The work functions of metal $A$ and $B$ are in the ratio $1: 2$. If light of frequency $f$ and $2f$ is incident on surface $A$ and $B$ respectively,then the ratio of the maximum kinetic energies of the emitted photoelectrons is:

Light of frequency $4\nu_0$ is incident on a metal surface with threshold frequency $\nu_0$. The maximum kinetic energy of the emitted photoelectrons is:

$A$ photon incident on a metal of work function $2 \text{ eV}$ produces photoelectrons of maximum kinetic energy of $2 \text{ eV}$. The wavelength associated with the photon is: (in $\text{ Å}$)

When a photon of energy $4.25 \, eV$ strikes the surface of a metal $A$,the ejected photoelectrons have a maximum kinetic energy $T_A \, eV$ and de-Broglie wavelength ${\lambda _A}$. The maximum kinetic energy of photoelectrons liberated from another metal $B$ by a photon of energy $4.70 \, eV$ is $T_B = (T_A - 1.50) \, eV$. If the de-Broglie wavelength of these photoelectrons is ${\lambda _B} = 2{\lambda _A}$,then:

The figure shows different graphs between stopping potential $(V_0)$ and frequency $(\nu)$ for photosensitive surfaces of cesium,potassium,sodium,and lithium. The plots are parallel. The correct ranking of the targets according to their work function,with the greatest first,is:

When a photosensitive metal surface is illuminated with radiation of wavelength $\lambda_1$,the stopping potential is $V_1$. If the same surface is illuminated with radiation of wavelength $3\lambda_1$,the stopping potential is $\frac{V_1}{6}$. The threshold wavelength for the photosensitive metal surface is: