Mercury violet light $(\lambda = 4558 \mathring{A})$ is falling on a photosensitive material $(\phi = 2.5 \text{ eV})$. The speed of the ejected electrons is,in $\text{m/s}$,about:

$K_1$ and $K_2$ are the maximum kinetic energies of photoelectrons emitted from a surface of a given material for light of wavelengths $\lambda_1$ and $\lambda_2$,respectively. If $\lambda_1 = 2\lambda_2$,then the work function of the material is given by:

When a metallic surface is illuminated with radiation of wavelength $\lambda$,the stopping potential is $V$. If the same surface is illuminated with radiation of wavelength $2\lambda$,the stopping potential is $\frac{V}{4}$. The threshold wavelength for the metallic surface is:

Consider the following statements regarding the photoelectric effect experiment:
$(I)$ Photoelectrons are emitted as soon as the metal is exposed to light.
$(II)$ There is a minimum frequency below which no photocurrent is observed.
$(III)$ The stopping potential is proportional to the frequency of light.
$(IV)$ The photocurrent varies linearly with the intensity of the light.
Which of the above statements indicate that light consists of quanta (photons) with energy proportional to frequency?

The figure represents the graph of photocurrent $I$ versus applied voltage $(V)$. The maximum kinetic energy of the emitted photoelectrons is:

When light of wavelength $300 \; nm$ falls on a photoelectric emitter,photoelectrons are liberated. For another emitter,light of $600 \; nm$ wavelength is sufficient for creating photoemission. What is the ratio of the work functions of the two emitters?