The maximum kinetic energy of photoelectrons emitted from a photocell is independent of:

Radiation of two photon energies,twice and five times the work function of a metal,are incident successively on the metal surface. The ratio of the maximum velocity of photoelectrons emitted in the two cases is:

In an experiment on photoelectric emission from a metallic surface,the wavelength of incident light is $2 \times 10^{-7} \,m$ and the stopping potential is $2.5 \,V$. The threshold frequency of the metal (in $Hz$) is approximately (charge of electron $e=1.6 \times 10^{-19} \,C$,Planck's constant $h=6.6 \times 10^{-34} \,J-s$):

$A$ metal is exposed to light of wavelength $800 \, nm$ and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of the photoelectrons doubles when light of wavelength $500 \, nm$ is used. The work function of the metal is $... \, eV$ (Take $hc = 1230 \, eV \cdot nm$).

The threshold wavelength for photoelectric emission from a material is $5500\,\mathring A$. Photoelectrons will be emitted when this material is illuminated with monochromatic radiation from a:
$A.$ $75\,W$ infra-red lamp
$B.$ $10\,W$ infra-red lamp
$C.$ $75\,W$ ultra-violet lamp
$D.$ $10\,W$ ultra-violet lamp
Choose the correct answer from the options given below:

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.