Light rays of wavelength `6000A^(@)` and of photon intensity `39.6Wm^-2` is incident on a metal surface. If only one percent of photons incident on the surface of electrons emitted per second unit area from the surface will be [Planck constant =`6.64xx10^(-34) J-S`,Velocity of light =`3xx10^(8) ms^(-1)`]

Options

A. `12xx 10^(18) `
B. `10 xx 10^(18) `
C. `12xx 10^(7) `
D. `12xx 10^(16) `

Answer

C

Explanation

Useful intensity for the emission of electron is
I . = 1% of `1/100 xx 39.6 xx = 0.396 " watt" //m^2`
Energy of each photon ` = (hc)/lamda`
` = ((6.64 xx 10^(-34))xx (3xx10^8))/(600x 10^(-10)) = 3.23 xx 10^(-19)J`
No. of photoelectrons emitted per second photoelectric equation
`= (0.396 ) /(3.32 xx 10 ^(-19)) ~~ 12 xx 10^(17)`