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

Using `v=c//lamda` , let us first calculate corresponding frequencies.
`{:(lamda,v=c//lamda,V_0),("3,650"Å,8.22xx10^(14)Hz,1.28V),("4,047"Å,7.41xx10^(14)Hz,0.95V),("4,358"Å,7.41xx10^(14)Hz,0.74V),("5,461"Å,5.49xx10^(14)Hz,0.16V),("6,907"Å,4.34xx10^(14)Hz,0.0V):}`
We know that `eV_0=hv-W_0`
Or `V_0=(hv)/e=W_0/e`
(a) The graph between ` v and v_0` is a straight line as shown below.

The slope of the plot is h/e and its Intercept on the V- axis is threshold frequency `(v_0)` . The first four points lie nearly on a straight line which intercepts the v-axis at `v_0=5.0 × 10^(14) Hz` . The fifth point corresponds to `vltv _0` no photoelectric emission occurs, so no stopping voltage is required to stop the current .
Slope of teh plot is found to be
`m=h/e=(V_(01)-V_(04))/(v_1-v_4)=((1.28-0.16)V)/((8.22-5.49)xx10^(14))Hz`
`impliesh=(1.12xx1.6xx10^(-19))/(2.73xx10^(14))=6.56xx10^(-34)Js`
(b) From the graph, it us clear that the threshold frequency for the given material is
`v_0=5.0xx10^(14)Hz.`
Thus the work function is given be
`W_0=(6.574xx10^(-34))xx(5.0xx10^(14))`
`=32.87xx10^(-20)J`
Or `W_0=(32.87xx10^(-20))/(1.6xx10^(-19))=2.05eV`