Chapter 29

(a) 0.070 eV

(b) 14

(a) $2\text{.}\text{21}\times {\text{10}}^{\text{34}}\text{J}$

(b) $2\text{.}\text{26}\times {\text{10}}^{\text{34}}$

(c) No

263 nm

3.69 eV

0.483 eV

2.25 eV

(a) 264 nm

(b) Ultraviolet

$1.95\times {\text{10}}^6\text{m/s}$

(a) $4.02\times {\text{10}}^{\text{15}}\text{/s}$

(b) 0.256 mW

(a) $\mathrm{–1.90\; eV}$

(b) Negative kinetic energy

(c) That the electrons would be knocked free.

$6.34\times {\text{10}}^{-9}\text{eV}$, $1.01\times {\text{10}}^{-27}\text{J}$

$2\text{.}\text{42}\times {\text{10}}^{\text{20}}\text{Hz}$

${\displaystyle \begin{array}{lll}\text{hc}& =& \left(\text{6.62607}\times {\text{10}}^{-\text{34}}\mathrm{J}\cdot \mathrm{s}\right)\left(\text{2.99792}\times {\text{10}}^8\text{m/s}\right)\left(\frac{{\text{10}}^9\text{nm}}{\mathrm{1\; m}}\right)\left(\frac{\text{1.00000 eV}}{\text{1.60218}\times {\text{10}}^{-\text{19}}\text{J}}\right)\\ & =& \text{1239.84 eV}\cdot \text{nm}\\ & \approx & \text{1240 eV}\cdot \text{nm}\end{array}}$

(a) 0.0829 eV

(b) 121

(c) 1.24 MeV

(d) $1\text{.}\text{24}\times {\text{10}}^5$

(a) $\text{25.0}\times {\text{10}}^3\text{eV}$

(b) $\text{6}\text{.}\text{04}\times {\text{10}}^{\text{18}}\text{Hz}$

(a) 2.69

(b) 0.371

(a) $\text{1}\text{.}\text{25}\times {\text{10}}^{\text{13}}\text{photons/s}$

(b) 997 km

$\text{8.33}\times {\text{10}}^{\text{13}}\text{photons/s}$

181 km

(a) $\text{1.66}\times {\text{10}}^{-\text{32}}\text{kg}\cdot \text{m/s}$

(b) The wavelength of microwave photons is large, so the momentum they carry is very small.

(a) 13.3 μm

(b) $9\text{.}\text{38}\times {\text{10}}^{-2}$ eV

(a) $2\text{.}\text{65}\times {\text{10}}^{-\text{28}}\text{kg}\cdot \text{m/s}$

(b) 291 m/s

(c) electron $3\text{.}\text{86}\times {\text{10}}^{-\text{26}}\text{J}$, photon $7\text{.}\text{96}\times {\text{10}}^{-\text{20}}\text{J}$, ratio $2\text{.}\text{06}\times {\text{10}}^6$

(a) $1\text{.}\text{32}\times {\text{10}}^{-\text{13}}\text{m}$

(b) 9.39 MeV

(c) $4.70\times {\text{10}}^{-2}\text{MeV}$

$E={\mathrm{\gamma mc}}^2$ and $P=\mathrm{\gamma mu}$, so

$\frac{E}{P}=\frac{{\text{γmc}}^2}{\text{γmu}}=\frac{{\text{c}}^2}{\text{u}}.$

As the mass of particle approaches zero, its velocity $u$ will approach $c$, so that the ratio of energy to momentum in this limit is

${\lim }_{m\mathrm{\to 0}}\frac{E}{P}=\frac{c^2}{c}=c$

which is consistent with the equation for photon energy.

(a) $3\text{.}\text{00}\times {\text{10}}^6\text{W}$

(b) Headlights are way too bright.

(c) Force is too large.

$7.28\times {\text{10}}^{–4}\text{m}$

$6.62\times {\text{10}}^7\text{m/s}$

$1.32\times {\text{10}}^{\mathrm{–13}}\text{m}$

(a) $6.62\times {\text{10}}^7\text{m/s}$

(b) $\mathrm{22.9\; MeV}$

15.1 keV

(a) 5.29 fm

(b) $4\text{.}\text{70}\times {\text{10}}^{-\text{12}}\text{J}$

(c) 29.4 MV

(a) $7.28\times {\text{10}}^{\text{12}}\text{m/s}$

(b) This is thousands of times the speed of light (an impossibility).

(c) The assumption that the electron is non-relativistic is unreasonable at this wavelength.

(a) 57.9 m/s

(b) $9\text{.}\text{55}\times {\text{10}}^{-9}\text{eV}$

(c) From Table 29.1, we see that typical molecular binding energies range from about 1eV to 10 eV, therefore the result in part (b) is approximately 9 orders of magnitude smaller than typical molecular binding energies.

29 nm,

290 times greater

$1\text{.}\text{10}\times {\text{10}}^{-\text{13}}\text{eV}$

$3\text{.}3\times {\text{10}}^{-\text{22}}\text{s}$

$2.66\times {\text{10}}^{-\text{46}}\text{kg}$

0.395 nm

(a) $1.3\times {\text{10}}^{-\text{19}}\text{J}$

(b) $2\text{.}1\times {\text{10}}^{\text{23}}$

(c) $1\text{.}4\times {\text{10}}^2\text{s}$

(a) $3\text{.}\text{35}\times {\text{10}}^5\text{J}$

(b) $1\text{.}\text{12}\times {\text{10}}^{\text{–3}}\text{kg}\cdot \text{m/s}$

(c) $1\text{.}\text{12}\times {\text{10}}^{\text{–3}}\text{m/s}$

(d) $6.23\times {\text{10}}^{\text{–7}}\text{J}$

(a) $1\text{.}\text{06}\times {\text{10}}^3$

(b) $5\text{.}\text{33}\times {\text{10}}^{-\text{16}}\text{kg}\cdot \text{m/s}$

(c) $1\text{.}\text{24}\times {\text{10}}^{-\text{18}}\text{m}$

(a) $1\text{.}\text{62}\times {\text{10}}^3\text{m/s}$

(b) $4\text{.}\text{42}\times {\text{10}}^{-\text{19}}\text{J}$ for photon, $1\text{.}\text{19}\times {\text{10}}^{-\text{24}}\text{J}$ for electron, photon energy is $3\text{.}\text{71}\times {\text{10}}^5$ times greater

(c) The light is easier to make because 450-nm light is blue light and therefore easy to make. Creating electrons with $\mathrm{7.43\; \mu eV}$ of energy would not be difficult, but would require a vacuum.

(a) $2\text{.}\text{30}\times {\text{10}}^{-6}\text{m}$

(b) $3\text{.}\text{20}\times {\text{10}}^{-\text{12}}\text{m}$

$3\text{.}\text{69}\times {\text{10}}^{-4}\mathrm{ºC}$

(a) 2.00 kJ

(b) $1\text{.}\text{33}\times {\text{10}}^{-5}\text{kg}\cdot \text{m/s}$

(c) $1\text{.}\text{33}\times {\text{10}}^{-5}\text{N}$

(d) yes

(b)

(c)

(b)

(c)

(c)

(a)

(a)

(c)

(d)