The work function of a photosensitive material is $4.0 \ eV$. The longest wavelength of light that can cause photon emission from the substance is approximately $...... \ nm$.

When the light source is kept $20 \ cm$ away from a photo cell,a stopping potential of $0.6 \ V$ is obtained. When the source is kept $40 \ cm$ away,the stopping potential will be .......... $V$.

$A$ metal surface is illuminated by light of a given intensity and frequency, causing photoelectric emission. If the intensity of the incident light is reduced to $1/4$ of its original value, what will be the kinetic energy of the emitted photoelectrons?

In the graph given below,if the slope is $4.12 \times 10^{-15} \ V-sec$,then the value of $h$ should be:

If the work function for a certain metal is $3.2 \times 10^{-19} \ J$ and it is illuminated with light of frequency $8 \times 10^{14} \ Hz$,the maximum kinetic energy of the photoelectrons would be in joule:

The threshold frequency of a photosensitive material is $v$. When photons of frequency $2v$ are incident on the material,photoelectrons are emitted with a maximum linear momentum $P$. To get photoelectrons with maximum linear momentum $2P$,the frequency of the incident photons is: (in $v$)