If ultraviolet radiation of $6.2 eV$ falls on an aluminium surface,then the kinetic energy of the fastest emitted electron is (work-function $= 4.2 eV$)

All electrons ejected from a metallic surface by incident light of wavelength $400 \,nm$ travelled $1 \,m$ in the direction of a uniform electric field of $2 \,N/C$ and came to rest. The work function of the surface is (in $\,eV$)

$A$ certain metallic surface is illuminated by monochromatic radiation of wavelength $\lambda$. The stopping potential for photoelectric current for this radiation is $3V_{0}$. If the same surface is illuminated with a radiation of wavelength $2\lambda$,the stopping potential is $V_{0}$. The threshold wavelength of this surface for the photoelectric effect is $n\lambda$. Find the value of $n$.

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays $(v_1 > v_2)$. If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of $1:k$,then the threshold frequency of the metallic surface is

Light of two different frequencies whose photons have energies $1.3 eV$ and $2.8 eV$ respectively,successfully illuminate a metallic surface whose work function is $0.8 eV$. The ratio of maximum speeds of emitted electrons will be

Light of wavelength $\lambda$ strikes a photoelectric surface and electrons are ejected with kinetic energy $K$. If $K$ is to be increased to twice its original value,the wavelength must be changed to $\lambda'$ such that :-