When photons of energy $hf$ fall on an aluminium plate (of work function $E_0$),photoelectrons of maximum kinetic energy $K$ are ejected. If the frequency of radiation is doubled,the maximum kinetic energy of the ejected photoelectrons will be:

$A$ 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:

The figure represents the graph of photocurrent $I$ versus applied voltage $(V)$. The maximum kinetic energy of the emitted photoelectrons is:

The work function of a metal is $1.6 \times 10^{-19} \, J$. When the metal surface is illuminated by light of wavelength $6400 \, \mathring{A}$,the maximum kinetic energy of the emitted photo-electrons will be. (Planck's constant $h = 6.4 \times 10^{-34} \, Js$)

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:

Photoelectric emission is observed from a metallic surface for frequencies $\nu_1$ and $\nu_2$ of the incident light rays $(\nu_1 > \nu_2)$. If the ratio of the maximum kinetic energy of the photoelectrons emitted in the first case to that in the second case is $3 : K$, then the threshold frequency of the metallic surface is: