On a photosensitive material,when the frequency of incident radiation is increased by $20 \%$,the maximum kinetic energy of emitted photoelectrons increases from $0.4 \ eV$ to $0.7 \ eV$. The work function of the material is (in $eV$)

Photoelectric emission takes place from a certain metal at threshold frequency $\nu$. If the radiation of frequency $2\nu$ is incident on the metal plate, the maximum velocity of the emitted photoelectron will be ($m = \text{mass of electron}$, $h = \text{Planck's constant}$)

If a photocell is illuminated with a radiation of $1240 \, Å$, the stopping potential is found to be $8 \, V$; then the work function of the emitter and the threshold wavelength are

Photoelectric effect can be explained by

When light of a given wavelength is incident on a metallic surface,the minimum potential needed to stop the emitted photoelectrons is $6.0 \ V$. This potential drops to $0.6 \ V$ if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal,respectively? $\left[\text{Take } hc = 1.24 \times 10^{-6} \ eV \ m\right]$

The threshold wavelengths for photoelectric emission from two metals $A$ and $B$ are $400 \ nm$ and $800 \ nm$ respectively. The ratio of their work functions,$\phi_{A} : \phi_{B}$ is: