A `25 watt` bulb, which is producing monochromatic light of wavelength `6600Å` is used to illuminated a metal surface. If the surface has `3%` efficiency for photoelectric effect, then the photoelectric current produced in deci ampere is (use `h=6.6xx10^(-34)Js`):

Explanation

No. of photons emitted per second by 25 watt source will be
`n=E/(hc//lambda)=(Elambda)/(hc)`
`=(25xx(6600xx10^(-10)))/((6.6xx10^(-34))xx(3xx10^(8)))=25/(3xx10^(-19))`
Current, `I=3%` of no. of photoelectrons emitted per sec x charge of electron
`=3/100ne=3/100xx25/(3xx10^(-19))xx1.6xx10^(-19)`
`=0.4A=4`deciampere