If the work function of a metal is $3 \ eV$,then the threshold wavelength will be ............ $\mathring{A}$.

$A$ photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $\frac{\lambda}{2}$. If the maximum kinetic energy of the emitted photoelectrons in the first case is one-fourth that in the second case,the work function of the surface of the material is ($c=$ speed of light,$h=$ Planck's constant).

The ratio of work functions of two metals is $1:2$. If light of frequencies $f$ and $2f$ are incident on them respectively,what is the ratio of the maximum kinetic energy of the emitted photoelectrons?

According to Einstein's photoelectric equation,the graph of the maximum kinetic energy of emitted photoelectrons versus the frequency of incident radiation is a straight line. Its slope . . . . . .

Let $K_1$ be the maximum kinetic energy of photoelectrons emitted by light of wavelength $\lambda_1$ and $K_2$ be the maximum kinetic energy corresponding to wavelength $\lambda_2$. If $\lambda_1 = 2\lambda_2$,then:

$A$ cobalt $(Co)$ plate is placed at a distance of $1 \,m$ from a point source of power $1 \,W$. Assume a circular area of the plate of radius $r = 1 \,Å$ is exposed to the radiation and ejects photoelectrons. The light energy is considered to be spread uniformly and the work function of cobalt is $5 \,eV$. The minimum time the target should be exposed to the light source to eject a photoelectron (assuming no reflection losses) is: (in $\,s$)