Ultraviolet light of wavelength `2271 Å` from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is `-1.3 V`, estimate the work function of the metal. How would the photo-cell respond to a high intensity `(~10^(5) W m^(2))` red light of wavelength `6328 Å` produced by a He-Ne laser ?

Explanation

Here `lambda=2271Å=2271xx10^(-10)m`
`V_(0)=1.3V,h=6.63xx10^(-34)Js`,
`c=3xx10^(8)ms^(-1),1eV=1.6xx10^(-19)J`
`implies` From Einstein.s equation,
`eV_(0)=hv-phi_(0)`
`therefore phi_(0)=hv-eV_(0)=(hc)/(lambda)-eV_(0)`
`therefore phi_(0)=(hc)/(elambda)-V_(0)`
`therefore phi_(0)=(6.63xx10^(-34)xx3xx10^(8))/(1.6xx10^(-19)xx2271xx10^(-10))-1.3`
`therefore phi_(0)=0.00547xx10^(3)-1.3`
`therefore phi_(0)~~5.47-1.3eV`
`therefore phi_(0)=4.17eV`
or `phi_(0)=6.672xx10^(-19)J`
`implies`Threshold frequency ,
`v_(0)=(phi)/(h)=(6.672xx10^(-19))/(6.63xx10^(-34))`
`therefore V_(0)=1.0063xx10^(15)Hz`
`therefore v_(0)=1xx10^(15)`Hz
`implies`For red light `lambda=6328 Å=6328xx10^(-10)`m
`therefore` Frequency of red light,
`v=(c)/(lambda)`
`therefore v=(3xx10^(8))/(6328xx10^(-10))=0.000474xx10^(18)`
`therefore v=4.74xx10^(14)`Hz
`implies` Thus,`vltv_(0)` hence photoelectric emission will not take place for red light of any intensity.