JEE Physics Atoms Notes Set 04

Atomic Models and Bohr's Postulates

Thomson pictured the atom as a uniform sphere of positive charge with electrons embedded in it. Rutherford's alpha-particle scattering showed that nearly all the mass and positive charge are concentrated in a tiny nucleus (about \( 10^{-15} \) m) with the electrons moving around it. Most alpha particles pass straight through, a few are deflected at large angles, and about one in \( 10^4 \) to \( 10^5 \) returns almost backwards.
The distance of closest approach of an alpha particle of kinetic energy K to a nucleus of charge Ze is \( r_0=\frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{K} \). The impact parameter b and scattering angle \( \theta \) are related by \( b=\frac{r_0}{2}\cot\frac\theta2 \); small b means large deflection.
The classical model fails because an orbiting charge would radiate and spiral into the nucleus. Bohr (for hydrogen and hydrogen-like ions) therefore postulated: (1) the electron moves in certain stable circular orbits without radiating; (2) in these orbits the angular momentum is quantised, \( mvr=\frac{nh}{2\pi} \); (3) a photon is emitted or absorbed when the electron jumps between orbits, with \( hf=E_i-E_f \).

Orbit Radius, Speed and Energy

Combining the Coulomb force with the quantisation rule gives, for a one-electron ion of atomic number Z:
Radius \( r_n=0.529\,\frac{n^2}{Z}\ \text{Å} \)
Speed \( v_n=2.19\times10^{6}\,\frac{Z}{n}\ \text{m/s} \)
Total energy \( E_n=-13.6\,\frac{Z^2}{n^2}\ \text{eV} \); kinetic energy \( K=-E_n \); potential energy \( U=2E_n \)
Time period \( T\propto\frac{n^3}{Z^2} \), frequency \( f\propto\frac{Z^2}{n^3} \), centripetal acceleration \( \propto\frac{Z^3}{n^4} \), and the current of the orbiting electron \( \propto\frac{Z^2}{n^3} \).
Ionisation energy from level n is \( 13.6\frac{Z^2}{n^2} \) eV. The first excitation energy of hydrogen (n = 1 to 2) is 10.2 eV. A sample of atoms excited to level N can emit up to \( \frac{N(N-1)}{2} \) different spectral lines; the nth excited state is level n + 1.

Spectral Series of Hydrogen

The wavelength emitted in a jump from \( n_2 \) to \( n_1 \) is given by \( \frac1\lambda=RZ^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \), where \( R=1.097\times10^{7}\ \text{m}^{-1} \).
Lyman series: ends at n = 1 (ultraviolet). Balmer: ends at n = 2 (visible, 6563 Å to 3646 Å). Paschen: ends at n = 3 (infrared). Brackett: n = 4, Pfund: n = 5. The longest wavelength of a series comes from the jump from the next level; the shortest (series limit) from n = infinity. At room temperature atoms are in the ground state, so only the Lyman series appears in absorption.

Excitation by Collisions and Photon Recoil

An atom can absorb only photons whose energy equals the gap between two levels. A bombarding electron, in contrast, can transfer any energy up to its own kinetic energy, but the atom takes only the amount that matches an energy gap, and the electron keeps the rest.
For a heavy projectile of mass m hitting an atom of mass M at rest, the largest energy that can be absorbed in the inelastic collision is \( \frac{M}{m+M}K \). With equal masses (neutron on hydrogen) it is K/2, so a neutron needs at least 20.4 eV to excite hydrogen.
When an atom emits a photon of energy E it recoils with momentum \( p=\frac{E}{c} \) and speed \( v=\frac{E}{Mc} \), where M is the mass of the atom.

X-rays: Continuous and Characteristic Spectra, Moseley's Law

In an X-ray tube, electrons accelerated through a potential difference V strike a heavy target. The braking of the electrons gives a continuous spectrum with a sharp short-wavelength limit \( \lambda_{min}=\frac{hc}{eV}=\frac{12400}{V\ (\text{volt})}\ \text{Å} \), which depends only on V and not on the target. On top of this appear sharp characteristic lines produced when a vacancy in an inner shell is filled by an electron from a higher shell: \( K_\alpha \) (L to K), \( K_\beta \) (M to K), \( L_\alpha \) (M to L). The wavelengths of these lines depend only on the target material.
Moseley's law: \( \sqrt{f}=a(Z-b) \); for \( K_\alpha \) lines \( f=\frac{3Rc}{4}(Z-1)^2 \), so \( \sqrt f \) against Z is a straight line.
Increasing the tube voltage lowers \( \lambda_{min} \) but leaves characteristic wavelengths unchanged. Increasing the filament current raises intensity but not the energy of the photons. Higher-frequency X-rays are more penetrating (hard X-rays).

Worked Examples

Worked Example 1. Find the wavelength of the first line of the Balmer series and the energy needed to ionise hydrogen from n = 2.
Answer:
First Balmer line is 3 to 2: \( \frac1\lambda=1.097\times10^7\left(\frac14-\frac19\right)=1.524\times10^6 \), so \( \lambda=6563\ \text{Å} \).
Ionisation energy from n = 2 is \( \frac{13.6}{4}=3.4\ \text{eV} \).

Worked Example 2. A hydrogen-like ion has its ground state energy -54.4 eV. Identify the ion and its first orbit radius.
Answer:
\( 13.6Z^2=54.4\Rightarrow Z=2 \), the ion is He⁺.
Radius \( =\frac{0.529}{2}=0.265\ \text{Å} \).

Worked Example 3. Find the minimum wavelength of X-rays from a tube operating at 31 kV.
Answer:
\( \lambda_{min}=\frac{12400}{31000}=0.40\ \text{Å} \).

Worked Example 4. An electron in the n = 4 level of hydrogen can return to the ground state. How many different photons can be emitted?
Answer:
Number of lines \( =\frac{4\times3}{2}=6 \).