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

According to Einstein's photoelectric equation,the graph between the kinetic energy of photoelectrons ejected and the frequency of incident radiation is

When a metallic surface is illuminated by light of wavelength $\lambda$,the stopping potential for the photoelectric current is $3 \text{ V}$. When the same surface is illuminated by light of wavelength $2 \lambda$,the stopping potential is $1 \text{ V}$. The threshold wavelength for this surface is:

Light of frequency $4v_0$ is incident on a metal surface with a threshold frequency $v_0$. The maximum kinetic energy of the emitted photoelectrons is:

According to Einstein's photoelectric equation, the graph of the kinetic energy of the emitted photoelectrons versus the frequency of incident radiation gives a straight line whose slope

In the experiment of $P.E.E.$,the $KE_{max}$ of an electron is $K_0$. If the frequency is increased by a factor of $n_1$,then the $KE_{max}$ becomes $n_2K_0$. Find the work function.