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

Explanation

Wavelength of ultraviolet light, `lambda` =2271Å=`2271xx10^(-10)m`
Stopping potential of the metal, `V_(0)=1.3V`
Plank's constant, `h=6.6xx10^(-34)J`
Charge on an electron,`e=1.6xx10^(-19)C`
Work function of the metal `=phi_(0)`
Frequency of light =v
We have the photo-energy relation from the photoelectric effect as: `phi_(0)=hv-eV_(0)`
`=(hc)/(lambda)-eV_(0)`
`=(6.6xx10^(-34)xx3xx10^(8))/(2271xx10^(-10))-1.6xx10^(-19)xx1.3`
`=8.72xx10^(-19)-2.08xx10^(-19)`
`=6.64xx10^(-19)J`
`=(6.64xx10^(-19))/(1.6xx10^(-19))=4.15eV`
Let `v_(0)` be the threshold frequency of the metal.
`therefore phi_(0)hv_(0)`
`v_(0)=(phi_(0))/(h)`
`=(6.64xx10^(-19))/(6.6xx10^(-34))=1.006xx10^(15)Hz`
Wavelength of red light, `lambda _(r)=6328Å=6328xx10^(-10)m`
`therefore` Frequency of red light, `v_(r)=(c)/(lambda_(r))`
`=(3xx10^(8))/(6328xx10^(-10))=4.74xx10^(14)Hz`
Since `v_(0)gt v_(r)`, the photocell will not respond to the red light produced by the laser.