Quantization of Energy

(a) Bohr's postulates

1. Stationary orbits: An electron revolves round the nucleus only in certain fixed circular orbits. The centripetal force is provided by the electrostatic attraction of the nucleus. While moving in these orbits the electron does not radiate energy, so these orbits are called stationary (non-radiating) orbits.
2. Quantisation of angular momentum: Only those orbits are allowed in which the angular momentum of the electron is an integral multiple of $\dfrac{h}{2\pi}$:

$$mvr = \frac{nh}{2\pi}, \qquad n = 1, 2, 3, \ldots$$

3. Frequency condition: Energy is emitted or absorbed only when the electron jumps from one allowed orbit to another. The energy of the emitted (or absorbed) photon equals the difference of the energies of the two orbits:

$$h\nu = E_{n_2} - E_{n_1}$$

(b) Radius of the $n$th orbit of the hydrogen atom

Consider an electron (mass $m$, charge $-e$) moving with speed $v$ in an orbit of radius $r$ round the hydrogen nucleus (charge $+e$).

1. The electrostatic force provides the centripetal force:

$$\frac{mv^2}{r} = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2} \quad\Rightarrow\quad mv^2 r = \frac{e^2}{4\pi\varepsilon_0} \qquad \ldots(1)$$

2. Bohr's quantum condition:

$$mvr = \frac{nh}{2\pi} \quad\Rightarrow\quad v = \frac{nh}{2\pi m r} \qquad \ldots(2)$$

3. Put $v$ from (2) into (1):

$$m\left(\frac{nh}{2\pi m r}\right)^2 r = \frac{e^2}{4\pi\varepsilon_0} \quad\Rightarrow\quad \frac{n^2h^2}{4\pi^2 m r} = \frac{e^2}{4\pi\varepsilon_0}$$

4. Solve for $r$:

$$\boxed{r_n = \frac{\varepsilon_0 n^2 h^2}{\pi m e^2}}$$

So $r_n \propto n^2$. For $n = 1$, $r_1 = 0.53 \times 10^{-10}$ m $= 0.53$ Å (Bohr radius).

(c) de Broglie wavelength of an electron accelerated through 500 V

Given: $V = 500$ V, $m = 9.1 \times 10^{-31}$ kg, $h = 6.62 \times 10^{-34}$ J s, $e = 1.6 \times 10^{-19}$ C.

Kinetic energy gained: $\tfrac{1}{2}mv^2 = eV$, so momentum $p = mv = \sqrt{2meV}$.

$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}$$ $$\lambda = \frac{6.62 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 500}} = \frac{6.62 \times 10^{-34}}{\sqrt{1.456 \times 10^{-46}}} = \frac{6.62 \times 10^{-34}}{1.207 \times 10^{-23}}$$ $$\boxed{\lambda \approx 5.49 \times 10^{-11}\ \text{m} = 0.0549\ \text{nm}}$$

(a) Why X-ray production is called the inverse photoelectric effect

In the photoelectric effect, a photon of energy $h\nu$ falls on a metal and its energy is converted into the kinetic energy of an emitted electron:

$$h\nu = \phi + \tfrac{1}{2}mv^2$$

In X-ray production (Coolidge tube), fast electrons accelerated through a high voltage $V$ strike a heavy-metal target. They are suddenly stopped and their kinetic energy is converted into the energy of X-ray photons:

$$eV = \tfrac{1}{2}mv^2 \;\rightarrow\; h\nu \qquad \left(h\nu_{\max} = eV\right)$$

So X-ray production is the reverse of the photoelectric effect: there, light energy becomes electron kinetic energy; here, electron kinetic energy becomes electromagnetic (photon) energy. Hence it is called the inverse photoelectric effect.

(b) Bragg's law of X-ray diffraction

Bragg's law: When X-rays of wavelength $\lambda$ fall on a crystal, the rays reflected from successive parallel lattice planes (spacing $d$) interfere constructively only when

$$\boxed{2d\sin\theta = n\lambda}, \qquad n = 1, 2, 3, \ldots$$

where $\theta$ is the glancing angle (angle between the incident ray and the crystal plane) and $n$ is the order of diffraction.

Explanation:

        incident          reflected
           1 \   θ     θ  / 1'
              \          /
   ------------A--------------------  plane 1
              /|\
             / | \        d (spacing)
          M /  |  \ N
   --------------B------------------  plane 2
           2 ray reflected at B -> 2'
   Extra path of ray 2 = MB + BN = 2 d sinθ

1. A crystal has atoms arranged in parallel planes separated by distance $d$. Each plane partly reflects the X-rays.
2. Ray 1 is reflected at A on the first plane; ray 2 goes deeper and is reflected at B on the second plane.
3. Draw AM and AN perpendicular to the incident and reflected parts of ray 2. Ray 2 travels an extra distance

$$MB + BN = d\sin\theta + d\sin\theta = 2d\sin\theta$$

4. The reflected rays reinforce (give a bright maximum) when this path difference is a whole number of wavelengths, $2d\sin\theta = n\lambda$.

Bragg's law is used to find the wavelength of X-rays (if $d$ is known) or the spacing of atomic planes in a crystal (if $\lambda$ is known).

a) Bohr's postulates

1. Stationary orbits: An electron revolves round the nucleus only in certain fixed circular orbits. The centripetal force is provided by the electrostatic attraction of the nucleus. While moving in these orbits the electron does not radiate energy, so these orbits are called stationary (non-radiating) orbits.
2. Quantisation of angular momentum: Only those orbits are allowed in which the angular momentum of the electron is an integral multiple of $\dfrac{h}{2\pi}$:

$$mvr = \frac{nh}{2\pi}, \qquad n = 1, 2, 3, \ldots$$

3. Frequency condition: Energy is emitted or absorbed only when the electron jumps from one allowed orbit to another. The energy of the emitted (or absorbed) photon equals the difference of the energies of the two orbits:

$$h\nu = E_{n_2} - E_{n_1}$$

b) (i) de Broglie wavelength of a neutron of kinetic energy 150 eV

Given: $K = 150$ eV $= 150 \times 1.6 \times 10^{-19}$ J $= 2.4 \times 10^{-17}$ J, $m = 1.675 \times 10^{-27}$ kg, $h = 6.62 \times 10^{-34}$ J s.

Since $K = \dfrac{p^2}{2m}$, $p = \sqrt{2mK}$ and

$$\lambda = \frac{h}{\sqrt{2mK}} = \frac{6.62 \times 10^{-34}}{\sqrt{2 \times 1.675 \times 10^{-27} \times 2.4 \times 10^{-17}}} = \frac{6.62 \times 10^{-34}}{\sqrt{8.04 \times 10^{-44}}} = \frac{6.62 \times 10^{-34}}{2.836 \times 10^{-22}}$$ $$\boxed{\lambda \approx 2.33 \times 10^{-12}\ \text{m}}$$

b) (ii) Proton and electron with the same kinetic energy

$$\lambda = \frac{h}{\sqrt{2mK}} \quad\Rightarrow\quad \lambda \propto \frac{1}{\sqrt{m}} \text{ for the same } K$$

The electron has a much smaller mass than the proton ($m_p \approx 1836\, m_e$), so the electron has the greater de Broglie wavelength (about $\sqrt{1836} \approx 43$ times that of the proton).

Consider an electron (mass $m$, charge $-e$) moving with speed $v_n$ in the $n$th orbit of radius $r_n$ round the hydrogen nucleus (charge $+e$).

1. The electrostatic attraction provides the centripetal force:

$$\frac{mv_n^2}{r_n} = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r_n^2} \quad\Rightarrow\quad mv_n^2\, r_n = \frac{e^2}{4\pi\varepsilon_0} \qquad \ldots(1)$$

2. Bohr's quantum condition:

$$mv_n r_n = \frac{nh}{2\pi} \qquad \ldots(2)$$

3. Divide (1) by (2):

$$\frac{mv_n^2 r_n}{mv_n r_n} = \frac{e^2/4\pi\varepsilon_0}{nh/2\pi} \quad\Rightarrow\quad v_n = \frac{e^2}{4\pi\varepsilon_0}\cdot\frac{2\pi}{nh}$$

4. Simplify:

$$\boxed{v_n = \frac{e^2}{2\varepsilon_0 n h}}$$

So $v_n \propto \dfrac{1}{n}$. For the ground state ($n = 1$), $v_1 = \dfrac{e^2}{2\varepsilon_0 h} \approx 2.18 \times 10^6$ m s$^{-1}$.

First excited state: this is $n = 2$. Putting $n = 2$:

$$\boxed{v_2 = \frac{e^2}{4\varepsilon_0 h}} = \frac{v_1}{2}$$

Numerically,

$$v_2 = \frac{(1.6 \times 10^{-19})^2}{4 \times 8.85 \times 10^{-12} \times 6.63 \times 10^{-34}} \approx 1.09 \times 10^6\ \text{m s}^{-1}$$

Given: $V = 300$ V, $m = 9.1 \times 10^{-31}$ kg, $h = 6.6 \times 10^{-34}$ J s, $e = 1.6 \times 10^{-19}$ C (standard value).

The electron gains kinetic energy $eV$: $\tfrac{1}{2}mv^2 = eV$, so its momentum is $p = \sqrt{2meV}$.

$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}$$ $$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 300}} = \frac{6.6 \times 10^{-34}}{\sqrt{8.736 \times 10^{-47}}} = \frac{6.6 \times 10^{-34}}{9.35 \times 10^{-24}}$$ $$\boxed{\lambda \approx 7.06 \times 10^{-11}\ \text{m} = 0.0706\ \text{nm}}$$

Heisenberg's uncertainty principle: It is impossible to measure both the position and the momentum of a particle simultaneously with perfect accuracy. The product of the uncertainty in position $\Delta x$ and the uncertainty in momentum $\Delta p$ is at least of the order of $h/4\pi$:

$$\Delta x \, \Delta p \geq \frac{h}{4\pi}$$

Consider an electron (mass $m$, charge $-e$) moving with speed $v_n$ in the $n$th orbit of radius $r_n$ round the hydrogen nucleus (charge $+e$).

1. The electrostatic attraction provides the centripetal force:

$$\frac{mv_n^2}{r_n} = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r_n^2} \quad\Rightarrow\quad mv_n^2\, r_n = \frac{e^2}{4\pi\varepsilon_0} \qquad \ldots(1)$$

2. Bohr's quantum condition:

$$mv_n r_n = \frac{nh}{2\pi} \qquad \ldots(2)$$

3. Divide (1) by (2):

$$\frac{mv_n^2 r_n}{mv_n r_n} = \frac{e^2/4\pi\varepsilon_0}{nh/2\pi} \quad\Rightarrow\quad v_n = \frac{e^2}{4\pi\varepsilon_0}\cdot\frac{2\pi}{nh}$$

4. Simplify:

$$\boxed{v_n = \frac{e^2}{2\varepsilon_0 n h}}$$

So $v_n \propto \dfrac{1}{n}$. For the ground state ($n = 1$), $v_1 = \dfrac{e^2}{2\varepsilon_0 h} \approx 2.18 \times 10^6$ m s$^{-1}$.

Excitation potential: The minimum accelerating potential (in volts) through which an electron must be accelerated so that, on colliding with an atom, it can raise the atom from its ground state to an excited (higher) state. For hydrogen, the first excitation potential is $10.2$ V (energy needed to go from $n = 1$ to $n = 2$ is $10.2$ eV).

Ionization potential: The minimum accelerating potential through which an electron must be accelerated so that, on colliding with an atom, it can completely remove an electron from the ground state of the atom (take it to $n = \infty$). For hydrogen, the ionization potential is $13.6$ V (ionization energy $13.6$ eV).

Energy values for n = 4 and n = 5

For hydrogen, $E_n = -\dfrac{13.6}{n^2}$ eV (check: $E_2 = -13.6/4 = -3.4$ eV, $E_3 = -13.6/9 = -1.51$ eV, as in the figure).

$$E_4 = -\frac{13.6}{4^2} = -\frac{13.6}{16} = \boxed{-0.85\ \text{eV}}$$ $$E_5 = -\frac{13.6}{5^2} = -\frac{13.6}{25} = \boxed{-0.544\ \text{eV}}$$

Number of possible transitions

With 5 levels, an electron can jump down from any level to any lower level. Number of different transitions:

$$\frac{n(n-1)}{2} = \frac{5 \times 4}{2} = 10$$

 n=5  -0.544 eV ------------------------------
 n=4  -0.85  eV ------------------------------
 n=3  -1.51  eV ------------------------------
 n=2  -3.40  eV ------------------------------
 n=1  -13.6  eV ------------------------------
 Transitions: 5->4, 5->3, 5->2, 5->1, 4->3,
              4->2, 4->1, 3->2, 3->1, 2->1   (10 in all)

So 10 different transitions (10 spectral lines) are possible among the five levels. (Among the first three levels alone, only the 3 shown, $\lambda_{31}$, $\lambda_{32}$, $\lambda_{21}$, are possible.)

Bragg's law: When X-rays of wavelength $\lambda$ fall on a crystal, the rays reflected from successive parallel lattice planes (spacing $d$) interfere constructively only when

$$\boxed{2d\sin\theta = n\lambda}, \qquad n = 1, 2, 3, \ldots$$

where $\theta$ is the glancing angle (angle between the incident ray and the crystal plane) and $n$ is the order of diffraction.

Explanation:

        incident          reflected
           1 \   θ     θ  / 1'
              \          /
   ------------A--------------------  plane 1
              /|\
             / | \        d (spacing)
          M /  |  \ N
   --------------B------------------  plane 2
           2 ray reflected at B -> 2'
   Extra path of ray 2 = MB + BN = 2 d sinθ

1. A crystal has atoms arranged in parallel planes separated by distance $d$. Each plane partly reflects the X-rays.
2. Ray 1 is reflected at A on the first plane; ray 2 goes deeper and is reflected at B on the second plane.
3. Draw AM and AN perpendicular to the incident and reflected parts of ray 2. Ray 2 travels an extra distance

$$MB + BN = d\sin\theta + d\sin\theta = 2d\sin\theta$$

4. The reflected rays reinforce (give a bright maximum) when this path difference is a whole number of wavelengths, $2d\sin\theta = n\lambda$.

Bragg's law is used to find the wavelength of X-rays (if $d$ is known) or the spacing of atomic planes in a crystal (if $\lambda$ is known).

(i) Postulates of Bohr's model

1. An electron revolves round the nucleus in certain stable circular orbits without radiating energy (stationary orbits).
2. The allowed orbits are those in which the angular momentum of the electron is an integral multiple of $h/2\pi$: $mvr = \dfrac{nh}{2\pi}$, $n = 1,2,3,\dots$
3. Radiation is emitted or absorbed only when the electron jumps between orbits, with the photon energy $hf = E_i - E_f$.

(ii) Radius of the $n$th orbit

For an electron of mass $m$ in the $n$th orbit of radius $r_n$ around a proton, the Coulomb force provides the centripetal force:

$$\frac{mv^2}{r_n} = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r_n^2} \quad\Rightarrow\quad mv^2 r_n = \frac{e^2}{4\pi\varepsilon_0}$$

From the quantum condition, $v = \dfrac{nh}{2\pi m r_n}$. Substituting,

$$m\,\frac{n^2h^2}{4\pi^2m^2r_n^2}\,r_n = \frac{e^2}{4\pi\varepsilon_0}$$ $$r_n = \frac{n^2h^2\varepsilon_0}{\pi m e^2}$$

For $n=1$ this gives the Bohr radius $r_1 = 0.529$ Å, so $r_n = 0.529\,n^2$ Å, i.e. $r_n \propto n^2$.

(i) Bragg's law

Strong reflection (constructive interference) of X-rays from the planes of a crystal occurs when the path difference between rays reflected from successive planes is a whole number of wavelengths:

$$2d\sin\theta = n\lambda, \quad n = 1,2,3,\dots$$

where $d$ is the spacing of the planes and $\theta$ is the glancing angle.

(ii) Upper limit of wavelength

Since $\sin\theta \le 1$, we need $n\lambda \le 2d$. The largest wavelength is for $n = 1$ and $\sin\theta = 1$:

$$\lambda_{\max} = 2d = 2\times10\ \text{Å} = 20\ \text{Å} = 2\times10^{-9}\ \text{m}$$

X-rays of wavelength greater than 20 Å cannot show diffraction from this crystal.

(a) Significance of negative energy

The negative energy shows that the electron is bound to the nucleus by electrostatic attraction. Zero energy is taken for a free electron at infinity, so energy equal to $|E_n|$ must be supplied to remove the electron from the atom. The more negative the energy, the more tightly the electron is bound.

(b) Transition that emits a 275 nm photon

Given: $\lambda = 275$ nm $= 275 \times 10^{-9}$ m, $h = 6.64 \times 10^{-34}$ J s, $c = 3.0 \times 10^8$ m s$^{-1}$.

Energy of the photon:

$$E = \frac{hc}{\lambda} = \frac{6.64 \times 10^{-34} \times 3.0 \times 10^8}{275 \times 10^{-9}} = 7.24 \times 10^{-19}\ \text{J}$$ $$E = \frac{7.24 \times 10^{-19}}{1.6 \times 10^{-19}}\ \text{eV} \approx 4.5\ \text{eV}$$

Energy released in each transition:

Only transition B releases 4.5 eV, so transition B emits the photon of wavelength 275 nm.

(c) Wavelength emitted when the electron jumps from $n_2$ to $n_1$

1. Energy of the electron in the $n$th orbit of hydrogen (Bohr theory):

$$E_n = -\frac{me^4}{8\varepsilon_0^2 n^2 h^2}$$

2. By Bohr's frequency condition, $h\nu = E_{n_2} - E_{n_1}$:

$$h\nu = \frac{me^4}{8\varepsilon_0^2 h^2}\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

3. Put $\nu = c/\lambda$:

$$\frac{1}{\lambda} = \frac{me^4}{8\varepsilon_0^2 h^3 c}\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

4. Writing $R = \dfrac{me^4}{8\varepsilon_0^2 h^3 c}$ (Rydberg constant, $\approx 1.097 \times 10^7$ m$^{-1}$):

$$\boxed{\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)}$$

(d) Wavelength of the second line of the Balmer series

For the Balmer series $n_1 = 2$; the second line is the jump from $n_2 = 4$ to $n_1 = 2$.

$$\frac{1}{\lambda} = R\left(\frac{1}{2^2} - \frac{1}{4^2}\right) = 1.09 \times 10^7 \times \frac{3}{16} = 2.044 \times 10^6\ \text{m}^{-1}$$ $$\boxed{\lambda = \frac{1}{2.044 \times 10^6} \approx 4.89 \times 10^{-7}\ \text{m} = 489\ \text{nm}}$$

(i) Bohr's postulates

1. Stationary orbits: An electron revolves round the nucleus only in certain fixed circular orbits. The centripetal force is provided by the electrostatic attraction of the nucleus. While moving in these orbits the electron does not radiate energy, so these orbits are called stationary (non-radiating) orbits.
2. Quantisation of angular momentum: Only those orbits are allowed in which the angular momentum of the electron is an integral multiple of $\dfrac{h}{2\pi}$:

$$mvr = \frac{nh}{2\pi}, \qquad n = 1, 2, 3, \ldots$$

3. Frequency condition: Energy is emitted or absorbed only when the electron jumps from one allowed orbit to another. The energy of the emitted (or absorbed) photon equals the difference of the energies of the two orbits:

$$h\nu = E_{n_2} - E_{n_1}$$

(ii) Frequency of the photon emitted from the second excited state to the ground level

The ground level is $n = 1$ ($E_1 = -13.6$ eV). The first excited state is $n = 2$ ($-3.40$ eV), so the second excited state is $n = 3$, with $E_3 = -1.51$ eV.

Energy of the photon:

$$h\nu = E_3 - E_1 = -1.51 - (-13.6) = 12.09\ \text{eV}$$ $$h\nu = 12.09 \times 1.6 \times 10^{-19} = 1.934 \times 10^{-18}\ \text{J}$$

Taking $h = 6.62 \times 10^{-34}$ J s:

$$\nu = \frac{1.934 \times 10^{-18}}{6.62 \times 10^{-34}}$$ $$\boxed{\nu \approx 2.92 \times 10^{15}\ \text{Hz}}$$

This is a line of the Lyman series (ultraviolet region).