If the work function of a metal is $3 \ eV$,then the threshold wavelength will be ............ $\mathring{A}$.

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 :

An electron in a hydrogen atom jumps from the second Bohr orbit to the ground state,and the energy difference of the two states is radiated in the form of photons. These photons are then allowed to fall on a metal surface having a work function equal to $4.2 \ eV$. Calculate the stopping potential. [Energy of electron in $n^{\text{th}}$ orbit $= -\frac{13.6}{n^2} \ eV$] (in $V$)

Photons of wavelength $\lambda$ emitted by a source of power $P$ are incident on a photocell. If the current produced in the cell is $I$,then the percentage of incident photons which produce current in the photocell is: (where $h$ is Planck's constant and $c$ is the speed of light in vacuum)

The work function of a metallic surface is $5.01\ eV$. Photoelectrons are emitted when light of wavelength $2000\ \mathring{A}$ falls on it. The minimum potential difference required to stop the fastest photoelectrons is ................. $V$.

The threshold frequency of a photoelectric metal is $\nu_0$. If light of frequency $4\nu_0$ is incident on this metal,then the maximum kinetic energy of emitted electrons will be $.......h\nu_0$.