The work function of a photoelectric material is $3.3 \text{ eV}$. The threshold frequency will be equal to

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 difference between threshold wavelengths for two metal surfaces $A$ and $B$ having work functions $\phi_A = 9 \, eV$ and $\phi_B = 4.5 \, eV$ in $nm$ is: (Given,$hc = 1242 \, eV \, nm$)

$A$ photon of $5.5 \ eV$ energy falls on the surface of a metal,emitting photoelectrons with a maximum kinetic energy of $4.0 \ eV$. The stopping voltage required for these electrons is .......... $V$.

Light of frequency $7.21 \times 10^{14} \; Hz$ is incident on a metal surface. Electrons with a maximum speed of $6.0 \times 10^{5} \; m/s$ are ejected from the surface. What is the threshold frequency for photoemission of electrons?

Photoelectric effect is observed on a surface for frequencies $5.5 \times 10^8 \text{ MHz}$ and $4.5 \times 10^8 \text{ MHz}$. If the ratio of the maximum kinetic energy of the emitted photoelectrons is $1:5$,then the threshold frequency for the metal surface is: