When a photosensitive surface is illuminated with light of wavelength $\lambda$,the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2\lambda$,the stopping potential is $V/3$. The threshold wavelength for the surface is:

Explain Einstein's photoelectric effect and derive Einstein's photoelectric equation.

In a photoelectric experiment,a monochromatic light is incident on the emitter plate $E$,as shown in the figure. When switch $S_1$ is closed and switch $S_2$ is open,the photoelectrons strike the collector plate $C$ with a maximum kinetic energy of $1 eV$. If switch $S_1$ is open and switch $S_2$ is closed and the frequency of the incident light is doubled,the photoelectrons strike the collector plate with a maximum kinetic energy of $20 eV$. The threshold wavelength of the emitter plate is (in $Å$)

When the light of frequency $2v_0$ (where $v_0$ is threshold frequency) is incident on a metal plate,the maximum velocity of electrons emitted is $v_1$. When the frequency of the incident radiation is increased to $5v_0$,the maximum velocity of electrons emitted from the same plate is $v_2$. The ratio of $v_1$ to $v_2$ is

If the maximum kinetic energy of emitted electrons in the photoelectric effect is $3.2 \times 10^{-19} \text{ J}$ and the work function for the metal is $6.63 \times 10^{-19} \text{ J}$,then the stopping potential and threshold wavelength respectively are:
[Planck's constant $h = 6.63 \times 10^{-34} \text{ J} \cdot \text{s}$]
[Velocity of light $c = 3 \times 10^{8} \text{ m/s}$]
[Charge on electron $e = 1.6 \times 10^{-19} \text{ C}$]

The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength $6630 \; \mathring{A}$ is $0.42 \; V$. If the threshold frequency is $x \times 10^{13} \; s^{-1}$,where $x$ is ... (nearest integer).
(Given: speed of light $= 3 \times 10^{8} \; m/s$,Planck's constant $= 6.63 \times 10^{-34} \; J \cdot s$)