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:

In a photoelectric experiment for $4000 \, \mathring{A}$ incident radiation,the potential difference to stop the ejection is $2 \, V$. If the incident light is changed to $3000 \, \mathring{A}$,then the potential required to stop the ejection of electrons will be

In photoelectric effect, the graph of stopping potential $(V_0)$ versus frequency $(\nu)$ is a straight line. The slope of this graph is . . . . . . .

Given below are two statements:
Statement $I$: Stopping potential in photoelectric effect does not depend on the power of the light source.
Statement $II$: For a given metal,the maximum kinetic energy of the photoelectron depends on the wavelength of the incident light.
In the light of the above statements,choose the most appropriate answer from the options given below.

$A$ photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $\lambda /2$. If the maximum kinetic energy of the emitted photoelectrons in the second case is $3$ times that in the first case,the work function of the surface of the material is
$(h =$ Planck's constant,$c =$ speed of light $)$

The work function of a metal is $1 \ eV$. If light of wavelength $3000 \ \mathring{A}$ is incident on the surface of this metal, the maximum velocity of the emitted photoelectrons is ........