$A$ metal surface of work function $1.13 eV$ is irradiated with light of wavelength $310 nm$. The retarding potential required to stop the escape of photoelectrons is [Take $hc = 1240 eV \cdot nm$] (in $V$)

$A$ certain metallic surface is illuminated by monochromatic radiation of wavelength $\lambda$. The stopping potential for photoelectric current for this radiation is $3V_{0}$. If the same surface is illuminated with a radiation of wavelength $2\lambda$,the stopping potential is $V_{0}$. The threshold wavelength of this surface for the photoelectric effect is $n\lambda$. Find the value of $n$.

When a photon of energy $hv$ is incident on an aluminum plate with a work function $E_0$,the maximum kinetic energy of the emitted photoelectrons is $K$. If the frequency of the incident radiation is doubled,the maximum kinetic energy of the emitted photoelectrons will be .......

When a metallic surface is illuminated with a radiation of wavelength $\lambda$, the stopping potential is $V$. If the same surface is illuminated with radiation of wavelength $6\lambda$, the stopping potential is $V/12$. The threshold wavelength for the surface is

When a photon of energy $4.0 \; eV$ strikes the surface of a metal $A$,the ejected photoelectrons have 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 $4.50 \; eV$ is $T_{B} = (T_{A} - 1.5) \; eV$. If the de-Broglie wavelength of these photoelectrons is $\lambda_{B} = 2 \lambda_{A}$,then the work function of metal $B$ is ............. $eV$.

Mercury violet light $(\lambda = 4558 \mathring{A})$ is falling on a photosensitive material $(\phi = 2.5 \text{ eV})$. The speed of the ejected electrons is,in $\text{m/s}$,about: