When radiation of frequency $8 \times 10^{15} \ Hz$ is incident on a metal surface with a work function of $6.125 \ eV$,what is the kinetic energy of the emitted photoelectrons in $eV$?

If the electron in a hydrogen atom jumps from the third Bohr orbit to the ground state directly and the difference between the energies of the two states is radiated in the form of photons. If the work function of the material is $4.1 \text{ eV}$, then the stopping potential is nearly:
$\left[\text{Energy of electron in } n^{\text{th}} \text{ orbit} = \frac{-13.6}{n^2} \text{ eV}\right]$ (in $\text{ V}$)

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 :

Photoelectrons are emitted with maximum velocity $v$ when light of frequency $3f$ is incident on a photosensitive material of work function $2hf$. If the frequency of the incident light is $4.25f$,the maximum velocity of the emitted photoelectrons is ($h$ = Planck's constant).

When radiation of wavelength $\lambda$ is incident on a metallic surface,the stopping potential of ejected photoelectrons is $4.8 \, V$. If the same surface is illuminated by radiation of double the previous wavelength,then the stopping potential becomes $1.6 \, V$. The threshold wavelength of the metal is $... \lambda$.

Light from the Paschen series of a hydrogen atom is able to eject photoelectrons from a metal. Then the work function of the metal is: