Light of frequency $\nu$ is incident on a substance of threshold frequency $\nu_0$ (where $\nu_0 < \nu$). The maximum kinetic energy of the emitted photo-electron will be:

The threshold frequency of potassium is $3 \times 10^{14} \ Hz$. The work function is ...... .

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:

The maximum velocity of the photoelectron emitted by the metal surface is $V$. The charge and mass of the photoelectron are denoted by $e$ and $m$ respectively. The stopping potential in volts is:

Light of wavelength $\lambda$ strikes a photoelectric surface and electrons are ejected with kinetic energy $K$. If $K$ is to be increased to twice its original value,the wavelength must be changed to $\lambda'$ such that :-

When light of frequency twice the threshold frequency is incident on a metal plate,the maximum velocity of the emitted electron is $v_{1}$. When the frequency of incident radiation is increased to five times the threshold value,the maximum velocity of the emitted electron becomes $v_{2}$. If $v_{2} = x v_{1}$,the value of $x$ will be...........