If the work function of a photometal is $6.825 \; eV$,its threshold wavelength will be ............. $\mathring A$ $(c = 3 \times 10^8 \; m/s)$.

Light comprising three wavelengths $310 \ nm$,$455 \ nm$,and $620 \ nm$ is incident on a surface separating two media at an angle of $45^o$. The refractive index for the $455 \ nm$ light is $\sqrt{2}$. If a metal plate with a work function of $1.2 \ eV$ is placed in the second medium,what is the maximum kinetic energy of the electrons emitted from the metallic plate in $eV$?

Radiation of monochromatic waves with a wavelength of $400 \ nm$ is incident on the surfaces of $Zn$,$Fe$,and $Ni$ metals,which have work functions of $3.4 \ eV$,$4.8 \ eV$,and $5.9 \ eV$ respectively. (Take $hc = 1242 \ eV \ nm$)
$(a)$ The maximum $KE$ of photoelectrons emitted from any metal surface is $0.3 \ eV$.
$(b)$ No photoelectrons are emitted from the surface of $Ni$.
$(c)$ If the frequency of the radiation source is doubled,the $KE$ of the photoelectrons also doubles.
$(d)$ If the wavelength of the incident radiation is less than $200 \ nm$,photoelectrons will be emitted from the surfaces of all three metals.
The correct statements are:

How does the maximum kinetic energy of a photoelectron depend on the frequency of incident radiation? On which factors does it not depend?

Light of frequency $v$ is incident on a certain photoelectric material with a threshold frequency $v_0$. What is the work function for the material?

When a photon of energy $4.0 \; eV$ strikes the surface of a metal $A$,the ejected photoelectrons have 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 $4.50 \; eV$ is $T_{B} = (T_{A} - 1.5) \; eV$. If the de-Broglie wavelength of these photoelectrons is $\lambda_{B} = 2 \lambda_{A}$,then the work function of metal $B$ is ............. $eV$.