Define the term 'intensity of radiation in photon picture of light.
Ultraviolet light of wavelength `2270Å` from 100W mercury source irradiates a photo cell made of a given metal. If the stopping potential is -1.3V, estimate the work function of the metal. How would the photon cell respond to a high intensity `(~10^(5) W.m^(-2))` red light of wavelength `6300Å` produced by a laser?

Explanation

1st part: Intensity of radiation can be defined as the amount of light energy incident per square metre per second.
2nd part: `V_(0) = 1.3 V, lamda = 2270 xx 10^(-10)m`
We know that `hf = hf_(0) + (1)/(2) mv_("max")^(2)`
Also, `hf = W_(0) + eV_(0)`
`:. W_(0) = hf - eV_(0) or, W_(0) = (hc)/(lamda) - eV_(0)`
`:. W_(0) = (6.62 xx 10^(-34) xx 3 xx 10^(8))/(2270 xx 10^(-10)) - 1.3 xx 1.6 xx 10^(-19)`
`= 6.669 xx 10^(-19)J = (6.669 xx 10^(-19))/(1.6 xx 10^(-19)) eV = 4.168 eV`
Again, `W_(0) = hf_(0) = (hc)/(lamda_(0))`
or, `lamda_(0) = (hc)/(W_(0)) = (6.62 xx 10^(-34) xx 3 xx 10^(8))/(6.669 xx 10^(-19)) m`
`= 2.98 xx 10^(-7) m = 2980 Å`
The photo cell would not respond to a high intensity `(~10^(5) W.m^(-2))` red light of wavelength `6300 Å` produced by a laser because this wavelength is greater than `2980 Å`