In a photoelectric experiment,the wavelength of the light incident on a metal is changed from $300\, nm$ to $400\, nm$. The decrease in the stopping potential is close to ................ $V$ $\left( \frac{hc}{e} = 1240\, nm \cdot V \right)$

When light of frequency $v$ is incident on two metallic plates $A$ and $B$,photoelectrons are emitted. If the work function of $A$ is greater than that of $B$,which of the following curves correctly represents the relationship between the stopping potential $V$ and the incident frequency $v$?

Two identical photo-cathodes receive light of frequencies $f_1$ and $f_2$. If the velocities of the photo-electrons (of mass $m$) coming out are $v_1$ and $v_2$ respectively,then:

Ultraviolet radiations of $6.2 \, eV$ fall on an aluminium surface (work function $4.2 \, eV$). The kinetic energy in joules of the fastest electron emitted is approximately:

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 stopping potential doubles when the frequency of the incident light changes from $v$ to $\frac{3v}{2}$. Then the work function of the metal must be