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 $R$.
$Assertion$ $A$ : Number of photons increases with increase in frequency of light.
$Reason$ $R$ : Maximum kinetic energy of emitted electrons increases with the frequency of incident radiation.
In the light of the above statements,choose the most appropriate answer from the options given below :

The work functions for tungsten and sodium are $4.5 \text{ eV}$ and $2.3 \text{ eV}$ respectively. If the threshold wavelength $\lambda_0$ for sodium is $5460 \text{ \AA}$, the value of $\lambda_0$ for tungsten is ............ $\text{ \AA}$.

In an experiment on the photoelectric effect,the frequency $\nu$ of the incident light is plotted against the stopping potential $V_0$. The work function of the photoelectric surface is given by ($e$ is the electronic charge):

In a photoemissive cell with exciting wavelength $\lambda$,the fastest electron has speed $v$. If the exciting wavelength is changed to $\frac{3\lambda}{4}$,the speed of the fastest emitted electron will be

When light of wavelength $300 \ nm$ falls on a photoelectric emitter,photoelectrons are just liberated. 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