Chapter 26, Problem 27E

(a)

To determine

The threshold wavelength for photoemission.

Answer to Problem 27E

Thethreshold wavelength is $\underset{\_}{\text{276}\text{nm}}$.

Explanation of Solution

Given that the work function is $4.49\text{eV}$.

Write the expression for the work function in terms of the threshold wavelength.

  $W=\frac{hc}{{\lambda }_0}$        (I)

Here, $W$ is the work function is, $h$ is the Planck’s constant, $c$ is the speed of light, and ${\lambda }_0$ is the threshold wavelength.

Solve Equation (I) for ${\lambda }_0$.

  ${\lambda }_0=\frac{hc}{W}$        (II)

Conclusion:

Substitute $6.62\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, and $4.49\text{eV}$ for $W$ in Equation (II) to find the threshold wavelength.

  $\begin{array}{c}{\lambda }_0=\frac{\left(6.62\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(4.49\text{eV}\times \frac{1.6\times {10}^{-19}\text{J}}{1\text{eV}}\right)}\\ =2.76\times {10}^{-7}\text{m}\\ =\text{276}\text{nm}\end{array}$

Therefore, the threshold wavelengthis $\underset{\_}{\text{276}\text{nm}}$.

(b)

To determine

The maximum kinetic energy of the emitted electrons.

Answer to Problem 27E

The maximum kinetic energy of the emitted electrons is $\underset{\_}{0.47\text{eV}}$.

Explanation of Solution

Given that the wavelength of ultraviolet light is $250\text{nm}$.

Write the expression for the maximum kinetic energy of the ejected electrons.

  $K_{\max }=E-W$        (III)

Here, $K{E}_{\max }$ is the maximum kinetic energy with which the electrons are ejected, and $E$ is the energy of the incident photons.

Write the expression relating energy, wavelength, speed of light and Planck’s constant.

  $E=\frac{hc}{\lambda }$        (IV)

Here, $\lambda$ is the wavelength of the photon.

Use Equation (IV) in (III).

  $K_{\max }=\frac{hc}{\lambda }-W$        (V)

Conclusion:

Substitute $4.1357\times {10}^{-15}\text{eV}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, $250\text{nm}$ for $\lambda$, and $4.49\text{eV}$ for $W$ in Equation (V) to find the maximum kinetic energy of the emitted electrons.

  $\begin{array}{c}{K}_{\max }=\frac{\left(4.1357\times {10}^{-15}\text{eV}\cdot \text{s}\right)\left(3\times {10}^8\text{m/s}\right)}{250\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}}-\left(4.49\text{eV}\right)\\ =0.47\text{eV}\end{array}$

Therefore, the maximum kinetic energy of the emitted electrons is $\underset{\_}{0.47\text{eV}}$.

(c)

To determine

The stopping potential.

Answer to Problem 27E

The stopping potentialis $\underset{\_}{\text{0}\text{.48}\text{eV}}$.

Explanation of Solution

Write the expression for Einstein’s photoelectric equation.

  $e{V}_0=hf-W$        (VI)

Here, $e$ is the charge of electron, $V_0$ is the stopping potential, and $f$ is the frequency of the incident light.

Write the expression for frequency.

  $f=\frac{c}{\lambda }$        (VII)

Use Equation (VII) in (VI) and solve for $V_0$.

  $\begin{array}{c}e{V}_0=\frac{hc}{\lambda }-W\\ V_0=\frac{1}{e}\left(\frac{hc}{\lambda }-W\right)\end{array}$        (VIII)

Conclusion:

Substitute $1.6\times {10}^{-19}\text{C}$ for $e$, $6.62\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, $250\text{nm}$ for $\lambda$, and $4.49\text{eV}$ for $W$.

  $\begin{array}{c}{V}_0=\frac{1}{\left(1.6\times {10}^{-19}\text{C}\right)}\left(\frac{\left(6.62\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m/s}\right)}{250\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}}-\left(\left(4.49\text{eV}\right)\left(\frac{1.6\times {10}^{-19}\text{J}}{1\text{eV}}\right)\right)\right)\\ =0.475\text{eV}\\ \approx \text{0}\text{.48}\text{eV}\end{array}$

Therefore, the stopping potentialis $\underset{\_}{\text{0}\text{.48}\text{eV}}$.