When light of frequency $v$ is incident on a metal,photoelectrons are emitted. This is because...

When a light of wavelength $300 \ nm$ falls on a photoelectric emitter,photoelectrons are emitted. For another emitter,light of wavelength $600 \ nm$ is just sufficient for liberating photoelectrons. The ratio of the work function of the two emitters is

$A$ light source of wavelength $\lambda$ illuminates a metal surface and electrons are ejected with maximum kinetic energy of $2 \ \text{eV}$. If the same surface is illuminated by a light source of wavelength $\frac{\lambda}{2}$, then the maximum kinetic energy of ejected electrons will be (The work function of the metal is $1 \ \text{eV}$). (in $\text{eV}$)

The maximum kinetic energy of emitted electrons in a photoelectric effect does not depend upon

In the photoelectric effect,the stopping potential $(V_0)$ versus frequency $(\nu)$ curve is plotted. ($h$ is Planck's constant and $\phi_0$ is the work function of the metal)
$(A)$ $V_0$ versus $\nu$ is linear.
$(B)$ The slope of the $V_0$ versus $\nu$ curve $= \frac{\phi_0}{h}$.
$(C)$ Planck's constant $h$ is related to the slope of the $V_0$ versus $\nu$ line.
$(D)$ The value of the electric charge of an electron is not required to determine $h$ using the $V_0$ versus $\nu$ curve.
$(E)$ The work function can be estimated without knowing the value of $h$.
Choose the correct answer from the options given below:

The kinetic energy of an emitted electron is $E$ when the light incident on the metal has wavelength $\lambda$. To double the kinetic energy,the incident light must have a wavelength of: