Chapter 40, Problem 65AP

(a)

To determine

The work function of sodium using the graph.

Answer to Problem 65AP

The work function of sodium is $1.7\text{ eV}$ .

Explanation of Solution

Write the equation for the maximum kinetic energy of the ejected photoelectrons.

  $K_{\text{max}}=e\Delta V_s$

Here, $K_{\text{max}}$ is the maximum kinetic energy of the ejected electron, $e$ is the magnitude of the charge of the electron and $\Delta V_s$ is the stopping potential.

Rewrite the above equation for $\Delta V_s$ .

  $\Delta V_s=\frac{K_{\text{max}}}{e}$                                                                                                            (I)

Write the Einstein’s photoelectric effect equation.

  $K_{\text{max}}=hf-\varphi$

Here, $h$ is the Plank’s constant, $f$ is the frequency of the incident photon and $\varphi$ is the work function of the metal.

Put the above equation in equation (I).

  $\begin{array}{c}\Delta V_s=\frac{hf-\varphi }{e}\\ =\frac{h}{e}f-\frac{\varphi }{e}\end{array}$                                                                                                          (II)

Consider the two points $\left(0\text{ V},4.1\times {10}^{14}\text{ Hz}\right)$ and $\left(3.3\text{ V},12\times {10}^{14}\text{ Hz}\right)$ in the graph in figure P40.65.

Substitute $0\text{ V}$ for $\Delta V_s$ and $4.1\times {10}^{14}\text{ Hz}$ for $f$ in equation (II) to form an equation.

  $0=\frac{h}{e}\left(4.1\times {10}^{14}\text{ Hz}\right)-\frac{\varphi }{e}$                                                                                        (III)

Substitute $3.3\text{ V}$ for $\Delta V_s$ and $12\times {10}^{14}\text{ Hz}$ for $f$ in equation (II) to form second equation.

  $3.3\text{ V}=\frac{h}{e}\left(12\times {10}^{14}\text{ Hz}\right)-\frac{\varphi }{e}$                                                                                  (IV)

Conclusion:

Add equations (III) and (IV) and rewrite it for $\varphi$ .

  $\begin{array}{c}0\text{ V}+3.3\text{ V}=\frac{h}{e}\left(4.1\times {10}^{14}\text{ Hz}\right)+\frac{h}{e}\left(12\times {10}^{14}\text{ Hz}\right)-2\frac{\varphi }{e}\\ 3.3\text{ V}=\frac{h}{e}\left(4.1\times {10}^{14}\text{ Hz}+12\times {10}^{14}\text{ Hz}\right)-\frac{2\varphi }{e}\\ \frac{2\varphi }{e}=\frac{h}{e}\left(4.1\times {10}^{14}\text{ Hz}+12\times {10}^{14}\text{ Hz}\right)-3.3\text{ V}\\ \varphi =\frac{1}{2}\left[h\left(4.1\times {10}^{14}\text{ Hz}+12\times {10}^{14}\text{ Hz}\right)-e\left(3.3\text{ V}\right)\right]\end{array}$                                  (V)

Substitute $6.626\times {10}^{-34}\text{ J}\cdot \text{s}$ for $h$ and $1.60\times {10}^{-19}\text{ C}$ for $e$ in equation (V) to find $\varphi$ .

  $\begin{array}{c}\varphi =\frac{1}{2}\left[\left(6.626\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(4.1\times {10}^{14}\text{ Hz}+12\times {10}^{14}\text{ Hz}\right)-\left(1.60\times {10}^{-19}\text{ C}\right)\left(3.3\text{ V}\right)\right]\\ =1.7\text{ eV}\end{array}$

Therefore, the work function of sodium is $1.7\text{ eV}$ .

(b)

To determine

The value of the ratio $h/e$ .

Answer to Problem 65AP

The value of the ratio $h/e$ is $4.2\times {10}^{-15}\text{ V}\cdot \text{s}$ .

Explanation of Solution

Subtract equation (III) from equation (IV).

  $\begin{array}{c}3.3\text{ V}-0\text{ V}=\left(\frac{h}{e}\left(12\times {10}^{14}\text{ Hz}\right)-\frac{\varphi }{e}\right)-\left(\frac{h}{e}\left(4.1\times {10}^{14}\text{ Hz}\right)-\frac{\varphi }{e}\right)\\ 3.3\text{ V}=\frac{h}{e}\left(12\times {10}^{14}\text{ Hz}-4.1\times {10}^{14}\text{ Hz}\right)\\ \frac{h}{e}=\frac{3.3\text{ V}}{\left(12\times {10}^{14}\text{ Hz}-4.1\times {10}^4\text{ Hz}\right)}\\ =4.2\times {10}^{-15}\text{ V}\cdot \text{s}\end{array}$

Conclusion:

Therefore, the value of the ratio $h/e$ is $4.2\times {10}^{-15}\text{ V}\cdot \text{s}$ .

(c)

To determine

The cutoff wavelength.

Answer to Problem 65AP

The cutoff wavelength is $7.3\times {10}^2\text{ nm}$ .

Explanation of Solution

Rewrite Einstein’s photoelectric equation for $\varphi$ .

  $\varphi =hf-K_{\text{max}}$

At the cutoff wavelength, kinetic energy of the photoelectrons becomes zero.

Substitute $0$ for $K_{\text{max}}$ in the above equation.

  $\begin{array}{l}\varphi =hf-0\\ \varphi =hf\end{array}$                                                                                                                 (V)

Write the relation between cutoff frequency and cutoff wavelength.

  $f=\frac{c}{{\lambda }_c}$

Here, $c$ is the speed of light in vacuum and ${\lambda }_c$ is the cutoff wavelength.

Put the above equation in equation (V).

  $\varphi =\frac{hc}{{\lambda }_c}$

Multiply numerator and denominator of right hand side of the above equation by $e$ .

  $\begin{array}{c}\varphi =\left(\frac{h}{e}\right)\frac{ec}{{\lambda }_c}\\ {\lambda }_c=\left(\frac{h}{e}\right)\frac{ec}{\varphi }\end{array}$                                                                                                             (VI)

Conclusion:

The value of $e$ is $1.60\times {10}^{-19}\text{ C}$ and the value of $c$ is $3.00\times {10}^8\text{ m/s}$ .

Substitute $4.2\times {10}^{-15}\text{ V}\cdot \text{s}$ for $h/e$ , $1.60\times {10}^{-19}\text{ C}$ for $e$ , $3.00\times {10}^8\text{ m/s}$ for $c$ and $1.7\text{ eV}$ for $\varphi$ in equation (VI) to find ${\lambda }_c$ .

  $\begin{array}{c}{\lambda }_c=\left(4.2\times {10}^{-15}\text{ V}\cdot \text{s}\right)\frac{\left(1.60\times {10}^{-19}\text{ C}\right)\left(3.00\times {10}^8\text{ m/s}\right)}{\left(1.7\text{ eV}\frac{1.60\times {10}^{-19}\text{ J}}{1\text{ eV}}\right)}\\ =7.3\times {10}^{-7}\text{ m}\frac{1\text{ nm}}{{10}^{-9}\text{ m}}\\ =7.3\times {10}^2\text{ nm}\end{array}$

Therefore, the cutoff wavelength is $7.3\times {10}^2\text{ nm}$ .