Question

In a photoelectric-effect experiment, the stopping potential was measured for several different wavelengths of incident light. The data are as follows:

Use an appropriate graph of the data to determine

(a) the metal used for the cathode and

(b) an experimental value for Planck’s constant .

Short answer

(a) The answer is work function corresponds to the Potassium.

(b) An experimental value for Planck’s constant is$6.97\times {10}^{-34}Js$.

Step-by-step solution

Part (a) step 1: Given Information

We need to find the metal used for the cathode.

Part (a) step 2: Simplify

Now, start expression for stopping potential:

$\begin{array}{rr}{V}_{stop}& =\frac{hf-h{f}_0}{e}=\frac{h}{e}f-\frac{h}{e}{f}_0\\ f& =\frac{c}{\lambda }\end{array}$

we can use the equation of a line through two points:

$y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$

Now we can use two pair of points from table in the text of the problem and find slope and -intercept. For example $Tl(500,0.19)andT2(450,0.48),$but before substituting into equation for the line we must calculate frequency from wavelength$f=\frac{c}{\lambda }\times f_1=\frac{3\times {10}^{\text{*}}}{500\times {10}^{-9}}=6\times {10}^{14}\text{Hz}$and on the same way $f_2=6\cdot 66667\cdot {10}^{14}\text{Hz}$T1 and T2 become $\text{Tl}\left(6\times {10}^{14},0.19\right)$and $T2\left(6.66667\times {10}^{14},0.48\right)$.

$\begin{array}{rr}y-0.19& =\frac{0.48-0.19}{6.66667\cdot {10}^{14}-6\cdot {10}^{14}}(x-6\times {10}^{14})\\ y-0.19& =4.35\cdot {10}^{-15}\left(x-6\times {10}^{14}\right)\\ y& =4.35\cdot {10}^{-15}\times x-2.42\end{array}$

We can compare equation of the line and general equation for stopping potential and conclude:

$\begin{array}{rr}\frac{h}{e}{f}_0& =2.42\text{V}\\ \frac{h}{e}& =4.35\cdot {10}^{-15}\text{Vs}\\ E_0& =h{f}_0=2.42\cdot e\\ E_0& =2.42\text{eV}\end{array}$

This work function corresponds to the Potassium.

Part (b) step 1: Given Information

We need to find an experimental value for Planck’s constant.

Part (b) step 2: Simplify

Easy we can get the experimental value for Planck's constant from the slope of the equation for stopping potential:

$$\begin{gathered} \frac{h}{e}=4.35\cdot {10}^{-15} \\ h=4.35\cdot {10}^{-15}\cdot e \\ h=4.35\cdot {10}^{-15}\cdot 1.6\cdot {10}^{-19} \\ h=6.97\cdot {10}^{-34}\text{Js} \end{gathered}$$