The maximum kinetic energy of photoelectrons emitted from a photocell is independent of:

When a photon of energy $4.0 \; eV$ strikes the surface of a metal $A$,the ejected photoelectrons have maximum kinetic energy $T_{A} \; eV$ and de-Broglie wavelength $\lambda_{A}$. The maximum kinetic energy of photoelectrons liberated from another metal $B$ by a photon of energy $4.50 \; eV$ is $T_{B} = (T_{A} - 1.5) \; eV$. If the de-Broglie wavelength of these photoelectrons is $\lambda_{B} = 2 \lambda_{A}$,then the work function of metal $B$ is ............. $eV$.

The stopping potential $V_0$ (in $volt$) as a function of frequency $(\nu)$ for a sodium emitter is shown in the figure. The work function of sodium,from the data plotted in the figure,will be: ................. $eV$
(Given: Planck's constant $(h) = 6.63 \times 10^{-34} \, Js$,electron charge $e = 1.6 \times 10^{-19} \, C$)

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

$A$ beam of light of wavelength $\lambda$ falls on a metal having work function $\phi$ placed in a magnetic field $B$. The most energetic electrons, moving perpendicular to the field, are bent in circular arcs of radius $R$. If the experiment is performed for different values of $\lambda$, then the $B^2$ vs. $\frac{1}{\lambda}$ graph will look like (keeping all other quantities constant):

The photoelectric threshold wavelength for silver is $\lambda_{0}$. The energy of the electron ejected from the surface of silver by an incident wavelength $\lambda$ (where $\lambda < \lambda_{0}$) will be: