The magnetic field associated with a light wave is given,at the origin,by $B = B_0 [\sin(3.14 \times 10^7 ct) + \sin(6.28 \times 10^7 ct)]$. If this light falls on a silver plate having a work function of $4.7 \ eV$,what will be the maximum kinetic energy of the photoelectrons? (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:

The ratio of work functions of two metals is $1:2$. If light of frequencies $f$ and $2f$ are incident on them respectively,what is the ratio of the maximum kinetic energy of the emitted photoelectrons?

In the following diagram,if $V_2 > V_1$,then:

When light falls on a metal surface,the maximum kinetic energy of the emitted photo-electrons depends upon

In an experiment on the photoelectric effect,the slope of the cut-off voltage versus frequency of incident light is found to be $4.12 \times 10^{-15} \; V \cdot s$. Calculate the value of Planck's constant.