Chapter 6

Bunsen’s burner

The wavelength of the radiation maximum decreases with increasing temperature.

$T_{\alpha }\text{/}{T}_{\beta }=1\text{/}\sqrt{3}\cong 0.58,$ so the star $\beta$ is hotter.

$3.3\times {10}^{\mathrm{-19}}\text{J}$

No, because then $\text{Δ}E\text{/}E\approx {10}^{\mathrm{-21}}$

$\mathrm{-0.91}$ V; 1040 nm

$h=6.40\times {10}^{\mathrm{-34}}\text{J}·\text{s}=4.0\times {10}^{\mathrm{-15}}\text{eV}·\text{s;}\text{−}3.5\%$

${(\text{Δ}\lambda )}_{\text{min}}=0\text{m}$ at a $0\text{°}$ angle; $71.0\text{pm}+0.5{\lambda }_c=72.215\text{pm}$

121.5 nm and 91.1 nm; no, these spectral bands are in the ultraviolet

$v_2=1.1\times {10}^6\text{m}\text{/}\text{s}\cong 0.0036c;$ $L_2=2\hslash K_2=3.4\text{eV}$

29 pm

$\lambda =2\pi n{a}_0=2\text{(}3.324\text{Å}\text{)}=6.648\text{Å}$

$\lambda =2.14\text{pm;}$ $K=261.56\text{keV}$

$0.052\text{°}$

doubles it

yellow

goes from red to violet through the rainbow of colors

would not differ

human eye does not see IR radiation

No

from the slope

Answers may vary

the particle character

Answers may vary

no; yes

no

right angle

no

They are at ground state.

Answers may vary

increase

for larger n

Yes, the excess of 13.6 eV will become kinetic energy of a free electron.

no

X-rays, best resolving power

proton

negligibly small de Broglie’s wavelengths

to avoid collisions with air molecules

Answers may vary

Answers may vary

yes

yes

a. 0.81 eV; b. $2.1\times {10}^{23};$ c. 2 min 20 sec

a. 7245 K; b. 3.62 μm

about 3 K

$4.835\times {10}^{18}$ Hz; 0.620 Å

263 nm; no

3.68 eV

4.09 eV

5.60 eV

a. 1.89 eV; b. 459 THz; c. 1.21 V

264 nm; UV

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

$1.66\times {10}^{\text{−}32}\text{kg}·\text{m/s}$

56.21 eV

$6.63\times {10}^{\text{−}23}\text{kg}·\text{m/s};$ 124 keV

82.9 fm; 15 MeV

(Proof)

$\text{Δ}{\lambda }_{30}\text{/}\text{Δ}{\lambda }_{45}=45.74\%$

121.5 nm

a. 0.661 eV; b. –10.2 eV; c. 1.511 eV

3038 THz

97.33 nm

a. h/$\pi ;$ b. 3.4 eV; c. – 6.8 eV; d. – 3.4 eV

$n=4$

365 nm; UV

no

7

145.5 pm

20 fm; 9 fm

a. 2.103 eV; b. 0.846 nm

80.9 pm

$2.21\times {10}^{\text{−}19}\text{m/s}$

$9.929\times {10}^{32}$

$\gamma =1060;$ 0.00124 fm

24.11 V

a. $P=2I\text{/}c=8.67\times {10}^{\mathrm{-6}}\text{N}\text{/}{\text{m}}^2;$ b. $a=PA\text{/}m=8.67\times {10}^{-4}\text{m}\text{/}{\text{s}}^2;$ c. 74.91 m/s

$x=4.965$

$7.124\times {10}^{16}{\text{W/m}}^3$

1.034 eV

$5.93\times {10}^{18}$

387.8 nm

a. $4.02\times {10}^{15};$ b. 0.508 mW

a. $4.02\times {10}^{15};$ b. 0.533 mW; c. 0.644 mA; d. 2.57 ns

a. 0.132 pm; b. 9.39 MeV; c. 0.047 MeV

a. 2 kJ; b. $1.33\times {10}^{\text{−}5}\text{kg}·\text{m/s};$ c. $1.33\times {10}^{\text{−}5}\text{N};$ d. yes

a. 0.003 nm; b. $105.56\text{°}$

$n=3$

a. $a_0\text{/}2;$ b. $\mathrm{-54.4}\text{eV}\text{/}{n}^2;$ c. $a_0\text{/}3,\mathrm{-122.4}\text{eV}\text{/}{n}^2$

a. 36; b. 18.2 nm; c. UV

396 nm; 5.23 neV

7.3 keV

728 m/s; $1.5\mu \text{V}$

$\lambda =hc\text{/}\sqrt{K(2E_0+K)}=3.705\times {10}^{\text{−}12}\text{m,}K=100\text{keV}$

$\text{Δ}{\lambda }_c^{(\text{electron})}\text{/}\text{Δ}{\lambda }_c^{(\text{proton})}=m_p\text{/}{m}_e=1836$

(Proof)

$5.1\times {10}^{17}\text{Hz}$