In the photoelectric effect,if the intensity of light is doubled,then the maximum kinetic energy of photoelectrons will become

Two monochromatic light beams $A$ and $B$ of equal intensity are incident normally on a metallic surface per unit area. Their wavelengths are $\lambda_A$ and $\lambda_B$ respectively. The incident light is effective in ejecting photoelectrons. The ratio of the number of photoelectrons emitted by beam $A$ to that by beam $B$ is:

When photons of energies twice and thrice the work function of a metal are incident on the metal surface one after other,the maximum velocities of the photoelectrons emitted in the two cases are $v_1$ and $v_2$ respectively. The ratio $v_1: v_2$ is

$A$ certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$. The stopping potential for the photoelectric current for this light is $3V_0$. If the same surface is illuminated with light of wavelength $2\lambda$,the stopping potential is $V_0$. The threshold wavelength for this surface for the photoelectric effect is:

When photons of energy $hv$ fall on a metal plate of work function $W_0$,photoelectrons of maximum kinetic energy $K$ are ejected. If the frequency of the radiation is doubled,the maximum kinetic energy of the ejected photoelectrons will be

When a metallic surface is illuminated with light of wavelength $\lambda$,the stopping potential is $x \ V$. When the same surface is illuminated by light of wavelength $2 \lambda$,the stopping potential is $\frac{x}{3} \ V$. The threshold wavelength for the metallic surface is ..........