In a photoelectric effect experiment,the threshold wavelength of incident light is $380 \, nm$. If the wavelength of the incident light is $260 \, nm$,the maximum kinetic energy of emitted electrons will be: .............. $eV$. Given $E \text{ (in } eV) = \frac{1237}{\lambda \text{ (in } nm)}$.

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

In a photoelectric experiment,a graph is drawn with stopping potential along $Y$-axis and the frequency of the incident light along $X$-axis. If the graph is a straight line which makes an angle $\theta$ with $Y$-axis,then $\tan \theta=$ ($h$-Planck's constant,$e$-charge of electron).

$A$ 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?

$A$ photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $\frac{\lambda}{2}$. If the maximum kinetic energy of the emitted photoelectrons in the first case is one-third that in the second case,the work function of the surface of the material is ($c=$ speed of light,$h=$ Planck's constant).

The maximum kinetic energy of the photo-electrons depends only on