The maximum velocity of the photoelectrons emitted by a metal surface is $9 \times 10^5 \ m/s$. The value of the ratio of charge $(e)$ to mass $(m)$ of the photoelectron is $1.8 \times 10^{11} \ C/kg$. The value of the stopping potential in volts is:

Light of wavelength $5000 \; \mathring A$ falls on a sensitive plate with photoelectric work function of $1.9 \; eV$. The kinetic energy of the photoelectron emitted will be ............ $eV$.

Why can wave theory not explain the change in the kinetic energy of an electron with a change in the frequency of incident light?

The ratio of work functions of two metals $A$ and $B$ is $1:2$. If radiations of frequencies $f$ and $2f$ are incident on the surfaces of $A$ and $B$ respectively,the ratio of the maximum kinetic energies of the emitted photoelectrons will be: (Given $f >$ threshold frequency of $A$ and $2f >$ threshold frequency of $B$)

$A$ metallic surface is illuminated with radiation of wavelength $\lambda$,the stopping potential is $V_0$. If the same surface is illuminated with radiation of wavelength $2 \lambda$,the stopping potential becomes $\frac{V_0}{4}$. The threshold wavelength for this metallic surface will be -

If the threshold wavelength for sodium is $6800 \mathring{A}$,then the work function will be ............. $eV$.