Chapter 27, Problem 11P

(a)

To determine

The work function in Joules.

Answer to Problem 11P

The work function in Joules is $1.02\times {10}^{-18}\text{J}$.

Explanation of Solution

The work function in joules is,

${\varphi }_{\text{J}}={\varphi }_{\text{eV}}\left(\frac{1.60\times {10}^{-19}\text{J}}{1\text{eV}}\right)$

• • ${\varphi }_{\text{J}}$ is the work function in joules
  • ${\varphi }_{\text{eV}}$ is the work function in electron volt

Substitute $6.35\text{eV}$ for ${\varphi }_{\text{eV}}$.

$\begin{array}{c}{\varphi }_{\text{J}}=\left(6.35\text{eV}\right)\left(\frac{1.60\times {10}^{-19}\text{J}}{1\text{eV}}\right)\\ =1.02\times {10}^{-18}\text{J}\end{array}$

Conclusion:

Thus, the work function in Joules is $1.02\times {10}^{-18}\text{J}$.

(b)

To determine

The cutoff frequency for platinum.

Answer to Problem 11P

The cutoff frequency for platinum is $1.53\times {10}^{15}\text{Hz}$.

Explanation of Solution

The energy of the photon is,

$E_{\text{photon}}=h{f}_c$

• • $h$ is the Plank’s constant
  • $f_c$ is the cutoff frequency

The cutoff frequency is,

$f_c=\frac{\varphi }{h}$

Substitute $6.35\text{eV}$ for $\varphi$ and $6.63\times {10}^{-34}\text{J}\text{.s}$ for $h$.

$\begin{array}{c}{f}_c=\frac{6.35\text{eV}}{6.63\times {10}^{-34}\text{J}\text{.s}}\left(\frac{1\text{eV}}{1.60\times {10}^{-19}\text{J}}\right)\\ =1.53\times {\text{10}}^{15}\text{Hz}\end{array}$

Conclusion:

Thus, the cutoff frequency for platinum is $1.53\times {10}^{15}\text{Hz}$.

(c)

To determine

The maximum wavelength of the light incident on the platinum surface.

Answer to Problem 11P

The maximum wavelength of the light incident on the platinum surface is $\text{196}\text{nm}$.

Explanation of Solution

The equation for cutoff wavelength is,

$\begin{array}{c}{\lambda }_c=\frac{hc}{\varphi }\\ =\frac{c}{f_c}\end{array}$

Substitute $3.00\times {10}^8\text{m/s}$ for $c$ and $1.53\times {10}^{15}\text{Hz}$ for $f_c$.

$\begin{array}{c}{\lambda }_c=\frac{3.00\times {10}^8\text{m/s}}{1.53\times {10}^{15}\text{Hz}}\\ =1.96\times {10}^{-7}\text{m}\\ =\text{196}\text{nm}\end{array}$

Conclusion:

Thus, the maximum wavelength of the light incident on the platinum surface is $\text{196}\text{nm}$.

(d)

To determine

The maximum kinetic energy of the ejected photoelectrons.

Answer to Problem 11P

The maximum kinetic energy of the ejected photoelectrons is $\text{2}\text{.15}\text{eV}$.

Explanation of Solution

The equation for maximum kinetic energy is,

$K{E}_{\text{max}}=E_{\text{photon}}-\varphi$

• • $E_{\text{photon}}$ is the energy of the photon

Substitute $8.50\text{eV}$ for $E_{\text{photon}}$ and $6.35\text{eV}$ for $\varphi$.

$\begin{array}{c}K{E}_{\text{max}}=8.50\text{eV}-6.35\text{eV}\\ =\text{2}\text{.15}\text{eV}\end{array}$

Conclusion:

Thus, the maximum kinetic energy of the ejected photoelectrons is $\text{2}\text{.15}\text{eV}$.

(e)

To determine

The stopping potential required to arrest the current of photoelectrons.

Answer to Problem 11P

The stopping potential required to arrest the current of photoelectrons is $\text{2}\text{.15}\text{V}$.

Explanation of Solution

We have the relation,

$\begin{array}{c}e{V}_s=K{E}_{\max }\\ V_s=\frac{K{E}_{\max }}{e}\end{array}$

Substitute $2.15\text{eV}$ for $K{E}_{\max }$ and $1.60\times {10}^{-19}\text{C}$.

$\begin{array}{c}{V}_s=\frac{2.15\text{eV}}{1.60\times {10}^{-19}\text{C}}\\ =2.15\text{V}\end{array}$

Conclusion:

Thus, the maximum kinetic energy of the ejected photoelectrons is $\text{2}\text{.15}\text{eV}$