Chapter 12: Atoms (Modern Physics)

• (a) no different from (both are about $10^{-10}$ m).
• (b) Thomson's model; Rutherford's model (the orbiting electron always feels the nucleus's pull).
• (c) Rutherford's model (an accelerating, orbiting electron radiates energy and spirals in).
• (d) Thomson's model; Rutherford's model (mass concentrated in a tiny nucleus).
• (e) both the models.

(a) no different from; (b) Thomson's, Rutherford's; (c) Rutherford's; (d) Thomson's, Rutherford's; (e) both the models.

The hydrogen nucleus is a single proton, about four times lighter than an alpha particle, and its charge is only $+e$ (gold has $+79e$).

• A heavy alpha particle hitting a lighter proton cannot bounce back; it just keeps going with a small deflection (like a heavy ball hitting a light one).
• So there would be no large-angle scattering and no back-scattering. The alphas would pass through almost undeflected, and some protons would be knocked forward.

Such an experiment could not reveal the small, heavy nucleus the way the gold foil did.

Almost no large-angle scattering: the light proton cannot turn back the heavier alpha particle, so the nucleus would not be revealed.

$$\nu =\frac{\Delta E}{h}=\frac{2.3\times 1.6\times 10^{-19}}{6.63\times 10^{-34}}\approx 5.6\times 10^{14}\text{Hz}$$

About $5.6\times 10^{14}$ Hz.

In the Bohr model, $K=-E$ and $U=2E$:

• Kinetic energy $K=+13.6$ eV
• Potential energy $U=-27.2$ eV

(Check: $K+U=13.6-27.2=-13.6$ eV.)

K = 13.6 eV; U = −27.2 eV.

$$\Delta E=E_4-E_1=-\frac{13.6}{16}-(-13.6)=13.6\left(1-\frac{1}{16}\right)=12.75\text{eV}$$

$$\nu =\frac{\Delta E}{h}=\frac{12.75\times 1.6\times 10^{-19}}{6.63\times 10^{-34}}\approx 3.1\times 10^{15}\text{Hz},\lambda =\frac{c}{\nu }\approx 9.7\times 10^{-8}\text{m}=97\text{nm}$$

λ ≈ 97 nm; ν ≈ $3.1\times 10^{15}$ Hz.

(a) $v_n=\frac{e^2}{2{\epsilon }_{0}nh}=\frac{2.18\times 10^6}{n}$ m/s:

• $v_1=2.18\times 10^6$ m/s
• $v_2=1.09\times 10^6$ m/s
• $v_3=7.27\times 10^5$ m/s

(b) $T=\frac{2\pi r_n}{v_n}$ with $r_n=0.53\times 10^{-10}{n}^2$ m:

$$T_1=\frac{2\pi \times 0.53\times 10^{-10}}{2.18\times 10^6}\approx 1.53\times 10^{-16}\text{s}$$

Since $r\propto n^2$ and $v\propto \frac{1}{n}$, $T\propto n^3$:

• $T_2=8T_1\approx 1.22\times 10^{-15}$ s
• $T_3=27T_1\approx 4.12\times 10^{-15}$ s

(a) $2.18\times 10^6$, $1.09\times 10^6$, $7.27\times 10^5$ m/s. (b) $1.53\times 10^{-16}$, $1.22\times 10^{-15}$, $4.12\times 10^{-15}$ s.

$r_n=n^2{r}_1$:

• $r_2=4\times 5.3\times 10^{-11}=2.12\times 10^{-10}$ m
• $r_3=9\times 5.3\times 10^{-11}=4.77\times 10^{-10}$ m

$2.12\times 10^{-10}$ m and $4.77\times 10^{-10}$ m.

At room temperature the atoms are in the ground state ($-13.6$ eV). With 12.5 eV they can reach at most

$$E=-13.6+12.5=-1.1\text{eV}$$

$E_3=-1.51$ eV is reachable, but $E_4=-0.85$ eV is not. So atoms can be excited up to $n=3$, and they emit:

• $3\to 1$: $\lambda =\frac{1240}{12.09}\approx 102.6$ nm (Lyman series)
• $2\to 1$: $\lambda =\frac{1240}{10.2}\approx 121.6$ nm (Lyman series)
• $3\to 2$: $\lambda =\frac{1240}{1.89}\approx 656$ nm (Balmer series, red H-alpha)

Two Lyman lines (102.6 nm and 121.6 nm) and one Balmer line (656 nm).

Bohr's condition: $mvr=\frac{nh}{2\pi }$, so

$$n=\frac{2\pi mvr}{h}=\frac{2\pi \times 6.0\times 10^{24}\times 3\times 10^4\times 1.5\times 10^{11}}{6.63\times 10^{-34}}\approx 2.6\times 10^{74}$$

Such an enormous quantum number means the allowed orbits are so close together that the Earth's motion looks completely continuous: quantisation does not show up for large objects.

$n\approx 2.6\times 10^{74}$.