A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the end of the red visible spectrum. In our experiment with rubidium photocell, the following lines from a mercury source were used: `lambda_(1)=3650A^(@), lambda_(2)=4047A^(@), lambda_(3)=4358A^(@), lambda_(4)=5461A^(@), lambda_(5)=6907A^(@)` The stopping voltages, respectively were measured to be:
`V_(01)=1.28V, V_(02)=0.95V, V_(03)=0.74V, V_(04)=0.16V, V_(05)=0V`.
(a) Determine the value of Planck's constant h.
(b) Estimate the threshold frequency and work function for the material.

Explanation

Given, `lambda _(1) = 3650 Å= 3650 xx 10^(10 ) m " " lambda _(4) - 5461 Å`
` lambda _(2) = 4047 Å = 4047 xx 10^(-10) m " " lambda _(5) = 6907 Å = 6907 xx 10^(-10) m `
` lambda _(3) = 4358 Å = 4358 xx 10^(-10) m `
` V_(01) =1.28 V, V_(02) = 0.95 , V_(03) = 0.74 V, V_(04) = 0.16 V, V_(05) = 0 `
(a) ` v_(1) = c/(lambda_(1)) = (3 xx10^(8))/(3650 xx10^(-10)) = 8.219 xx10^(14)" " v_(4) c/(lambda _(4)) = (3xx10^8)/(5461 xx 10^(-10)) = 5.493 xx10^(14)` Hz
` v_(2) = c/(lambda_(2)) = (3xx 10^(8))/(4047 xx 10^(-10)) = 7.412xx 10^(14) Hz " " v_(5) = c/(lambda_(5))= (3xx10^(8))/(6907xx10^(-10)) = 4.343 xx10^(14) Hz`
` v_(3) = c/(lambda _(3)) = (3xx10^(8))/(4358 xx10^(-10)) = 6.884 xx10^(14) Hz`
As we know that `eV_(0) = hv- phi _(0)`
` V_(0) = (hv)/e = (phi_(0))/e`
As the graph between `V_(0)` and frequency v is a straight line :
The slope ofthis gives the values of `h/e`
` h = (1.12 xx 1.6 xx 10^(-19))/(2.726 xx 10^(4)) = 6.674xx10^(-34) ` J-s
(b ) `phi _(0) = hv_(0) = 6.574 xx 10^(-34) xx 5xx 10^(14)`
` = 32 . 870 xx 10^(-20) J = (32 870 xx 10^(-20))/(1.6 xx 10^(-19))eV `
= 2.05 eV