If the work function of a metal is $\phi$ and the frequency of the incident light is $v$,there is no emission of photoelectrons for:

If light of wavelength $\lambda_1$ is allowed to fall on a metal,then the kinetic energy of the photoelectrons emitted is $E_1$. If the wavelength of light changes to $\lambda_2$,then the kinetic energy of the electrons changes to $E_2$. Then the work function of the metal is:

$A$ light whose frequency is equal to $6 \times 10^{14} \, Hz$ is incident on a metal whose work function is $2 \, eV$. $[h = 6.63 \times 10^{-34} \, Js, 1 \, eV = 1.6 \times 10^{-19} \, J]$. The maximum kinetic energy of the emitted electrons will be ............ $eV$. (in $.49$)

Light of two different frequencies whose photons have energies $1 \text{ eV}$ and $2.5 \text{ eV}$ respectively,successively illuminate a metallic surface whose work function is $0.5 \text{ eV}$. The ratio of the maximum speeds of the emitted electrons will be:

Photoelectric effect can be explained by

The work functions for sodium and copper are $2 \ eV$ and $4 \ eV$, respectively. Which of them is suitable for a photocell using $4000 \ \mathring{A}$ light?