The Line Spectra of the Hydrogen Atom

The Line Spectra of the Hydrogen Atom

The spectrum of H-atom studied by Lyman, Balmer, Paschen, Bracken and Pfund can now be explained on the basis of Bohr's Model. It is now clear that when an electron jumps from a higher energy state to a lower energy state, the radiation is emitted in form of photons. The radiation emitted in such a transition corresponds to the spectral line in the atomic spectra of H-atom.

Lyman Series

When an electron jumps from any of the higher states to the ground state or 1st state

(n = 1), the series of spectral lines emitted lies in ultra-violet region and are called as Lyman Series. The wavelength (or wave number) of any line of the series can be given by using the relation:

$\bar{v}=R{Z}^2\left(\frac{1}{1^2}-\frac{1}{n_2^2}\right)n_2=2,3,4,5,.....$

(For H atom Z = 1)

Series limit (for H - atom): $\infty \to 1i.e.\bar{v}=R$

$\alpha$ line: 2 $\to$ 1; also known as first line or first member

$\beta$ line: 3 $\to$ 2; also known as second line or second member

$\gamma$ line: 4 $\to$ 1; also known as third line or third member

Balmer Series

When an electron jumps from any of the higher states to the state with n = 2 (IInd state), the series of spectral lines emitted lies in visible region and are called as Balmer Series. The wave number of any spectral line can be given by using the relation:

$\bar{v}=R{Z}^2\left(\frac{1}{2^2}-\frac{1}{n_2^2}\right)n_2=3,4,5,6,...........$

Series limit (for H – atom) : $\infty \to 2i.e.\bar{v}=\frac{R}{4}$

$\alpha line:3\to 2;\beta line:4\to 2;\gamma line:5\to 2$

Paschen Series

When an electron jumps from any of the higher states to the state with n = 3 (IIIrd state), the series of spectral lines emitted lies in near infra-red region and are called as Paschen Series. The wave number of any spectral line can be given by using the relation:

$\bar{v}=R{Z}^2\left(\frac{1}{3^2}-\frac{1}{n_2^2}\right)n_2=4,5,6,7,..............$

Series limit (for H – atom) : $\infty \to 3i.e.\bar{v}=\frac{R}{9}$

$\alpha line:4\to 3;\beta line:5\to 3;\gamma line:6\to 3$

Brackett Series

When an electron jumps from any of the higher states to the state with n = 4 (IVth state), the series of spectral lines emitted lies in far infra-red region and called as Brackett Series. The wave number of any spectral line can be given by using the relation:

$\bar{v}=R{Z}^2\left(\frac{1}{4^2}-\frac{1}{n_2^2}\right)n_2=5,6,7,8,...........$

Pfund Series

When an electron jumps from any of the higher states to the state with n = 5 (Vth state), the series of spectral lines emitted lies in far infra-red region and are called as Pfund Series. The wave number of any spectral line can be given by using the relation:

$\bar{v}=R{Z}^2\left(\frac{1}{5^2}-\frac{1}{n_2^2}\right)n_2=6,7,8,...........$

Short Review of formulas (for one electron atom or ions):

1. Velocity of electron in nth orbit = v_n = 2.165 x 10⁶ Z/n m/s

2. Radius of nth orbit = r_n = 0.53 x 10^–10 n²/Z m

3. Binding energy of an electron in nth state = E_n = –13.6 Z²/n² eV/atom

$E_n=-2.17\times {10}^{-18}\frac{Z^2}{n^2}J/atom=-13.6\frac{Z^2}{n^2}eV/atom$

4. Kinetic energy $=KE=1/2m{v}_n^2=KZ{e}^2/2r_n$

5. Potential energy $=PE=-kZ{e}^2/r_n$

6. Total energy of an electron $=-E_n=-kZ{e}^2/2r_n$

PE = 2TE ; PE = –2KE ; TE = –KE

7. Binding energy of an electron in nth state

$E_n=-\frac{13.6}{n^2}{Z}^{2}eV$

8. Ionisation Energy = – B.E.

$I.E.=+\frac{13.6}{n^2}{Z}^{2}eV$

9. Ionisation Potential

Ionisation potential $=\frac{I.E.}{e}=\frac{13.6}{n^2}{Z}^{2}V$

10. Excitation Energy

The energy taken up by an electron to move from lower energy level to higher energy level. Generally it defined from ground state.

$\bullet$ Ist excitation energy = transition from n₁ = 1 to n₂ = 2

$\bullet$ IInd excitation energy = transition from n₁ = 1 to n₂ = 3

$\bullet$IIIrd excitation energy = transition from n₁ = 1 to n₂ = 4 and so on ....

$\bullet$ The energy level n = 2 is also called as Ist excited state.

$\bullet$The energy level n = 3 is also called as IInd excited state. & so on ...

In general, excitation energy (DE) when an electron is excited from a lower state n₁ to any higher state n₂ is given as:

$\Delta E=13.6Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)eV$

11. Energy released when an electron jumps from a higher energy level (n₂) to a lower energy level (n₁) is given as:

$\Delta E=13.6Z^2\left(\frac{1}{n_1^2-n_2^2}\right)eV$

If v be the frequency of photon emitted and l be the wavelength, then:

$\Delta E=hv=h\frac{c}{\lambda }$

The wavelength (l) of the light emitted an also be determined by using:

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

R = 1.096 x 10⁷ /m

Important: Also remember the value of 1/R = 911.5 Å for calculation of l to be used in objectives only).

12. The number of spectral lines when an electron falls from n₂ to n₁ = 1 (i.e. to the ground state) is given by:

$No.oflines=\frac{n_2\left(n_2-1\right)}{2}$

If the electron falls from n₂ to n₁, then the number of spectral lines is given by:

$No.oflines=\frac{\left(n_2-n_1+1\right)\left(n_2-n_1\right)}{2}$

Illustration 1: A doubly ionised Lithium atom is hydrogen like with atomic number 3.

(i) Find the wavelength of radiation required to excite the electron in Li⁺⁺ from the first to the third Bohr Orbit. (Ionization energy of the hydrogen atom equals 13.6 eV).

(ii) How many spectral lines are observed in the emission spectrum of the above excited system?

Solution:

(i) $E_n=\frac{-13.6Z^2}{n^2}$

\begin{align}  \text{Excitation energy }=\text{ }\Delta E={{E}_{3}}-{{E}_{1}}=-13.6\times {{\left( 3 \right)}^{2}}\left[ \dfrac{1}{{{3}^{2}}}-\dfrac{1}{{{1}^{2}}} \right] \\  \ \ \ \ \ =+13.6\times \left( 9 \right)\left[ 1-1/9 \right]=13.6\times \left( 9 \right)\left( 8/9 \right)=108.8\ eV. \\ \end{align}

\begin{align}  Wavelength\ \lambda =\dfrac{hc}{\Delta E}=\dfrac{\left( 6.63\times {{10}^{-34}} \right)\left( 3\times {{10}^{8}} \right)}{\left( 13.6\times 8 \right)\left( 1.6\times {{10}^{-19}} \right)} \\  =\left( \dfrac{6.63\ \times \left( 3 \right)}{\left( 13.6 \right)\left( 8 \right)\left( 1.6 \right)} \right){{10}^{-7}} \\ =  =114.26\times {{10}^{-10}}\ m \\  =114.3\ \text{{}\!\!\mathrm{\AA}\!\!\text{ }} \\ \end{align}

(ii) From the excited state (E₃), coming back to ground state, there can be ³C₂ = 3 possible radiations.

X-RAYS

X-rays were discovered accidentally by Rontgen in 1895. The first Nobel Prize was awarded to Rontgen in 1901. This highly penetrating electromagnetic radiation has proved to be a very powerful tool to study the crystal structure, in material research, in the radiography of metals and in medical sciences. Laue, Henry and Lawrence Bragg, Barkla, Siegbahn were some of the Nobel Laureates who have made contribution to these studied. $\gamma -$ray spectroscopy and electron-spectroscopy were some of the spin-offs of these studies apart from the discovery of elements.

Experimental production of X-rays and the Bragg spectrometer:

Electrons from a a heated element were accelerated by very high potential and made to impinge on the target (anode). The X-rays produced are collimated by parallel plates and are incident on a crystal (LiF, quartz, diamond, etc.) As the inter-atomic distance is of the same order as the wavelength of X-rays diffraction is produced and they are detected by counters or photographic plates.

X – Rays has following property

(i) Short wavelength (0.1 A° to 1 A°) electromagnetic radiation.

(ii) Are produced when a metal anode is bombarded by very high energy electrons.

(iii) Are not affected by electric and magnetic field.

(iv) They cause photoelectric emission.

Characteristics equation eV = hv_m

e = electron charge;

V = accelerating potential

v_m = maximum frequency of X radiation

(v) Intensity of X – rays depends on number of electrons hitting the target.

(vi) Cut off wavelength or minimum wavelength, where v(in volts) is the p.d. applied to the tube ${\lambda }_{\min }\cong \frac{12400}{V}$A°.

(vii) Continuous spectrum due to retardation of electrons.

(viii) Characteristic Spectrum due to transition of electron from higher to lower

$v\propto {\left(z-b\right)}^2;\upsilon =a{\left(z-b\right)}^2\left[\text{Moseley }{}^{\prime }\text{ s Law}\right]$

b = 1 for K$\alpha$ ; B = 7.4 for L$\alpha$

Where b is Shielding factor (different for different series).

(ix) Bragg's Law 2 d sin $\theta$ = n $\lambda$

($\theta$ = angle for max intensity)

Note: (a) Binding energy = – [Total Mechanical Energy]

(b) Vel. of electron in n^th orbit for hydrogen atom $\cong \frac{c}{137n}$;

c = speed of light.

(c) For x – rays $\frac{1}{\lambda }=R{\left(z-b\right)}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

(d) Series limit of series means minimum wave length of that series.

THEORETICAL EXPLANATION

Production of Continuous Spectrum

The accelerated electrons are suddenly stopped by the target. According to Maxwell's theory, wherever a charged particle is accelerated or decelerated, they emit radiation. This is called Bremsstrahlung or braking radiation.

When the whole of the energy of the electron is converted to radiation, one gets the maximum energy or$\lambda$ minimum. If V is the potential difference applied, it is converted to the kinetic energy of the electron. If Ve = E = hv, then the wave-length $\lambda =\frac{hc}{E}$ . hc = 12400 eVÅ, and if E is given in eV, ${\lambda }_{\min .}=\frac{hc}{E(eV)}\left(eV.\stackrel{\circ }{A}\right).$ ${\lambda }_{\min .}$ is directly given in Å.

When the electrons lose their energy by multiple collisions and penetration inside the target, the radiation, produced has less energy

The energy that is not converted to radiation only heats up the target. It has to be cooled property to prevent damage to the X-ray tube.

The maximum value of intensity is approximately at $\frac{3}{2}{\lambda }_{\min .}$ (This is often marked ${\lambda }_{\min .}$ Actually it is not the maximum value of $\lambda$ but $\lambda$ at intensity maximum).

The distribution of intensity depends on the material of the target, the current and the potential difference applied.

I $\alpha$ V² and I $\alpha$ current

The current is normally in milli-amperes and the voltage in kilovolts.

Illustration 2: If the wavelength of the $K_{\alpha }$ line of platinum is 0.2Å, what is the energy needed to excite this line? (the energy of the first level of platinum, $\left|E_{n-1}\right|$ = 81keV). What is the corresponding absorption energy? What is the ionization energy of platinum?

Solution: The energy corresponding to Q.2Å is $\Delta E=\frac{12400eV-\stackrel{\circ }{A}}{0.2\stackrel{\circ }{A}}$

= 62 x 10³ eV = 62 KeV

But if one gives 62 keV energy to the electron in the x-ray tube, one cannot et $K_{\alpha }$ x-ray line because the higher levels are full. $K_{\alpha }$line can be excited only when a vacancy is created in the ground state, i.e to remove the electron from the k-level. The energy required is obviously the ionization energy of the k-level. which is 81 keV. There is no absorption line corresponding to the $K_{\alpha }$emission line. In x-rays one gets only absorption edge (see the theory part).

The ionization energy of the k-level is same as the k-absorption edge which is the energy of excitation of the k-electron. This is 81 keV for platinum.