Question

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 red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used
${{\lambda }_{1}}=3650A,{{\lambda }_{2}}=4358A,{{\lambda }_{3}}=4358A,{{\lambda }_{4}}=5461A,{{\lambda }_{5}}=6907A.$

The stopping voltages, respectively, were measured to be:
${{V}_{01}}=1.28V,{{V}_{02}}=0.95V,{{V}_{03}}=0.74V,{{V}_{04}}=0.16V,{{V}_{05}}=0V,~$

Determine the value of Planck's constant h, the threshold frequency and work function for the material.

Answer 1

Plot a graph of Voltage vs Frequency with the values given above. The slope of the curve is $\dfrac{h}{e}$ and makes an intercept $\dfrac{h{{\nu }_{0}}}{e}$ on the negative. Substitute the charge of the electron to obtain Planck's constant and further find the threshold frequency and work function.

Complete step by step answer:
We have the values of Voltage but we need to find the values of frequency for the corresponding readings.
Formula to find frequency:$\nu =\dfrac{c}{{{\lambda }_{{}}}}$
Where,
c is speed of light,
$\lambda$ is the wavelength of light.

Substituting the values of c and $\lambda$ to obtain the values of $\nu$.
$\begin{aligned} & {{\nu }_{1}}=\dfrac{c}{{{\lambda }_{1}}}=\dfrac{3*{{10}^{8}}}{3650*{{10}^{-10}}}=8.22*{{10}^{14}}Hz \\\ & {{\nu }_{2}}=\dfrac{c}{{{\lambda }_{2}}}=\dfrac{3*{{10}^{8}}}{4047*{{10}^{-10}}}=7.41*{{10}^{14}}Hz \\\ & {{\nu }_{3}}=\dfrac{c}{{{\lambda }_{3}}}=\dfrac{3*{{10}^{8}}}{4358*{{10}^{-10}}}=6.88*{{10}^{14}}Hz \\\ & {{\nu }_{4}}=\dfrac{c}{{{\lambda }_{4}}}=\dfrac{3*{{10}^{8}}}{5461*{{10}^{-10}}}=5.49*{{10}^{14}}Hz \\\ & {{\nu }_{5}}=\dfrac{c}{{{\lambda }_{5}}}=\dfrac{3*{{10}^{8}}}{6907*{{10}^{-10}}}=4.34*{{10}^{14}}Hz \\\ \end{aligned}$
Plotting a graph of ${{V}_{0}}$vs $\nu$:

In the graph given, the X axis represents the values of
$\nu$(x${{10}^{14}}$), and the Y axis represents the value of ${{V}_{0}}$.

From Einstein's equations of photoelectric effect, we have
$h\nu =h{{\nu }_{0}}+\dfrac{1}{2}m{{v}^{2}}_{\max }$
If ${{V}_{0}}$ is the stopping potential, then we have
$e{{V}_{0}}$=$\dfrac{1}{2}m{{v}^{2}}_{\max }$
$h\nu =h{{\nu }_{0}}+e{{V}_{0}}$
Or ${{V}_{0}}=\dfrac{h\nu }{e}-\dfrac{h{{\nu }_{0}}}{e}$
The above equation represents a straight line whose slope is $\dfrac{h}{e}$ and makes and intercept $\dfrac{h{{\nu }_{0}}}{e}$ with negative y axis.

From the above formula we can calculate slope of the graph,
$\dfrac{1.28-0.16}{8.22*{{10}^{14}}-5.49*{{10}^{14}}}=4.1*{{10}^{-15}}Js{{C}^{-1}}$

& \dfrac{h}{e}=4.1*{{10}^{-15}} \\\ & h=e*4.1*{{10}^{-15}} \\\ & h=6.57*{{10}^{-34}}Js \\\ & {{\nu }_{0}}=5.15*{{10}^{14}}Hz \\\ \end{aligned}$$ Now that we found the value of Planck's constant, we can find the work function for the metal, $\begin{aligned} & W=h{{\nu }_{0}} \\\ & W=3.38*{{10}^{-19}}J=\dfrac{3.38*{{10}^{-19}}}{1.6*{{10}^{-19}}}=2.11eV \\\ \end{aligned}$ Therefore, The value of Planck's constant is $$6.57*{{10}^{-34}}Js$$, Threshold frequency is $$5.15*{{10}^{14}}Hz$$, Work Function of the metal is $3.38*{{10}^{-19}}J$ $=2.11eV$ **Note:** The symbol for voltage and frequency look similar but are different physical quantities. Always recheck the value before substituting it in an equation.