When a certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$,the stopping potential for photoelectric current is $6V_0$. When the same surface is illuminated with light of wavelength $2\lambda$,the stopping potential is $2V_0$. The threshold wavelength for the photoelectric effect for this surface 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

The maximum kinetic energy of a photoelectron is $E$ when the wavelength of incident radiation is $\lambda$. If the wavelength of the incident radiation is reduced to $\frac{\lambda}{3}$,the maximum kinetic energy becomes $4E$. The work function of the metal is:

The maximum wavelength of light which causes photoelectric emission from a photosensitive metal surface is $\lambda_0$. Two light beams of wavelengths $\frac{\lambda_0}{3}$ and $\frac{\lambda_0}{9}$ are incident on the metal surface. The ratio of the maximum velocities of the emitted photoelectrons is

Light of wavelength $\lambda$ strikes a photo-sensitive surface and electrons are ejected with kinetic energy $E$. If the kinetic energy is to be increased to $2E$,the wavelength must be changed to $\lambda'$ where

Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A) :$ Emission of electrons in photoelectric effect can be suppressed by applying a sufficiently negative electric potential to the photoemissive substance.
Reason $(R) :$ $A$ negative electric potential, which stops the emission of electrons from the surface of a photoemissive substance, varies linearly with frequency of incident radiation.
In the light of the above statements, choose the most appropriate answer from the options given below: