Two photons with energies three times and nine times the work function of a metal surface are incident on it. What is the ratio of the maximum velocities of the photoelectrons emitted in the two cases?

$A$ material $A$ exhibits the photoelectric effect. Its work function is $2.5 \ eV$ and the threshold wavelength is $\lambda$. Another material $B$ has a work function of $5 \ eV$. Find the threshold wavelength required to produce the photoelectric effect in $B$.

The wave equation of an electric field at a point is $E = 100 \frac{V}{m} [\sin(7 \omega t) + \cos(10 \omega t) + \cos(15 \omega t)]$ at instant $t$. If the work function of the photocell is $\phi$,then the stopping potential is:

If the frequency of incident radiation $(v)$ is increased,keeping other factors constant,the stopping potential ($v > v_0$,threshold frequency) will:

$A$ 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:

In the photoelectric effect, the stopping potential depends on: