The work function of a metal is $4.2 \ eV$. Its threshold wavelength will be .......... $ \mathring A $.

In a photoelectric experiment,a graph of maximum kinetic energy $(KE_{\max})$ against the frequency of incident radiation $(\nu)$ is plotted. If $A$ and $B$ are the intercepts on the $X$ and $Y$ axis respectively,the Planck's constant is given by:

When a photon of energy $hv$ is incident on an aluminum plate (work function $E_0$),photoelectrons with maximum kinetic energy $K$ are emitted. What will be the maximum kinetic energy of the emitted photoelectrons when a photon of energy $2hv$ is incident on the same aluminum plate?

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

An electromagnetic wave of wavelength $\lambda$ is incident on a photosensitive surface of negligible work function. If the photoelectron emitted from the surface has mass $m$ and de-Broglie wavelength $\lambda_{d}$,then

$A$ metal surface of work function $1.13 eV$ is irradiated with light of wavelength $310 nm$. The retarding potential required to stop the escape of photoelectrons is [Take $hc = 1240 eV \cdot nm$] (in $V$)