When a light of wavelength $4900 Å$ falls on a photosensitive metal, a negative $2 \,V$ potential is required to stop the emitted electrons. Then, the work-function of the material is nearly (given charge on electron $= 1.602 \times 10^{-19} C$ and Planck's constant $= 6.625 \times 10^{-34} Js$) (in $eV$)

$A$ metal surface is illuminated by light of two different wavelengths $248 \ nm$ and $310 \ nm$. The maximum speeds of the photoelectrons corresponding to these wavelengths are $u_1$ and $u_2$,respectively. If the ratio $u_1: u_2 = 2: 1$ and $hc = 1240 \ eV \ nm$,the work function of the metal is nearly: (in $eV$)

When a photosensitive surface is illuminated with light of wavelength $\lambda$,the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2\lambda$,the stopping potential is $V/3$. The threshold wavelength for the surface is:

Assertion : In the process of photoelectric emission, all emitted electrons do not have the same kinetic energy.
Reason : If radiation falling on the photosensitive surface of a metal consists of different wavelengths, then the energy acquired by electrons absorbing photons of different wavelengths shall be different.

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 :

The work function of a metal is $2.1 \text{ eV}$. Which of the following wavelengths will be able to emit photoelectrons from its surface?