For intensity $I$ of a light of wavelength $5000 \, Å$, the photoelectron saturation current is $0.40 \, μA$ and the stopping potential is $1.36 \, V$. The work function of the metal is ........... $eV$.

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 ............. $eV$.

Maximum velocity of the photoelectron emitted by a metal is $1.8 \times 10^{6} \ m/s$. Take the value of specific charge of the electron as $1.8 \times 10^{11} \ C/kg$. Then the stopping potential in volt is

According to Einstein's photoelectric equation,the graph of the maximum kinetic energy of emitted photoelectrons versus the frequency of incident radiation is a straight line. Its slope . . . . . .

Let $v$ be the frequency of the incident photon and $v_0$ be the threshold frequency. The condition for the emission of photoelectrons is:

$A$ metal surface is illuminated by light of a given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one-fourth of its original value, then the maximum kinetic energy of the emitted photoelectrons would become