The photoelectric cut-off voltage in a certain experiment is $1.5 \ V$. What is the maximum kinetic energy of the emitted photoelectrons?

In an experiment of photoelectric effect,the stopping potential was measured to be $V_{1}$ and $V_{2}$ with incident light of wavelength $\lambda$ and $\frac{\lambda}{2}$ respectively. The relation between $V_{1}$ and $V_{2}$ is:

$A$ photosensitive metallic surface is illuminated alternately with lights of wavelength $3100 \mathring A$ and $6200 \mathring A$. It is observed that the maximum speeds of the photoelectrons in the two cases are in the ratio $2:1$. The work function of the metal is $(hc = 12400 \, eV \mathring A)$.

The work function of caesium is $2.14\, eV$. Find the wavelength of the incident light if the photo current is brought to zero by a stopping potential of $0.60\, V$. (Result in $nm$)

In the graph of maximum kinetic energy of photoelectrons emitted from a metal surface versus the frequency of incident radiation based on Einstein's photoelectric equation,the slope is .......

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