Chapter 27, Problem 61P

(a)

To determine

The Planck’s constant from the graph given in question.

Answer to Problem 61P

The Planck’s constant from the graph given in question is $6.66\times {10}^{-34}\text{J}\cdot \text{s}$.

Explanation of Solution

Write Einstein’s photoelectric effect.

  $K_{\text{max}}=hf-\varphi$                                                                                                         (I)

Here, $K_{\text{max}}$ is the maximum kinetic energy of the electron, $h$ is the Planck’s constant, $f$ is the frequency of the light and $\varphi$ is the work function of the metal.

Write the expression for the kinetic energy of the electron.

  $K_{\text{max}}=e{V}_{\text{s}}$

Here, $e$ is the electronic charge, $V_{\text{s}}$ is the stopping potential .

Substitute $e{V}_{\text{s}}$ for $K_{\text{max}}$ in equation (I) to get linear equation between stopping potential and frequency of light.

  $\begin{array}{l}e{V}_{\text{s}}=hf-\varphi \\ V_{\text{s}}=\frac{h}{e}f-\frac{\varphi }{e}\end{array}$                                                                                                            (II)

Write the general equation of straight line in $x-y$ graph.

  $y=mx+c$                                                                                                             (III)

Here, $m$ is the slope of $x-y$ graph and $c$ is the y intercept.

Equation (II) and (III) indicates that, $V_{\text{s}}$ versus $f$ is a straight line with slope $\frac{h}{e}$ and $-\frac{\varphi }{e}$ as $y$ intercept of the graph.

Write the expression for the slope of straight line in stopping potential versus frequency graph.

  $m=\frac{V_{\text{2}}-V_{\text{1}}}{f_2-f_1}$                                                                                                          (IV)

Here, $V_{1,}{V}_2$ are the stopping potential at $f_1$ and $f_2$ respectively.

Conclusion:

The slope of stopping potential versus frequency graph given in question gives that value of $h$.

From graph, stopping potential at $800\text{THz}$ is $1.5\text{V}$ and stopping potential at $439\text{THz}$ is $0\text{V}$.

Substitute $1.5\text{V}$ for $V_2$, $0\text{V}$ for $V_1$ , $800\text{THz}$ for $f_2$ and $439\text{THz}$ for $f_1$ in equation (IV) to get $m$.

  $\begin{array}{c}m=\frac{1.5\text{V}-0\text{V}}{\left(800\text{THz}-439\text{THz}\right)\times \frac{{10}^{12}\text{Hz}}{1\text{THz}}}\\ =4.155\times {10}^{-15}\text{V}\cdot \text{s}\end{array}$

It is found that slope is equal to $\frac{h}{e}$.

Equate slope obtained with $\frac{h}{e}$ to get $h$.

  $\begin{array}{c}\frac{h}{e}=4.155\times {10}^{-15}\text{V}\cdot \text{s}\\ h=e\left(4.155\times {10}^{-15}\text{V}\cdot \text{s}\right)\end{array}$

Substitute $1.602\times {10}^{-19}\text{C}$ in above equation to get $h$.

  $\begin{array}{c}h=\left(1.602\times {10}^{-19}\text{C}\right)\left(4.155\times {10}^{-15}\text{V}\cdot \text{s}\right)\\ =6.66\times {10}^{-34}\text{J}\cdot \text{s}\end{array}$

Therefore, the Planck’s constant from the graph given in question is $6.66\times {10}^{-34}\text{J}\cdot \text{s}$.

(b)

To determine

The work function of the metal from the data.

Answer to Problem 61P

The work function of the metal from the data is $1.82\text{eV}$ .

Explanation of Solution

Rewrite equation (II) .

  $V_{\text{s}}=\frac{h}{e}f-\frac{\varphi }{e}$

In graph, the straight-line intercept the $x$ axis at its threshold frequency. At this frequency stopping potential is zero indicating that kinetic energy of the electron is zero.

The threshold frequency of the metal from the graph is $439\text{THz}$.

Substitute $0$ for $V_{\text{s}}$ , $f_0$ for $f$ and $m$ for $\frac{h}{e}$ in above equation to get equation of work function.

  $\begin{array}{c}0=m{f}_0-\frac{\varphi }{e}\\ \varphi =em{f}_0\end{array}$                                                                                                           (V)

Here,$f_0$ is the work function of the metal.

Conclusion:

Substitute $4.155\times {10}^{-15}\text{V}\cdot \text{s}$ for $m$ , $439\text{THz}$ for $f_0$ and $1.602\times {10}^{-19}\text{C}$ fro $e$ in equation (V) to get $\varphi$.

  $\begin{array}{c}\varphi =\left(1.602\times {10}^{-19}\text{C}\right)\left(4.155\times {10}^{-15}\text{V}\cdot \text{s}\right)\left(439\text{THz}\times \frac{{10}^{12}\text{Hz}}{1\text{THz}}\right)\times \frac{1\text{eV}}{1.602\times {10}^{-19}\text{J}}\\ =1.82\text{eV}\end{array}$

Therefore, the work function of the metal from the data is $1.82\text{eV}$ .