One milliwatt of light of wavelength $4560 \ \mathring{A}$ is incident on a cesium surface of work function $1.9 \ eV$. Given that the quantum efficiency of photoelectric emission is $0.5\%$. Planck's constant $h = 6.62 \times 10^{-34} \ J \cdot s$,velocity of light $c = 3 \times 10^8 \ m/s$,the corresponding photoelectric current is:

  • A
    $1.856 \times 10^{-6} \ A$
  • B
    $1.856 \times 10^{-7} \ A$
  • C
    $1.856 \times 10^{-5} \ A$
  • D
    $1.856 \times 10^{-4} \ A$

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