The stopping potential as a function of frequency of incident radiation is plotted for two different photoelectric surfaces $A$ and $B$. The graph shows that the work function of $A$ is

When light of frequency $v_{1}$ is incident on a metal with work function $W$ (where $h v_{1} > W$), then the photocurrent falls to zero at a stopping potential of $V_{1}$. If the frequency of light is increased to $v_{2}$, the stopping potential changes to $V_{2}$. Therefore, the charge of an electron $e$ is given by:

Light of wavelength $3500 \ \mathring A$ is incident on two metals $A$ and $B$. If their work functions are $4.2 \ eV$ and $1.9 \ eV$ respectively,which metal will emit photoelectrons?

The photoelectric effect was successfully explained first by

When a certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$,the stopping potential for photoelectric current is $4 V_0$. When the same surface is illuminated with light of wavelength $3 \lambda$,the stopping potential is $V_0$. The threshold wavelength for this surface for the photoelectric effect is

The threshold frequency for a certain metal is $v_0$. When radiation of frequency $2v_0$ is incident on this metal surface,the maximum velocity of the emitted photoelectrons is $2 \times 10^6 \ m/s$. If radiation of frequency $3v_0$ is incident on the same metal surface,the maximum velocity of the emitted photoelectrons is . . . . . .