory of relativity,the Nobel Committee specifically cited his contribution to theoretical physics and his discovery of the law of the photoelectric effect.

(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Using the approximation $hc \approx 1240 eV \cdot nm$,we get $E = \frac{1240 eV \cdot nm}{332 nm} \approx 3.735 eV$.
According to Einstein's photoelectric equation,$E = W_0 + K_{max}$,where $W_0$ is the work function and $K_{max}$ is the maximum kinetic energy.
$K_{max} = E - W_0 = 3.735 eV - 1.07 eV = 2.665 eV$.
The retarding potential $V_0$ required to stop the photo-electrons is given by $K_{max} = eV_0$.
Therefore,$V_0 = 2.665 V \approx 2.66 V$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{\max})$ of emitted photo-electrons is given by:
$K_{\max} = h\nu - W_0$
Where:
$h$ is Planck's constant,
$\nu$ is the frequency of the incident light,
$W_0$ is the work function of the metal.
Since $h$ and $W_0$ are constants for a given metal,$K_{\max}$ depends directly on the frequency $(\nu)$ of the incident light.

(C) The work function $W_0$ is related to the threshold wavelength $\lambda_0$ by the formula:
$W_0 = \frac{hc}{\lambda_0}$
Using the approximation $hc \approx 12400 \ eV \cdot \mathring A$ (or $12375 \ eV \cdot \mathring A$ for specific textbook conventions):
$\lambda_0 = \frac{12375}{W_0} \mathring A$
Given $W_0 = 4.2 \ eV$:
$\lambda_0 = \frac{12375}{4.2} \approx 2946.4 \mathring A$.
Rounding to the nearest provided option,we get $\lambda_0 \approx 2955 \mathring A$.

(D) According to Einstein's photoelectric equation: $E = W_0 + K_{\max}$.
First,calculate the energy of the incident photon: $E = \frac{hc}{\lambda} \approx \frac{12400 \; eV \cdot \mathring{A}}{3000 \; \mathring{A}} \approx 4.13 \; eV$.
Given the work function $W_0 = 1 \; eV$,the maximum kinetic energy is $K_{\max} = E - W_0 = 4.13 \; eV - 1 \; eV = 3.13 \; eV$.
Convert $K_{\max}$ to Joules: $K_{\max} = 3.13 \times 1.6 \times 10^{-19} \; J \approx 5 \times 10^{-19} \; J$.
Using $K_{\max} = \frac{1}{2} m v^2$,where $m = 9.1 \times 10^{-31} \; kg$:
$v = \sqrt{\frac{2 K_{\max}}{m}} = \sqrt{\frac{2 \times 5 \times 10^{-19}}{9.1 \times 10^{-31}}} \approx \sqrt{1.1 \times 10^{12}} \approx 1.05 \times 10^6 \; m/s$.
Thus,the velocity is approximately $1 \times 10^6 \; m/s$.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photo-electrons is given by $K_{max} = h\nu - \phi_0$,where $h$ is Planck's constant,$\nu$ is the frequency of incident light,and $\phi_0$ is the work function.
The retarding potential (stopping potential) $V_0$ is related to the maximum kinetic energy by the relation $eV_0 = K_{max}$.
Substituting this into the equation,we get $eV_0 = h\nu - \phi_0$,which simplifies to $V_0 = \frac{h}{e}\nu - \frac{\phi_0}{e}$.
Since $h$,$e$,and $\phi_0$ are constants,the stopping potential $V_0$ is a linear function of the frequency $\nu$. Therefore,as the frequency of the incident light increases,the retarding potential increases uniformly.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by:
$K_{\max} = \frac{hc}{\lambda} - W_0$
Given:
Work function $W_0 = 1.6 \times 10^{-19} \, J$
Wavelength $\lambda = 6400 \, \mathring{A} = 6400 \times 10^{-10} \, m$
Planck's constant $h = 6.4 \times 10^{-34} \, Js$
Speed of light $c = 3 \times 10^8 \, m/s$
Calculating the energy of the incident photon:
$E = \frac{hc}{\lambda} = \frac{6.4 \times 10^{-34} \times 3 \times 10^8}{6400 \times 10^{-10}}$
$E = \frac{19.2 \times 10^{-26}}{6.4 \times 10^{-7}} = 3 \times 10^{-19} \, J$
Now,calculating $K_{\max}$:
$K_{\max} = 3 \times 10^{-19} - 1.6 \times 10^{-19} = 1.4 \times 10^{-19} \, J$

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ of the emitted electron is given by:
$K_{\max} = E - W_0$
Where $E$ is the energy of the incident photon and $W_0$ is the work function of the metal.
Given: $E = 6.2 \, eV$ and $W_0 = 4.2 \, eV$.
$K_{\max} = 6.2 \, eV - 4.2 \, eV = 2.0 \, eV$.
To convert this energy into Joules,we multiply by the charge of an electron $(1.6 \times 10^{-19} \, C)$:
$K_{\max} = 2.0 \times 1.6 \times 10^{-19} \, J = 3.2 \times 10^{-19} \, J$.

(C) The work function $W_0$ is related to the threshold wavelength $\lambda_0$ by the formula: $W_0 = \frac{hc}{\lambda_0}$.
Since $h$ and $c$ are constants, the work function is inversely proportional to the threshold wavelength: $W_0 \propto \frac{1}{\lambda_0}$.
Therefore, we can write the ratio: $\frac{(W_0)_T}{(W_0)_{Na}} = \frac{(\lambda_0)_{Na}}{(\lambda_0)_T}$.
Rearranging for the threshold wavelength of tungsten $(\lambda_0)_T$: $(\lambda_0)_T = \frac{(\lambda_0)_{Na} \times (W_0)_{Na}}{(W_0)_T}$.
Substituting the given values: $(\lambda_0)_T = \frac{5460 \times 2.3}{4.5}$.
Calculating the result: $(\lambda_0)_T \approx 2791 \text{ \AA}$.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{max})$ of the emitted photo-electrons is given by the formula:
$K_{max} = E - \Phi$
Where:
$E$ is the energy of the incident photon = $3.4 \ eV$
$\Phi$ is the work function of the metal = $2 \ eV$
Substituting the given values into the equation:
$K_{max} = 3.4 \ eV - 2 \ eV$
$K_{max} = 1.4 \ eV$
Therefore,the maximum kinetic energy of the photo-electrons is $1.4 \ eV$.

(B) The work function $W_0$ is given by the formula $W_0 = \frac{hc}{\lambda_0}$.
Given,threshold wavelength $\lambda_0 = 6600 \, \mathring{A}$.
Using the relation $W_0 (\text{in } eV) = \frac{12375}{\lambda_0 (\text{in } \mathring{A})}$.
Substituting the value: $W_0 = \frac{12375}{6600} \, eV$.
$W_0 = 1.875 \, eV \approx 1.87 \, eV$.

(D) The correct answer is $D$.
Albert Einstein successfully explained the photoelectric effect in $1905$.
He proposed that light consists of discrete packets of energy called photons,where the energy of each photon is given by $E = h
u$.
While Max Planck introduced the concept of energy quantization for blackbody radiation,it was Einstein who applied this concept to explain the photoelectric effect,demonstrating that light behaves as particles in this phenomenon.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K_{\max} = \frac{hc}{\lambda} - W_0$.
Since the stopping potential $V_0$ is related to kinetic energy by $K_{\max} = eV_0$,we have $eV_0 = \frac{hc}{\lambda} - W_0$.
Rearranging this,we get $V_0 = \frac{hc}{e\lambda} - \frac{W_0}{e}$.
Here,$h$ is Planck's constant,$c$ is the speed of light,$e$ is the charge of an electron,$\lambda$ is the wavelength of incident light,and $W_0$ is the work function of the metal.
As the wavelength $\lambda$ decreases from $4000 \, \mathring{A}$ to $3000 \, \mathring{A}$,the term $\frac{hc}{e\lambda}$ increases.
Since $W_0$ is constant for a given metal,the stopping potential $V_0$ must increase.
Therefore,the potential required to stop the ejection of electrons will be greater than $2 \, V$.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by:
$K_{\max} = h\nu - W_0$
Where:
$h = 6.63 \times 10^{-34} \ J \cdot s$ (Planck's constant)
$\nu = 8 \times 10^{14} \ Hz$ (Frequency of incident light)
$W_0 = 3.2 \times 10^{-19} \ J$ (Work function of the metal)
Substituting the values:
$K_{\max} = (6.63 \times 10^{-34} \times 8 \times 10^{14}) - 3.2 \times 10^{-19}$
$K_{\max} = (53.04 \times 10^{-20}) - 3.2 \times 10^{-19}$
$K_{\max} = 5.304 \times 10^{-19} - 3.2 \times 10^{-19}$
$K_{\max} = 2.104 \times 10^{-19} \ J$
Therefore,the correct option is $A$.

(C) According to Einstein's photoelectric equation, $K_{max} = h\nu - \Phi_0$, where $K_{max} = eV_s$.
Here, $V_s$ is the stopping potential, $h$ is Planck's constant, $\nu$ is the frequency of incident light, and $\Phi_0$ is the work function of the material.
Rearranging the equation: $eV_s = h\nu - \Phi_0$, which gives $V_s = \frac{h}{e}\nu - \frac{\Phi_0}{e}$.
This shows that the stopping potential $V_s$ depends on the frequency $\nu$ of the incident light and the work function $\Phi_0$, which is a characteristic property of the cathode material.
Therefore, the stopping potential depends on both the frequency of the incident light and the nature of the cathode material.

(B) The threshold wavelength is $\lambda_0 = 200 \ nm = 2000 \ \mathring{A}$.
The incident wavelength is $\lambda = 100 \ nm = 1000 \ \mathring{A}$.
Using Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by:
$K_{\max} = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$
Using the shortcut formula $K_{\max} \ (eV) \approx 12400 \ \left[ \frac{1}{\lambda \ (\mathring{A})} - \frac{1}{\lambda_0 \ (\mathring{A})} \right]$:
$K_{\max} = 12400 \ \left[ \frac{1}{1000} - \frac{1}{2000} \right]$
$K_{\max} = 12400 \ \left[ \frac{2-1}{2000} \right] = \frac{12400}{2000} = 6.2 \ eV$.

(B) The stopping potential $(V_s)$ depends on the frequency of the incident light and the work function of the metal surface,as given by Einstein's photoelectric equation: $K_{max} = h\nu - \phi = eV_s$.
Changing the distance between the light source and the photocell changes the intensity of the light,which affects the number of photoelectrons emitted per second (photoelectric current),but it does not change the energy of individual photons or the work function of the metal.
Since the frequency of the incident light remains constant,the maximum kinetic energy of the emitted photoelectrons remains the same.
Therefore,the stopping potential remains independent of the distance of the light source.
Thus,the stopping potential remains $0.6 \ V$.

(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Using $h = 6.6 \times 10^{-34} \,Js$,$c = 3 \times 10^8 \,m/s$,and $\lambda = 4000 \times 10^{-10} \,m$:
$E = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{4000 \times 10^{-10}} \,J = 4.95 \times 10^{-19} \,J$.
Converting to $eV$ by dividing by $e = 1.6 \times 10^{-19} \,C$:
$E = \frac{4.95 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.09 \,eV$.
According to Einstein's photoelectric equation,$E = W_0 + K_{max}$,where $W_0$ is the work function and $K_{max}$ is the maximum kinetic energy.
Given the stopping potential $V_s = 2 \,V$,the maximum kinetic energy is $K_{max} = e \times V_s = 2 \,eV$.
Substituting the values: $3.09 \,eV = W_0 + 2 \,eV$.
Therefore,$W_0 = 3.09 - 2 = 1.09 \,eV$,which is approximately $1.1 \,eV$.

(D) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = W_0 + \frac{1}{2}mv_{\max}^2$.
Assuming the work function $W_0$ is negligible compared to the incident photon energy $\frac{hc}{\lambda}$,we have $\frac{1}{2}mv_{\max}^2 \approx \frac{hc}{\lambda}$.
This implies $v_{\max}^2 \propto \frac{1}{\lambda}$,or $v_{\max} \propto \frac{1}{\sqrt{\lambda}}$.
Let the initial wavelength be $\lambda_1 = \lambda$ and the final wavelength be $\lambda_2 = 4\lambda$.
The ratio of the maximum velocities is $\frac{v_{\max, 2}}{v_{\max, 1}} = \sqrt{\frac{\lambda_1}{\lambda_2}} = \sqrt{\frac{\lambda}{4\lambda}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus,the maximum velocity becomes half of its initial value.

(D) The work function $W_0$ is given by the relation $W_0 = h\nu_0$,where $h$ is Planck's constant and $\nu_0$ is the threshold frequency.
Given: $W_0 = 2.51 eV = 2.51 \times 1.6 \times 10^{-19} J$ and $h = 6.63 \times 10^{-34} J \cdot s$.
Rearranging the formula for threshold frequency: $\nu_0 = \frac{W_0}{h}$.
Substituting the values: $\nu_0 = \frac{2.51 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}}$.
$\nu_0 \approx 0.60588 \times 10^{15} Hz = 6.06 \times 10^{14} Hz$.
Rounding to the nearest provided option,we get $6.08 \times 10^{14} \text{ cycle/sec}$.

(A) The cut-off voltage (also known as stopping potential) depends on the maximum kinetic energy of the emitted photoelectrons.
According to Einstein's photoelectric equation,$K_{max} = h\nu - \Phi$,where $h\nu$ is the energy of the incident photon and $\Phi$ is the work function of the metal.
The stopping potential $V_s$ is given by $eV_s = K_{max}$.
Changing the distance of the light source changes the intensity of the light incident on the photocell,which affects the number of photoelectrons emitted (photoelectric current),but it does not change the frequency of the incident light.
Since the frequency of the incident light remains constant,the maximum kinetic energy of the emitted photoelectrons remains unchanged.
Therefore,the cut-off voltage remains $V$ regardless of the distance of the source.

(D) The work function $W_0$ is given by $W_0 = h \nu_0$,where $h$ is Planck's constant and $\nu_0$ is the threshold frequency.
Given $W_0 = 3.3 \text{ eV} = 3.3 \times 1.6 \times 10^{-19} \text{ J}$.
Planck's constant $h \approx 6.6 \times 10^{-34} \text{ J s}$.
Substituting the values:
$\nu_0 = \frac{W_0}{h} = \frac{3.3 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}}$
$\nu_0 = \frac{5.28 \times 10^{-19}}{6.6 \times 10^{-34}} = 0.8 \times 10^{15} \text{ Hz} = 8 \times 10^{14} \text{ Hz}$.

(A) According to Einstein's photoelectric equation,the energy of an incident photon is given by $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency of the light.
For the emission of photoelectrons to occur,the energy of the incident photon must be greater than or equal to the work function $\phi$ of the metal.
If the energy of the incident photon is less than the work function $(h\nu < \phi)$,no photoelectrons will be emitted.
Therefore,the condition for no emission is $\nu < \frac{\phi}{h}$.

(C) According to Einstein's photoelectric equation,the kinetic energy $E$ is given by: $E = \frac{hc}{\lambda} - W_0$,where $W_0$ is the work function.
For kinetic energy $2E$,the new wavelength $\lambda'$ satisfies: $2E = \frac{hc}{\lambda'} - W_0$.
From the first equation,$\frac{hc}{\lambda} = E + W_0$,and from the second,$\frac{hc}{\lambda'} = 2E + W_0$.
Dividing the two equations: $\frac{\lambda'}{\lambda} = \frac{E + W_0}{2E + W_0}$.
This can be written as: $\lambda' = \lambda \left( \frac{1 + W_0/E}{2 + W_0/E} \right)$.
Since $W_0 > 0$,the term $\frac{1 + W_0/E}{2 + W_0/E}$ is always greater than $\frac{1}{2}$ and less than $1$.
Therefore,$\frac{\lambda}{2} < \lambda' < \lambda$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of emitted electrons is given by $K_{max} = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function of the metal.
Since the stopping potential $V_0$ is related to kinetic energy by $K_{max} = eV_0$,we have $eV_0 = \frac{hc}{\lambda} - \phi$,which simplifies to $V_0 = \frac{hc}{e\lambda} - \frac{\phi}{e}$.
As the wavelength $\lambda$ of the incident radiation decreases from $6000 \, \mathring{A}$ to $4000 \, \mathring{A}$,the term $\frac{hc}{e\lambda}$ increases.
Since the work function $\phi$ is a constant property of the metal,the stopping potential $V_0$ must increase.
Therefore,the correct option is $B$.

(C) The work function $\Phi_0$ is related to the threshold wavelength $\lambda_0$ by the formula: $\Phi_0 = \frac{hc}{\lambda_0}$.
Using the approximation $hc \approx 12375 \ eV \cdot \mathring{A}$,we have:
$\lambda_0 = \frac{12375}{\Phi_0 \ (eV)} \ \mathring{A}$.
Given $\Phi_0 = 4.125 \ eV$,we calculate:
$\lambda_0 = \frac{12375}{4.125} \ \mathring{A} = 3000 \ \mathring{A}$.
Thus,the correct option is $C$.

(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Using the approximation $hc \approx 12400 \; eV \cdot \mathring A$,we get $E = \frac{12400}{5000} = 2.48 \; eV$.
According to Einstein's photoelectric equation,$E = W_0 + K_{\max}$,where $W_0$ is the work function and $K_{\max}$ is the maximum kinetic energy.
Substituting the values: $2.48 = 1.9 + K_{\max}$.
Therefore,$K_{\max} = 2.48 - 1.9 = 0.58 \; eV$.

(C) The work function $W_0$ is given as $4.0 \,eV$.
The threshold wavelength $\lambda_0$ is the longest wavelength capable of causing photoelectric emission.
It is calculated using the formula: $\lambda_0 = \frac{hc}{W_0}$.
Using the approximation $hc \approx 12400 \,eV \cdot \mathring{A}$:
$\lambda_0 = \frac{12400 \,eV \cdot \mathring{A}}{4.0 \,eV} = 3100 \,\mathring{A}$.
Since $1 \,nm = 10 \,\mathring{A}$,we have $\lambda_0 = 310 \,nm$.

(B) The relationship between the maximum kinetic energy $(K_{\max})$ of photoelectrons and the stopping potential $(V_s)$ is given by the equation: $K_{\max} = e V_s$.
Given that the maximum kinetic energy $K_{\max} = 4 eV$.
Substituting the value into the equation: $4 eV = e V_s$.
Therefore,the stopping potential $V_s = 4 V$.

(D) The threshold wavelength $\lambda_0$ is given by the formula $\lambda_0 = \frac{hc}{\phi}$, where $\phi$ is the work function.
Using the approximation $\lambda_0 \approx \frac{12400 \text{ eV Å}}{\phi \text{ (in eV)}}$, we get $\lambda_0 = \frac{12400}{2.1} \approx 5904.76 \text{ Å}$.
For photoelectric emission to occur, the incident wavelength $\lambda$ must be less than or equal to the threshold wavelength $\lambda_0$ (i.e., $\lambda \le \lambda_0$).
Comparing the given options, only $4000 \text{ Å}$ is less than $5904.76 \text{ Å}$.
Therefore, only $4000 \text{ Å}$ can emit photoelectrons.

(C) According to Einstein's photoelectric equation: $K_{\max} = h\nu - W_0$,where $K_{\max}$ is the maximum kinetic energy,$h$ is Planck's constant,$\nu$ is the frequency of incident light,and $W_0$ is the work function of the metal.
Let the initial frequency be $\nu$. The initial kinetic energy is $K_1 = h\nu - W_0$.
When the frequency is doubled,the new frequency becomes $2\nu$. The new kinetic energy is $K_2 = h(2\nu) - W_0 = 2h\nu - W_0$.
Comparing $K_2$ with $2K_1$:
$2K_1 = 2(h\nu - W_0) = 2h\nu - 2W_0$.
Since $W_0 > 0$,it follows that $2h\nu - W_0 > 2h\nu - 2W_0$.
Therefore,$K_2 > 2K_1$. The kinetic energy becomes more than doubled.

(B) The work function $W_0$ of a photoelectric emitter is given by the formula $W_0 = \frac{hc}{\lambda_0}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda_0$ is the threshold wavelength.
Assuming the given wavelengths are the threshold wavelengths for the respective emitters,we have $\lambda_{01} = 300 \; nm$ and $\lambda_{02} = 600 \; nm$.
The ratio of the work functions is $\frac{W_{01}}{W_{02}} = \frac{hc / \lambda_{01}}{hc / \lambda_{02}} = \frac{\lambda_{02}}{\lambda_{01}}$.
Substituting the values: $\frac{W_{01}}{W_{02}} = \frac{600 \; nm}{300 \; nm} = \frac{2}{1}$.
Thus,the ratio is $2:1$.

(C) The work function $W_0$ is given by the formula $W_0 = \frac{hc}{\lambda_0}$.
Here,$h = 6.625 \times 10^{-34}\;J\cdot s$,$c = 3 \times 10^8\;m/s$,and $\lambda_0 = 5000\;\mathring{A} = 5000 \times 10^{-10}\;m$.
Substituting the values:
$W_0 = \frac{6.625 \times 10^{-34} \times 3 \times 10^8}{5000 \times 10^{-10}}$
$W_0 = \frac{19.875 \times 10^{-26}}{5 \times 10^{-7}}$
$W_0 \approx 3.975 \times 10^{-19}\;J \approx 4 \times 10^{-19}\;J$.

(B) According to the photoelectric effect,the number of photoelectrons emitted $(N)$ is directly proportional to the intensity (illumination $L$) of the incident light,provided the frequency is above the threshold frequency.
According to Einstein's photoelectric equation,$E = \frac{hc}{\lambda} - W_0$,where $E$ is the maximum kinetic energy,$h$ is Planck's constant,$c$ is the speed of light,$\lambda$ is the wavelength,and $W_0$ is the work function of the metal.
From this equation,it is clear that $E$ depends on the wavelength $\lambda$ such that $E \propto \frac{1}{\lambda}$ (assuming $W_0$ is constant).
Therefore,$N \propto L$ and $E \propto \frac{1}{\lambda}$.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy of the emitted photoelectrons is given by $K_{max} = E - W_0$.
Here,the energy of the incident light $E = 2 \ eV$ and the work function of the metal $W_0 = 0.6 \ eV$.
Substituting these values,we get $K_{max} = 2 \ eV - 0.6 \ eV = 1.4 \ eV$.
The stopping potential $V_0$ is related to the maximum kinetic energy by the equation $K_{max} = e V_0$.
Therefore,$V_0 = \frac{K_{max}}{e} = \frac{1.4 \ eV}{e} = 1.4 \ V$.

(B) The condition for photoelectric emission is that the incident photon energy must be greater than the work function of the metal, or equivalently, the incident wavelength must be less than the threshold wavelength $(\lambda < \lambda_0)$.
The threshold wavelength $\lambda_0$ is given by $\lambda_0 = \frac{hc}{\phi} \approx \frac{12400 \ \text{eV} \cdot \mathring{A}}{\phi \text{ (in eV)}} \ \mathring{A}$.
For Sodium $(Na)$: $\phi_{Na} = 2 \ eV$. Therefore, $\lambda_{0,Na} = \frac{12400}{2} = 6200 \ \mathring{A}$.
Since the incident light wavelength $\lambda = 4000 \ \mathring{A}$ is less than $6200 \ \mathring{A}$, Sodium is suitable.
For Copper $(Cu)$: $\phi_{Cu} = 4 \ eV$. Therefore, $\lambda_{0,Cu} = \frac{12400}{4} = 3100 \ \mathring{A}$.
Since the incident light wavelength $\lambda = 4000 \ \mathring{A}$ is greater than $3100 \ \mathring{A}$, Copper is not suitable.
Thus, Sodium is the correct choice.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy of photoelectrons is given by $K_{\max} = h\nu - \phi$,where $h$ is Planck's constant,$\nu$ is the frequency of incident light,and $\phi$ is the work function of the metal.
This equation shows that $K_{\max}$ depends only on the frequency of the incident light and the nature of the material (work function).
Intensity of light affects the number of photons incident per unit area per unit time,which in turn affects the number of photoelectrons emitted (photoelectric current),but it does not affect the energy of individual photoelectrons.
Therefore,if the intensity of light is doubled,the maximum kinetic energy of the photoelectrons remains unchanged.

(B) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$. Using $hc \approx 12375 \text{ eV} \cdot \mathring{A}$,we get $E = \frac{12375}{4558} \approx 2.71 \text{ eV}$.
According to Einstein's photoelectric equation,$E = \phi + K_{\max}$,where $K_{\max} = \frac{1}{2}mv_{\max}^2$.
Substituting the values: $2.71 \text{ eV} = 2.5 \text{ eV} + K_{\max} \Rightarrow K_{\max} = 0.21 \text{ eV}$.
Converting $K_{\max}$ to Joules: $K_{\max} = 0.21 \times 1.6 \times 10^{-19} \text{ J} = 0.336 \times 10^{-19} \text{ J}$.
Using $v_{\max} = \sqrt{\frac{2K_{\max}}{m}}$,where $m = 9.1 \times 10^{-31} \text{ kg}$:
$v_{\max} = \sqrt{\frac{2 \times 0.336 \times 10^{-19}}{9.1 \times 10^{-31}}} \approx \sqrt{0.0738 \times 10^{12}} \approx 2.71 \times 10^5 \text{ m/s}$.
Rounding to the nearest provided option,the speed is approximately $2.65 \times 10^5 \text{ m/s}$.

(A) Einstein explained the phenomenon of the photoelectric effect based on Planck's theory.
According to this,the maximum kinetic energy $E_k$ of photo-electrons emitted from a metal surface is given by the equation:
$E_k = h\nu - \phi$ -- $(i)$
where $h\nu$ is the energy of the incident photon and $\phi$ is the work function of the metal.
For a given metal,the threshold frequency $\nu_0$ is the minimum frequency required to eject an electron. The energy required for this is the work function:
$\phi = h\nu_0$ -- $(ii)$
Substituting equation $(ii)$ into equation $(i)$,we get:
$E_k = h\nu - h\nu_0$
$E_k = h(\nu - \nu_0)$

(D) According to Einstein's photoelectric equation,$K_{max} = h\nu - \Phi_0$,where $K_{max} = eV_0$.
Thus,$eV_0 = h\nu - \Phi_0$,which implies $V_0 = \frac{h\nu}{e} - \frac{\Phi_0}{e}$.
Here,$V_0$ depends on the frequency of incident light $(\nu)$ and the work function $(\Phi_0)$ of the material.
It does not depend on the intensity of the incident light,as intensity only affects the number of photoelectrons emitted per second,not their maximum kinetic energy.
Therefore,the stopping potential is independent of the intensity of the incident light.

(A) The threshold wavelength $\lambda_0$ is related to the work function $W_0$ by the formula: $\lambda_0 = \frac{hc}{W_0}$.
Using the shortcut formula $\lambda_0 (\text{in } \mathring{A}) = \frac{12375}{W_0 (\text{in } eV)}$,we substitute $W_0 = 3 \ eV$.
$\lambda_0 = \frac{12375}{3} = 4125 \ \mathring{A}$.
Therefore,the correct option is $A$.

(B) According to Einstein's photoelectric equation, $K_{max} = h\nu - \Phi_0 = \frac{hc}{\lambda} - \Phi_0$.
As the wavelength $\lambda$ of the incident photon decreases, the energy of the incident photon $(E = \frac{hc}{\lambda})$ increases.
Since the work function $\Phi_0$ of the metal surface remains constant, the maximum kinetic energy $K_{max}$ of the emitted photoelectrons increases.
Because $K_{max} = \frac{1}{2}mv_{max}^2$, an increase in $K_{max}$ leads to an increase in the maximum velocity $v_{max}$ of the emitted photoelectrons.

(A) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = W_0 + K_{max}$,where $K_{max} = \frac{1}{2}mv^2$.
For $\lambda_1 = 400\; nm = 400 \times 10^{-9}\; m$,velocity is $v$: $\frac{hc}{400 \times 10^{-9}} = W_0 + \frac{1}{2}mv^2$ ... $(i)$
For $\lambda_2 = 250\; nm = 250 \times 10^{-9}\; m$,velocity is $2v$: $\frac{hc}{250 \times 10^{-9}} = W_0 + \frac{1}{2}m(2v)^2 = W_0 + 2mv^2$ ... $(ii)$
From $(i)$,$\frac{1}{2}mv^2 = \frac{hc}{400 \times 10^{-9}} - W_0$. Substituting this into $(ii)$:
$\frac{hc}{250 \times 10^{-9}} = W_0 + 4(\frac{hc}{400 \times 10^{-9}} - W_0)$
$\frac{hc}{250 \times 10^{-9}} = W_0 + \frac{hc}{100 \times 10^{-9}} - 4W_0$
$3W_0 = hc(\frac{1}{100 \times 10^{-9}} - \frac{1}{250 \times 10^{-9}}) = hc(\frac{2.5 - 1}{250 \times 10^{-9}}) = hc(\frac{1.5}{250 \times 10^{-9}})$
$3W_0 = hc(0.006 \times 10^9) = hc(6 \times 10^6)$
$W_0 = 2hc \times 10^6\; J$.

(A) According to Einstein's photoelectric equation,the energy of the incident photon $(E)$ is equal to the sum of the work function $(W_0)$ and the maximum kinetic energy $(K_{max})$ of the emitted photoelectrons.
$E = W_0 + K_{max}$
Given that $K_{max} = eV_0$,where $V_0$ is the stopping potential,the equation becomes:
$E = W_0 + eV_0$
Substituting the given values ($E = 4 \ eV$ and $W_0 = 2 \ eV$):
$4 \ eV = 2 \ eV + eV_0$
$eV_0 = 4 \ eV - 2 \ eV$
$eV_0 = 2 \ eV$
Therefore,the stopping potential $V_0 = 2 \ V$.

(B) The work function $(\Phi_0)$ is defined as the minimum amount of energy required to remove an electron from the surface of a metal.
Threshold frequency $(v_0)$ is the minimum frequency of incident radiation below which no photoelectric emission occurs.
The relationship between the work function and the threshold frequency is given by the formula $\Phi_0 = hv_0$,where $h$ is the Planck constant.