The kinetic energy of an emitted electron is $E$ when the light incident on the metal has wavelength $\lambda$. To double the kinetic energy,the incident light must have a wavelength of:

How does the maximum kinetic energy of a photoelectron depend on the frequency of incident radiation? On which factors does it not depend?

Sodium and copper have work functions of $2.3 \ eV$ and $4.5 \ eV$ respectively. The ratio of their threshold wavelengths is approximately .......

Photons of frequencies equal to the frequencies of $H_\beta$ and $H_{\infty}$ lines of hydrogen are incident on a photosensitive plate,whose threshold frequency is equal to the frequency of the $H_\alpha$ line of hydrogen. The ratio of the maximum kinetic energies of the emitted electrons is

$A$ metal plate of area $1 \times 10^{-4} \, m^2$ is illuminated by radiation of intensity $16 \, mW/m^2$. The work function of the metal is $5 \, eV$. The energy of the incident photons is $10 \, eV$ and only $10\%$ of the incident photons produce photoelectrons. The number of emitted photoelectrons per second and their maximum kinetic energy, respectively, will be: $[1 \, eV = 1.6 \times 10^{-19} \, J]$

Electrons ejected from the surface of a metal,when light of a certain frequency is incident on it,are stopped fully by a retarding potential of $3 \ V$. The photoelectric effect on this metallic surface begins at a frequency of $6 \times 10^{14} \ s^{-1}$. The frequency of the incident light in $s^{-1}$ is: [Planck's constant $= 6.4 \times 10^{-34} \ J \cdot s$,charge on the electron $= 1.6 \times 10^{-19} \ C$]