A mercury lamp is convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photocell, the following lines from a mercury source were used:
λ₁ = 3650 Å, λ₂ = 4047 Å, λ₃ = 4358 Å, λ₄ = 5461 Å, λ₅ = 6907 Å
The stopping voltages, respectively were measured to be:
V₀₁ = 1.28 V,V₀₂ = 0.95 V,V₀₃ = 0.74 V, V₀₄ = 0.16 V,V₀₅ = 0V
(a)    Determine the value of Planck's constant h.
(b)    Estimate the threshold frequency and work function for the material.

Using the given data, we first determine frequency in each case and then plot a graph between stopping potential V₀ and frequency v. 

$$\begin{gathered} {\mathrm{v}}_1 = \frac{\mathrm{c}}{{\mathrm{\lambda }}_1} = \frac{3 \times {10}^8}{3650 \times {10}^{-10}} = 8.219 \times {10}^{14}\mathrm{Hz} \\ {\mathrm{v}}_2 = \frac{\mathrm{c}}{{\mathrm{\lambda }}_2} = \frac{3 \times {10}^8}{4047 \times {10}^{-10}} = 7.412 \times {10}^{14}\mathrm{Hz} \\ {\mathrm{v}}_3 = \frac{\mathrm{c}}{{\mathrm{\lambda }}_3} = \frac{3 \times {10}^8}{4358 \times {10}^{-10}} = 6.884 \times {10}^{14}\mathrm{Hz} \\ {\mathrm{v}}_4 = \frac{\mathrm{c}}{{\mathrm{\lambda }}_4} = \frac{3 \times {10}^8}{5461 \times {10}^{-10}} = 5.493 \times {10}^{14}\mathrm{Hz} \\ {\mathrm{v}}_5 = \frac{\mathrm{c}}{{\mathrm{\lambda }}_5} = \frac{3 \times {10}^8}{6907 \times {10}^{-10}} = 4.343 \times {10}^{14}\mathrm{Hz} \end{gathered}$$ 

V_o versus v plot is shown below: 

 

The first four points lie nearly on a straight line which intercepts the frequency axis at threshold frequency v₀ = 5.0 x 10¹⁴ Hz.

The fifth point v (= 4.3 x 10¹⁴ Hz) corresponds to v < v₀, so there is no photoelectric emission and not stopping voltage is required to stop the current.

Slope of V₀ versus v graph is, 

$$\begin{gathered} \frac{∆\mathrm{V}}{∆ \mathrm{v}} = \frac{(1.28 - 0)\mathrm{V}}{(8.2-5.0) \times {10}^{14}{\mathrm{s}}^{-1}} \\ = 4.0 \times {10}^{-15} \mathrm{Vs} \end{gathered}$$ 

Now, from Einstein's photoelectric equation, 

Kinetic energy of photon, $\boxed{\mathrm{K}.\mathrm{E}. = \mathrm{eV} = \mathrm{hv} - {\mathrm{W}}_0}$ 

$\therefore$                     $\mathrm{e}∆\mathrm{V} = \mathrm{h}∆\mathrm{v}$                      $[{\mathrm{W}}_0 (work function) \mathrm{is} \mathrm{a} \mathrm{constant} ]$ 

$\Rightarrow$                    $\frac{∆\mathrm{V}}{∆\nu } = \frac{\mathrm{h}}{\mathrm{e}}$ 

Hence,             $\frac{\mathrm{h}}{\mathrm{e}} = 4.0 \times {10}^{-15} \mathrm{Vs}$ 

Planck's constant, $\mathrm{h} = \mathrm{e} \times 4.0 \times {10}^{-15}\mathrm{Js}$

$\Rightarrow$                          $\mathrm{h} = 1.6 \times {10}^{-19} \times 4.0 \times {10}^{-15}\mathrm{Js}$ 

                             $= 6.4 \times {10}^{-34}\mathrm{Js}$ 

(b) Threshold frequency, ${\mathrm{v}}_0 = 5.0 \times {10}^{14}\mathrm{Hz}$
     $$\begin{gathered} \therefore \mathrm{Work} \mathrm{function} of the metal, {\mathrm{\varphi }}_0 = {\mathrm{hv}}_0 \\ = 6.4 \times {10}^{34} \times 5.0 \times {10}^{14} \end{gathered}$$

i.e.,   ${\mathrm{\varphi }}_0 = \frac{6.4 \times 5\times {10}^{-20}}{1.6 \times {10}^{-19}}\mathrm{eV} = 2.00 \mathrm{eV}$ is the required work function of the metal.