The photoelectric work function for a metal surface is $4.125 \ eV$. The cut-off wavelength for this surface is .......... $\mathring{A}$.

Let $v$ be the frequency of the incident photon and $v_0$ be the threshold frequency. The condition for the emission of photoelectrons is:

The work functions of Silver and Sodium are $4.6 \ eV$ and $2.3 \ eV$,respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is :

Light of wavelength '$\lambda$' which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with some velocity, then the stopping potential:

The threshold wavelength of a metal surface is $5 \times 10^{-10} \, m$. When it is illuminated by light of wavelength $2 \times 10^{-10} \, m$,the stopping potential is $V_0$. What will be the stopping potential if the wavelength of the incident light is doubled?

$A$ 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 $\,s$)