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

Hint: Plot a graph of Voltage vs Frequency with the values given above. The slope of the curve is ${\displaystyle \frac{h}{e}}$ and makes an intercept ${\displaystyle \frac{h{\nu }_0}{e}}$ 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:$\nu ={\displaystyle \frac{c}{{\lambda }_{}}}$
Where,
c is speed of light,
$\lambda$ is the wavelength of light.

Substituting the values of c and $\lambda$ to obtain the values of $\nu$.
$\begin{array}{rl}& {\nu }_1={\displaystyle \frac{c}{{\lambda }_1}}={\displaystyle \frac{3\ast {10}^8}{3650\ast {10}^{-10}}}=8.22\ast {10}^{14}Hz\\ & {\nu }_2={\displaystyle \frac{c}{{\lambda }_2}}={\displaystyle \frac{3\ast {10}^8}{4047\ast {10}^{-10}}}=7.41\ast {10}^{14}Hz\\ & {\nu }_3={\displaystyle \frac{c}{{\lambda }_3}}={\displaystyle \frac{3\ast {10}^8}{4358\ast {10}^{-10}}}=6.88\ast {10}^{14}Hz\\ & {\nu }_4={\displaystyle \frac{c}{{\lambda }_4}}={\displaystyle \frac{3\ast {10}^8}{5461\ast {10}^{-10}}}=5.49\ast {10}^{14}Hz\\ & {\nu }_5={\displaystyle \frac{c}{{\lambda }_5}}={\displaystyle \frac{3\ast {10}^8}{6907\ast {10}^{-10}}}=4.34\ast {10}^{14}Hz\end{array}$
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
$h\nu =h{\nu }_0+{\displaystyle \frac{1}{2}}m{v^2}_{max}$
If $V_0$ is the stopping potential, then we have
$e{V}_0$=${\displaystyle \frac{1}{2}}m{v^2}_{max}$
$h\nu =h{\nu }_0+e{V}_0$
Or $V_0={\displaystyle \frac{h\nu }{e}}-{\displaystyle \frac{h{\nu }_0}{e}}$
The above equation represents a straight line whose slope is ${\displaystyle \frac{h}{e}}$ and makes and intercept ${\displaystyle \frac{h{\nu }_0}{e}}$ with negative y axis.

From the above formula we can calculate slope of the graph,
$${\displaystyle \frac{1.28-0.16}{8.22\ast {10}^{14}-5.49\ast {10}^{14}}}=4.1\ast {10}^{-15}Js{C}^{-1}$$
$$\begin{array}{rl}& {\displaystyle \frac{h}{e}}=4.1\ast {10}^{-15}\\ & h=e\ast 4.1\ast {10}^{-15}\\ & h=6.57\ast {10}^{-34}Js\\ & {\nu }_0=5.15\ast {10}^{14}Hz\end{array}$$

Now that we found the value of Planck's constant, we can find the work function for the metal,
$\begin{array}{rl}& W=h{\nu }_0\\ & W=3.38\ast {10}^{-19}J={\displaystyle \frac{3.38\ast {10}^{-19}}{1.6\ast {10}^{-19}}}=2.11eV\end{array}$

Therefore,
The value of Planck's constant is $$6.57\ast {10}^{-34}Js$$,
Threshold frequency is $$5.15\ast {10}^{14}Hz$$,
Work Function of the metal is $3.38\ast {10}^{-19}J$ $=2.11eV$

Note: The symbol for voltage and frequency look similar but are different physical quantities. Always recheck the value before substituting it in an equation.