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

The maximum kinetic energy of emitted photoelectrons depends on

$A$ light source is placed at a distance of $20 \ cm$ from a photocell,and the measured stopping potential is $0.6 \ V$. If the distance of the source is changed to $40 \ cm$,what will be the stopping potential in $V$?

In an experiment on the photoelectric effect,the wavelength of the incident radiation is $\lambda$. The wavelength of the incident radiation is reduced to $\frac{1}{3}$ of the initial value,and the maximum kinetic energy of the photoelectron is observed to be $n$ times the previous value. The threshold wavelength for the metal plate is:

When a photon of wavelength $3000 \ \mathring{A}$ is incident on a metal surface,the energy of the emitted electron is $0.5 \ eV$. If a photon of wavelength $2000 \ \mathring{A}$ is incident on the same metal surface,the energy of the emitted electron will be:

In the photoelectric effect,a graph is plotted between the maximum kinetic energy of electrons and the frequency of incident photons. The slope of this curve will be .......