One milliwatt of light of wavelength $4560\ \mathring A $ is incident on a cesium surface of work function $1.9\ eV$. Given that quantum efficiency of photoelectric emission is $0.5\%$. Planck's constant $h = 6.62 \times 10^{-34}\ J-s$, velocity of light $= 3 \times 10^8\ ms^{-1}$, the corresponding photo electric current is
- A$1.856 × 10^{-6}\ A$
- B$1.856 × 10^{-7}\ A$
- C$1.856 × 10^{-5}\ A$
- D$1.856 × 10^{-4}\ A$
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