$A$ metal has a work function of $2.5 eV$. If radiation of frequency $3.2 \times 10^{15} Hz$ is incident on this metal surface,then the maximum kinetic energy of the ejected photoelectrons is (Planck's constant,$h = 6.6 \times 10^{-34} J-s$) (in $eV$)

Sodium and copper have work functions of $2.3 \ eV$ and $4.5 \ eV$ respectively. The ratio of their threshold wavelengths is approximately .......

Two identical capacitors are arranged as shown. The work function of plate $1$ is $\phi \ eV$. The $emf$ of the battery is $\frac{\phi}{e}$. If the energy of the incident photon is $hv$,then what is the maximum kinetic energy of the $e^-$ reaching plate $2$?

If the wavelengths of incident radiation are $2500 \ \mathring A$ and $5000 \ \mathring A$ respectively,and the work function of the metal surface is $2 \ eV$,find the approximate ratio of the stopping potentials for the emitted photoelectrons. (in $:1$)

In the photoelectric effect,if the intensity of light is doubled,then the maximum kinetic energy of the photoelectrons will become:

Assertion : When ultraviolet light is incident on a photocell,its stopping potential is $V_0$ and the maximum kinetic energy of the photoelectrons is $K_{max}$. When the ultraviolet light is replaced by $X-$ rays,both $V_0$ and $K_{max}$ increase.
Reason : Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.