The threshold wavelengths for photoelectric emission from two metals $A$ and $B$ are $400 \ nm$ and $800 \ nm$ respectively. The ratio of their work functions,$\phi_{A} : \phi_{B}$ is:

When radiation of the wavelength $\lambda$ is incident on a metallic surface,the stopping potential is $4.8 \ V$. If the same surface is illuminated with radiation of double the wavelength,then the stopping potential becomes $1.6 \ V$. Then,the threshold wavelength for the surface is :

When photons of energy $1 \ eV$ and $2.5 \ eV$ are incident on a metal surface with a work function of $0.5 \ eV$,what is the ratio of the maximum kinetic energies of the emitted photoelectrons?

$A$ photon of energy $3.4 \ eV$ is incident on a metal having work function $2 \ eV$. The maximum kinetic energy $(K.E.)$ of photo-electrons is equal to ........... $eV$. (in $eV$)

The electric field of a light wave is given as $\vec E = 10^{-3} \cos \left( \frac{2\pi x}{5 \times 10^{-7}} - 2\pi \times 6 \times 10^{14} t \right) \hat x \, N/C$. This light falls on a metal plate with a work function of $2 \, eV$. The stopping potential of the photoelectrons is ................ $V$.

When a photosensitive surface is irradiated by light of wavelength $\lambda_1$ and $\lambda_2$,the maximum kinetic energies of the emitted photoelectrons are $E_1$ and $E_2$ respectively. The work function of the photosensitive surface is: