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A light source emitting three wavelength 5000Å, 6000Å, 7000Å has a total power of 10^(-3) m. W and the beam of diameter2xx10^(-3)m. The power density is distributed equally amongst the three wavelengths. The beam shines normally on a metallic surface of area 10^(4)m^(2) which has a work function of 1.9 eV. Assuming that each proton liberates an electron, calculate the charge emitted per unit area in one second : |
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Answer» 94 C `W=(hc)/(lambda_(0)) :. Lambda_(0)=6513Å` For PHOTOELECTRIC emission `lambda lt lambda_(0)`, so cut of the given wavelength `lambda_(1)=5000Å and lambda_(2)=6000Å` will causephoto emission. Power of source `=10^(-3)W` Diameter of beam `D=2xx10^(-3)m` Radius of beam `r=10^(-3)m` `:.` Area of cross section of beam `A=pi r^(2)=pi xx10^(6)m^(2)` Power DENSITY of beam `=(P)/(A)` `=(10^(-3))/(pixx10^(-6))=(1)/(pi)xx10^(3)W//m^(3)` Power density of each wavelength, `I.=(10^(3))/(3pi)W//m^(3)` Energy of photon of wavelength `lambda_(1)` `E_(1)=(hc)/(lambda_(1))=3.97xx10^(-19)J` Number of photons of wavelength `lambda_(1)` incident on metallic surface per unit area per sec. `N_(1)=(I)/(E_(1))=(10^(3))/(3pi xx3.97xx10^(-19))=2.67xx10^(20)` Energy of photon of wavelength `lambda_(2)` `E_(2)=(hc)/(lambda_(2))=3.31xx10^(-19)J` Then `N_(2)=(I.)/(E_(2))=3.21xx10^(20)` As each photon liberates an electron, the number of electrons EMITTED per sec unit area `N=N_(1)+N_(2)` `=2.67xx10^(20)+3.21xx10^(20)` `=5.88xx10^(20)` `:.` Charge emitted per sec. per unit area is `q=Ne` `=5.88xx10^(20)xx1.6xx10^(-19)` `or q=94C` |
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