Ultraviolet light of wavelength $300 \ nm$ and intensity $1.0 \ W/m^2$ is incident on a photosensitive surface. If only $1 \%$ of the incident photons emit photoelectrons,calculate the number of photoelectrons emitted per second from a surface area of $1 \ cm^2$.

  • A
    $9.61 \times 10^{14} \ s^{-1}$
  • B
    $4.12 \times 10^{13} \ s^{-1}$
  • C
    $1.51 \times 10^{12} \ s^{-1}$
  • D
    $2.13 \times 10^{11} \ s^{-1}$

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Similar Questions

In a photoelectric experiment,three different lights are incident on a metal with a work function of $1.5 \ eV$. Light $A$ has a wavelength of $200 \ nm$ and an intensity of $1.8 \ W/m^2$,light $B$ has a wavelength of $400 \ nm$ and an intensity of $1 \ W/m^2$,and light $C$ has a wavelength of $600 \ nm$ and an intensity of $0.5 \ W/m^2$. The photocurrent versus voltage is measured. Which graphs correspond to which light?

The maximum kinetic energy of a photoelectron liberated from the surface of lithium with work function $2.35 \text{ eV}$ by electromagnetic radiation whose electric component varies with time as: $E=a[1+\cos(2 \pi f_1 t)] \cos(2 \pi f_2 t)$ (where $a$ is a constant) is (given $f_1=3.6 \times 10^{15} \text{ Hz}$,$f_2=1.2 \times 10^{15} \text{ Hz}$ and Planck's constant $h=6.6 \times 10^{-34} \text{ Js}$) (in $\text{ eV}$)

In a photocell circuit,the stopping potential $V_0$ is a measure of the maximum kinetic energy of the photoelectrons. The following graph shows experimentally measured values of stopping potential versus frequency $\nu$ of incident light. The values of Planck's constant and the work function as determined from the graph are (taking the magnitude of electronic charge to be $e = 1.6 \times 10^{-19} \, C$):

$A$ 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 emitted photoelectrons is ............... $eV$ (Given $h = 6.6 \times 10^{-34} Js$).

$A$ metal surface is illuminated by light of a given intensity and frequency, causing photoelectric emission. If the intensity of the incident light is reduced to $1/4$ of its original value, what will be the kinetic energy of the emitted photoelectrons?

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