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

(C) According to Einstein's photoelectric equation, the maximum kinetic energy $(K_{max})$ of emitted electrons is given by:
$K_{max} = E - \Phi_0$
where $E$ is the energy of the incident photon and $\Phi_0$ is the work function of the metal.
Given the cutoff frequency (threshold frequency) is $\nu$, the work function is $\Phi_0 = h\nu$.
The energy of the incident radiation with frequency $3\nu$ is $E = h(3\nu) = 3h\nu$.
Substituting these values into the equation:
$K_{max} = 3h\nu - h\nu = 2h\nu$.
Since $K_{max} = \frac{1}{2}mv_{max}^2$, we have:
$\frac{1}{2}mv_{max}^2 = 2h\nu$
$v_{max}^2 = \frac{4h\nu}{m}$
$v_{max} = \sqrt{\frac{4h\nu}{m}} = 2\sqrt{\frac{h\nu}{m}}$.
Thus, the correct option is $C$.

(B) According to Einstein's photoelectric equation, the maximum kinetic energy $K$ is given by $K = \frac{hc}{\lambda} - \Phi$, where $\Phi$ is the work function of the metal.
For wavelength $\lambda_1$, $K_1 = \frac{hc}{\lambda_1} - \Phi$.
For wavelength $\lambda_2$, $K_2 = \frac{hc}{\lambda_2} - \Phi$.
Given $\lambda_1 = 3\lambda_2$, we substitute this into the equation for $K_1$:
$K_1 = \frac{hc}{3\lambda_2} - \Phi$.
Since $\frac{hc}{\lambda_2} = K_2 + \Phi$, we have $K_1 = \frac{K_2 + \Phi}{3} - \Phi = \frac{K_2}{3} + \frac{\Phi}{3} - \Phi = \frac{K_2}{3} - \frac{2\Phi}{3}$.
Since $\Phi > 0$, it follows that $K_1 < \frac{K_2}{3}$.

(C) Einstein's photoelectric equation is given by $K_{max} = h\nu - \Phi$, 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.
Comparing this equation with the straight-line equation $y = mx + c$, we get $y = K_{max}$, $x = \nu$, $m = h$, and $c = -\Phi$.
The slope of the graph of $K_{max}$ versus $\nu$ is $h$ (Planck's constant).
Since $h$ is a universal constant, the slope is the same for all metals and is independent of the intensity of the incident radiation.

(B) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ is given by $K_{max} = \frac{hc}{\lambda} - \phi$, where $\phi$ is the work function.
For the first case with wavelength $\lambda$: $K_1 = \frac{hc}{\lambda} - \phi$.
For the second case with wavelength $\lambda/3$: $K_2 = \frac{hc}{(\lambda/3)} - \phi = \frac{3hc}{\lambda} - \phi$.
Given that $K_2 = 4K_1$, we substitute the expressions:
$\frac{3hc}{\lambda} - \phi = 4(\frac{hc}{\lambda} - \phi)$.
$\frac{3hc}{\lambda} - \phi = \frac{4hc}{\lambda} - 4\phi$.
Rearranging the terms to solve for $\phi$:
$4\phi - \phi = \frac{4hc}{\lambda} - \frac{3hc}{\lambda}$.
$3\phi = \frac{hc}{\lambda}$.
$\phi = \frac{hc}{3\lambda}$.

(D) According to Einstein's photoelectric equation: $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$, where $\lambda_0$ is the threshold wavelength.
For the first case: $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$ --- $(1)$
For the second case: $e(\frac{V}{4}) = \frac{hc}{1.5\lambda} - \frac{hc}{\lambda_0}$ --- $(2)$
Multiply equation $(2)$ by $4$: $eV = \frac{4hc}{1.5\lambda} - \frac{4hc}{\lambda_0} = \frac{8hc}{3\lambda} - \frac{4hc}{\lambda_0}$ --- $(3)$
Equating $(1)$ and $(3)$: $\frac{hc}{\lambda} - \frac{hc}{\lambda_0} = \frac{8hc}{3\lambda} - \frac{4hc}{\lambda_0}$
$\frac{4hc}{\lambda_0} - \frac{hc}{\lambda_0} = \frac{8hc}{3\lambda} - \frac{hc}{\lambda}$
$\frac{3hc}{\lambda_0} = \frac{5hc}{3\lambda}$
$\frac{3}{\lambda_0} = \frac{5}{3\lambda}$
$\lambda_0 = \frac{9}{5}\lambda$.

(D) According to Einstein's photoelectric equation, the maximum kinetic energy of an emitted photoelectron is given by $K_{max} = h\nu - \phi$, where $\phi$ is the work function of the metal.
Since the photocathodes are identical, they have the same work function $\phi$.
For frequency $n_1$, the kinetic energy is $\frac{1}{2}mV_1^2 = hn_1 - \phi$ --- $(1)$
For frequency $n_2$, the kinetic energy is $\frac{1}{2}mV_2^2 = hn_2 - \phi$ --- $(2)$
Subtracting equation $(2)$ from equation $(1)$:
$\frac{1}{2}mV_1^2 - \frac{1}{2}mV_2^2 = (hn_1 - \phi) - (hn_2 - \phi)$
$\frac{1}{2}m(V_1^2 - V_2^2) = h(n_1 - n_2)$
$V_1^2 - V_2^2 = \frac{2h}{m}(n_1 - n_2)$.