$A$ mercury lamp is a convenient source for studying the frequency dependence of photoelectric emission,as it provides a number of spectral lines ranging from the $UV$ to the red end of the visible spectrum. In our experiment with a rubidium photocell,the following lines from a mercury source were used:
$\lambda_1 = 3650 \,\mathring{A}, \lambda_2 = 4047 \,\mathring{A}, \lambda_3 = 4358 \,\mathring{A}, \lambda_4 = 5461 \,\mathring{A}, \lambda_5 = 6907 \,\mathring{A}$
The stopping voltages,respectively,were measured to be:
$V_{01} = 1.28 \,V, V_{02} = 0.95 \,V, V_{03} = 0.74 \,V, V_{04} = 0.16 \,V, V_{05} = 0 \,V$
Determine the value of Planck's constant $h$,the threshold frequency,and the work function for the material.

(N/A) Einstein's photoelectric equation is given by:
$e V_0 = h \nu - \phi_0$
$V_0 = \frac{h}{e} \nu - \frac{\phi_0}{e} \dots (i)$
Where $V_0$ is the stopping potential,$h$ is Planck's constant,$e$ is the charge of an electron,$\nu$ is the frequency of radiation,and $\phi_0$ is the work function.
Using $\nu = \frac{c}{\lambda}$ (where $c = 3 \times 10^8 \,m/s$),we calculate the frequencies:
$\nu_1 = 8.219 \times 10^{14} \,Hz, \nu_2 = 7.412 \times 10^{14} \,Hz, \nu_3 = 6.884 \times 10^{14} \,Hz, \nu_4 = 5.493 \times 10^{14} \,Hz, \nu_5 = 4.343 \times 10^{14} \,Hz$

Frequency $(\times 10^{14} \,Hz)$	$8.219$	$7.412$	$6.884$	$5.493$	$4.343$
Stopping potential $V_0$ $(V)$	$1.28$	$0.95$	$0.74$	$0.16$	$0$

The graph of $V_0$ versus $\nu$ is a straight line. The threshold frequency $\nu_0$ is the intercept on the $\nu$-axis,which is $5 \times 10^{14} \,Hz$.
The slope of the line is $\frac{h}{e} = \frac{1.28 - 0.16}{(8.219 - 5.493) \times 10^{14}} = \frac{1.12}{2.726 \times 10^{14}} \approx 4.108 \times 10^{-15} \,V \cdot s$.
$h = (4.108 \times 10^{-15}) \times (1.6 \times 10^{-19}) \approx 6.573 \times 10^{-34} \,J \cdot s$.
Work function $\phi_0 = h \nu_0 = (6.573 \times 10^{-34}) \times (5 \times 10^{14}) = 3.286 \times 10^{-19} \,J = 2.054 \,eV$.