Find the number of photons emitted per second by a 25 W source of monochromatic light of wavelength `6600 Å`. What is the photoelectric current assuming 3% efficiency for photoelectric effect. Given `h = 6.6 xx 10^(-34) Js`.
Correct Answer - `1xx10^(16) Hz`
For frequency `2.5xx10^(16) Hz, h v=hv_(o)+KE_(max)`
`:. h(2.5xx10^(16))=hv_(o)+KE_(max) ...(i)`
For frequency `4xx10^(16)` Hz,
`h(4xx10^(16))=hv_(o)+2KE_(max) ...(2)`
Multiply `eq^(n)` (1) by `2` and substract `eq^(n)` (2) from it....
Correct Answer - 4
The number pf photon is `N=E/(hv)=(PDeltat)/(h(c//lambda))=(lambdaPDeltat)/(hc)`
Substitution of the data gives
`N=((5.60xx10^(-7)m)xx(100Js^(-1))xx(1.0s))/((6.626xx10^(-36)s)xx(3xx10^(8)ms^(-1)))=2.8xx10^(20)`
Correct Answer - B
`E=n(hc)/(lambda)`
`10^(-7)= (nxx(6.626 xx 10^(-34)Js)(3xx 10^(8) m//s))/(5000xx10^(-10)m)`
` n = 2.5 xx 10^(11)` photons per second