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Light rays of wavelengths 6000 \AA and of photon intensity 39.6 watt/m2 is incident on a metal surface. If only one percent of photons incident on surface emit photoelectrons, then the number of electrons emitted per second per unit area from the surface will be : (h = 6.64 \times 10^{-34} J-s, velocity of light = 3 \times 10^8 m/s )

Option: 1

12\times10^{18}


Option: 2

10\times10^{18}


Option: 3

12\times10^{17}


Option: 4

12\times10^{16}


Answers (1)

best_answer

\begin{aligned} & \mathrm{I}=\frac{\mathrm{nhc}}{\mathrm{At} \lambda} \\ & 39.6=\frac{\mathrm{n} \times 6.6 \times 10^{-34} \times 3 \times 10^8}{1 \times 1 \times 6000 \times 10^{-10}} \end{aligned}

n=12\times10^{17}

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