Monochromatic light of wavelength $\lambda = 4770 \ \mathring{A}$ is incident separately on the surface of four different metals $A, B, C$ and $D$. The work functions of $A, B, C$ and $D$ are $4.2 \ \text{eV}, 3.7 \ \text{eV}, 3.2 \ \text{eV}$ and $2.3 \ \text{eV}$,respectively. From which of these metals will electrons be emitted?

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:

Let $K_1$ be the maximum kinetic energy of photoelectrons emitted by light of wavelength $\lambda_1$ and $K_2$ be the maximum kinetic energy corresponding to wavelength $\lambda_2$. If $\lambda_1 = 2\lambda_2$,then:

When a certain metal 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 $\frac{V}{3}$. The threshold wavelength for the surface is:

In the photoelectric effect,the kinetic energy $(K.E.)$ of electrons emitted from the metal surface depends upon:

If light of wavelength $\lambda_1$ is allowed to fall on a metal,then the kinetic energy of the photoelectrons emitted is $E_1$. If the wavelength of light changes to $\lambda_2$,then the kinetic energy of the electrons changes to $E_2$. Then the work function of the metal is: