$A$ monochromatic light source of intensity $5 \, mW$ emits $8 \times 10^{15}$ photons per second. This light ejects photoelectrons from a metal surface. The stopping potential for this setup is $2.0 \, V$. The work function of the metal will be ............ $eV$. (in $.9$)

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:

$A$ photon of energy $8 \ eV$ is incident on a metal surface of threshold frequency $1.6 \times 10^{15} \ Hz$. The maximum kinetic energy of the photoelectrons emitted (in $eV$) is: (Take $h = 6 \times 10^{-34} \ J \cdot s$ and $1 \ eV = 1.6 \times 10^{-19} \ J$)

In a photoemissive cell with exciting wavelength $\lambda$,the fastest electron has speed $v$. If the exciting wavelength is changed to $\frac{3\lambda}{4}$,the speed of the fastest emitted electron will be

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:

In a photoelectric experiment,ultraviolet light of wavelength $280 \, nm$ is used with a lithium cathode having a work function $\phi = 2.5 \, eV$. If the wavelength of incident light is switched to $400 \, nm$,find out the change in the stopping potential (in $V$).
$(h = 6.63 \times 10^{-34} \, J \cdot s, c = 3 \times 10^8 \, m/s)$