Photoelectric emission occurs only when the incident light has more than a certain minimum

Ultraviolet light of wavelength $300 \ nm$ and intensity $1.0 \ W/m^2$ is incident on a photosensitive surface. If only $1 \%$ of the incident photons emit photoelectrons,calculate the number of photoelectrons emitted per second from a surface area of $1 \ cm^2$.

When a photon of energy $4.25 \, eV$ strikes the surface of a metal $A$,the ejected photoelectrons have a maximum kinetic energy $T_A \, eV$ and de-Broglie wavelength ${\lambda _A}$. The maximum kinetic energy of photoelectrons liberated from another metal $B$ by a photon of energy $4.70 \, eV$ is $T_B = (T_A - 1.50) \, eV$. If the de-Broglie wavelength of these photoelectrons is ${\lambda _B} = 2{\lambda _A}$,then:

In a photoelectric experiment,three different lights are incident on a metal with a work function of $1.5 \ eV$. Light $A$ has a wavelength of $200 \ nm$ and an intensity of $1.8 \ W/m^2$,light $B$ has a wavelength of $400 \ nm$ and an intensity of $1 \ W/m^2$,and light $C$ has a wavelength of $600 \ nm$ and an intensity of $0.5 \ W/m^2$. The photocurrent versus voltage is measured. Which graphs correspond to which light?

On a photosensitive material,when the frequency of incident radiation is increased by $30 \%$,the kinetic energy of emitted photoelectrons increases from $0.4 \ eV$ to $0.9 \ eV$. The work function of the surface is (in $eV$)

Ultraviolet radiations of $6.2 \, eV$ fall on an aluminium surface (work function $4.2 \, eV$). The kinetic energy in joules of the fastest electron emitted is approximately: