The threshold frequency of a metal is $f_0$. When light of frequency $2f_0$ is incident on the metal plate, the maximum velocity of the photoelectrons is $v_1$. When the frequency of the incident radiation is increased to $5f_0$, the maximum velocity of the photoelectrons emitted is $v_2$. The ratio of $v_1$ to $v_2$ is

When light of frequency $v$ is incident on two metallic plates $A$ and $B$,photoelectrons are emitted. If the work function of $A$ is greater than that of $B$,which of the following curves correctly represents the relationship between the stopping potential $V$ and the incident frequency $v$?

When light of wavelength $\lambda$ is incident on a photosensitive surface,the stopping potential is $V$. When light of wavelength $3 \lambda$ is incident on the same surface,the stopping potential is $\frac{V}{6}$. Then the threshold wavelength for the surface is:

The maximum kinetic energies of photoelectrons emitted are $K_1$ and $K_2$ when lights of wavelengths $\lambda_1$ and $\lambda_2$ are incident on a metallic surface. If $\lambda_1 = 3 \lambda_2$,then:

$A$ ray of light with wavelength $\lambda$ is incident on three different photoelectric cells namely $1$, $2$ and $3$. The threshold wavelengths of these photoelectric cells are $\lambda_1$, $\lambda_2$ and $\lambda_3$, respectively, and the magnitudes of the stopping potentials of these cells are $V_1$, $V_2$ and $V_3$, respectively. The relation between $\lambda$ and the threshold wavelengths is $\lambda_1 < \lambda$, $\lambda_2 > \lambda$ and $\lambda_3 >> \lambda$. The correct option is:

When a metallic surface is illuminated with monochromatic light of wavelength $\lambda$,the stopping potential is $5V_0$. When the same surface is illuminated with light of wavelength $3\lambda$,the stopping potential is $V_0$. Then the work function of the metallic surface is: