Light of frequency $4\nu_0$ is incident on a metal surface with threshold frequency $\nu_0$. The maximum kinetic energy of the emitted photoelectrons is:

An isolated lead ball is charged upon continuous irradiation by $EM$ radiation of wavelength, $\lambda = 221 \,nm$. The maximum potential attained by the lead ball, if its work function is $4.14 \,eV$, is (take, $h = 6.63 \times 10^{-34} \,J \cdot s$, $c = 3 \times 10^8 \,m/s$, $e = 1.6 \times 10^{-19} \,C$): (in $\,V$)

Photoemission from a metal surface takes place for frequencies $v_1$ and $v_2$ of incident rays. If the ratio of the maximum kinetic energy of photoelectrons is $1:K$,then the threshold frequency of the metallic surface is

When a certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$,the stopping potential for photoelectric current is $4 V_0$. When the same surface is illuminated with light of wavelength $3 \lambda$,the stopping potential is $V_0$. The threshold wavelength for this surface for the photoelectric effect is

Assertion : Photosensitivity of a metal is high if its work function is small.
Reason : Work function $= hf_0$ where $f_0$ is the threshold frequency.

For a photosensitive material, the work function is $W_0$ and the stopping potential is $V$. What is the wavelength of the incident radiation? ($h$ = Planck's constant, $c$ = velocity of light, $e$ = electronic charge)