BOHR MODEL & X-RAYS

BOHR MODEL & X-RAYS

Physics: Modern Physics

Neils Bohr gave forward his model of atom with answeres most of the flaws that Rutherford model had. This model introduced quantization principal and this model is regarded as first quantum mechanical model. We will explore in detail, this structure of atom.

We will also look into X - ray formation which is a converse process of photoclectric effect and will examine Moreley's law in pantium.

THE BOHR THEORY OF HYDROGE ATOM (AND HYDROGEN LIKE ATOM)
Bohr gave following postulates for electron in hydrogen atom :-
• An electron in an atom could resolve in certain stable orbits without the emission of radiant energy.

• An electron resolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of . Thus angular of electron is QUANTAIZED.
L = , where n principal quantum number.

• An electron might a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, a single photon is amitted having energy equal to the energy difference between the initial and final slater.
The frequency of emitted photon is given by
h = E_i - E_f

Dumb Question:- Are these equations valid for all atoms ?
Ans:- No, equations are valid only for single electron system, i.e. systems where only one electron is present. Ideally this equation applies only to Hydrogen atom, Singly ionized helium He⁺, and doubly ionized Lithium Li^{+ +}.

In Bohr's model, radius of nth orbit is given by r_n = and speed of electron is given by V_n =
if values of constant are evaluate then,
r_n = 0.529 n²/z and Vⁿ = 2.19 X 10⁶ m/sec.
where n Principal quantum number/shell number.
z atomic number of nucleus involved.

Why ??
from Bohr's postulate, angular momentum of electron in nth orbit is L = also from mechanics L = mv_nr_n
mv_nr_n = ............................................... (1)
for electron to resolve in its orbit, Electrostatic force is providing the required centripetal force.




Bohr Energy levels:- The energy of a revolving electron is :-
(i) Kinetic Energy, due to its revolution in the orbit.
(ii) Electrostatic Potential Energy.

The total energy E_u is the sum of kinetic and potential energies.
E_u = K_u + V_u = -
if the value of constants are placed then
E_u =

Dumb Question:- What is the significance of total energy of e^- as a negative quantity?
Ans:- Total energy as negative quantity sindicates that it is held in a stable state and energy needs to be supplied to it if electron has to be pulled out.

Dumb Question:- Where has the zero of potential energy been let ?
Ans:- The zero of electrostatic potential energy has been setup at infinity i.e. outside the bounds of atom.
If an electron moves from n_i orbit to lower n_f orbit then it lose, energy which is emitted in the form of radiations and the wavelength of such radiation is given by formula.
, Where R = Rydberg's constant = 1.097 x 10⁷ m^{- 1}

Why ??
energy of an orbit is given by E_n = -
from Bohr's postulate when an electron falls from higher orbit to lower orbit then energy of emitted radiation is given by
E = E_i - E_f =

if R = (constant)
then

Few named transitions :-
Lymen Series n = 2, 3, 4,...............
Balmen Series n = 3, 4, 5,...............
Paschen Series n = 4, 5, 6,...............
Brackett Series n = 5, 6, 7,...............
Pfund Series n = 6, 7, 8,...............
(In all these cases electron is moving down from nth orbit to concerned orbit)

• Total number of emission lines from same excited state n₁ to another state n₂ (<>1) is given by

Solving (1) & (2) simultaneously yields the required result.
few handy difinations:-
IONIZATION POTENTIAL:- The minimum energy needed to ionize an atom is called ionization energy, and the potential defference through which an electron should be accelerated to acquire this energy is called ionization potential. The ioniozation energy of hydrogen atom is ground state is 13.6 ev and ioniozation potentiation is 13.6 v.

BINDING ENERGY:- Binding enrgy of a system is defined as energy liberated when its constituents are brought from infinity to form the system. For hydrogen atom binding energy is same as its ionization energy.

EXCITATION ENERGY:- The energy required to take an atom from its ground state to an excited state is ncalled excitation energy of that excited state, and the potential.

ILLUSTRATION:- Find the longest and shortest wavelength in the Lyman Series for Lydrogen. In what region of the electromagnetic spectrum does each series lie ?
Solution:- The transition equation for Lyman Series is given by,
, n = 2, 3, ........................
The longest wavelength is corresponding to n = 0
= 1.099 x 10₇(1 - 1/4) = 0.823 x 1s₇
= 1.254 x 10_{- 7}m = 1215
The shortest wavelength corresponds to n =
= 1.097 x 10⁷(1 - )
_min = 0.911 x 10^-7 m = 911
AS can be seen both, the weavelength lie in ultraviolet (UV) region of electromagnetic spectrum.

X-RAYS
Electromagnetic radiations with wavelength from 0.1A⁰ to 100A⁰ falls into the category of X - rays, however the boundaries of this catygory are not sharp.
• The process of production of X-rays is reserve of photoelectric efffect. Here electromagnetic radiations are emitted by the bombordment of high speed thermionically emitted electrons on the metal surface.
X-rays produced are of two types:-

(a) CONTIMUOUS X-RAYS:- When electrons of vary high kinetic energy accelarates in an electric field then according to Maxwells theory, it radiates energy in the form of electromagnetics waves X-rays produced in this fashion are called continuous X-rays (also called Bremsstrahlung).
• Continuous X-rays produced at a given accelerating potential V very in wavelength, but none has wavelength shorten than a certain value _min.
, V potential aeron which electron is accelerated.

Dumb Question:- What is the significance of this minimum wavelength of the X-Ray ?
Ans:- This minimum wavelength corresponds to the maximum energy of the X-ray produced. This condition occures when striking electron looses all its energy and this energy gets converted into X-rays. This for it
Energy of X-ray = total kinetic energy of electron = energy gained when accelerated across potential barrier

Note That the maximum wavelength 0 , this is the case when striking electron looses very small fraction of its energy.

CHARACTERISTICS X-RAYS:- The sharp lines superimposed on the continuous spectrum are known as characteristic X-rays. because they are characteristic of the target material.

s

• Characteristic X-rays emission occures when a banbarding electron that collides with a target atom has sufficient energy to remove on inner shell electron from the atom.
• The vacancy created in the shell is filled when an electron from a higher level drops down into it. This transition is of a photon whose energy equals the difference in energy between the two levels.

Naming of characteristic X-rays:- Let the incoming electron dislodge an atomic electron from innermost shell - the K shell. If the vacancy is filled by an electron dropping from the next higher shell - the L shell emission is said to be of K_d series. If vacancy is filled by an electron dropping from M shell, line is produced and J_o an.

Dumb Question:- Why do occur earlier than K_d in the plot ?
Ans:- represents transition from higher shell then the k_d line, reperesents transition over larger energy range, thus would have lower wavelength as . thats why occurs earlier than K_d.

MOSELEY'S LAW FOR CHARACTERISTIC X-RAY:-
Moseley measured the frequency of characteristic X-rays from a large number of elements and plotted the Vs Z. The plot was close to a straight line (emission was for K_d line).
Moseley forwarded his law of the frequency of K_d lin:-
= a(Z - 1) = 4.98 x 10⁷(Z - 1)
where 'a' is a constant
for general characteristic X-rays, Moseley's law is
= a(Z - b) b is another constant.

ILLUSTRATION:- Use Moseley's law with b = 1 to find the frequency of the K_d X-rays of La(Z = 57) if the frequency of K_d X-rays of Cu(Z = 2a) is known to be 1.88 x 10¹⁸ Hz.
Solution:- Using equation = a(Z - b)



PROBLEM:-
Easy:- 1. An experiment measuring the K_d line of Fe yields 1.94 . Determine atomic number of iron.
Solution:- From Moseley law
= (4.97 X 10⁷)(Z - 1) or Z = i +
also f =
We get Z = 1 +
Z = 260.2 26 (atomic number of iron)

2. Stopping potential of 24, 100, 110 and KV are measured for electrons emitted from o certain element when it is irradiated with monochromatic X-ray. If this element is used as target in an X-ray tube, What will be the waavelength of the line.
Solution:- The stopping energy, eV_s is equal to the difference between the energy of the incident photon and the binding energy of the electron in a particular shell.
eV_s = E_P - E_B
The different stopping potential arise from electrons being emitted from different shells, with the smallest value (24 KV) corresponding to ejection of a K - shell electron. Subtrating the expression for the two smallest stopping potentials, we obtain
eV_SL - eV_SK = (E_P - E_BL) - (E_P - E_BK) = E_BK - E_BL
or 100 KeV - 24 KeV = E_BK - E_BL
The difference, 76 KeV is the energy of the K_d line. The corresponding wavelength is
= 0.163

3. The wavelength of the first member of the balmen series in hydrogen spectrum is 6563 . What is the wavelength of the first member of Lymen series ?
Solution:- Balmen series
Lymen series



4. The wavelength of K_d X-rays tungsten is 0.21 . If the enrgy of tungsten atom with an L electron knocked out in 11.3 KeV. What will be the energy of this atom when a K electron is knocked out ?
Solution:- Energy of K_d photon : E =
=> E = = 59.1 KeV
Let E_K = energy of the with a vacancy in the K - shell
E_L = energy of the atom with vacancy in the L - shell.
Then E_K - E_L = E => E_K + E_L = 59.1 + 11.3 = 70.4 KeV
5. Calculate the second excitation potential of helium. Given e = 1.6 X 10^-18C,
m = 9.1 X 10^-31Kg.
Solution:- Energy of first orbit, E₁ = 36.6 z²
Energy of third orbit, E₃ =
Third excilation potential = E₃ - E₁ = 13.6 z²(1 - 1/9)eV = 48.4 eV [for helium Z=2]


6. In Moseley's equation = a(z - b), a and b are constant. Find their values with the help of following data

Element	z	Wavelength of Kd X-ray.
M0 C0	42 27	0.71 1.785

Solution:- = a(Z - b) or = a(Z₁ - b) .................................... (i)
and = a(Z₂ - b) ........................................... (ii)
from (i) & (ii) we have
= a(z₁ - z₂) .......................................... (iii)
Solving the 3 equations with C = 3 X 10⁸m/s, = 0.71 X 10⁷⁰ and ₂ = 1.785 X 10^-10m and Z₁ = 42 and Z₂ = 27, we get
a = 5 X 10⁷(Hz)^1/2 and b = 1.37

7. The wavelength of the first line of Lymen series for hydrogen is identical to that of the second line of Balmen series for series for hydrogen like in x, find x.
Solutio:- Wavelength of the first of Lyman series for hydrogen atom will be given by the equation

The wavelength of the second Balmen line for hydrogen like i an X is

Given ₁ = ₂ or

Z = 2
i.e. x ian is He⁺

8. The p.d. across an X-ray tube is 100000 volts and the current through it is 2.5 mA. Calculate (a) the number of electrons striking the anode per second, (b) the speed with which they strike.
Solution :- (i) mv² = e.V => v = = 1.88 x 10⁸ m/s.

(ii) Number of electrons hitting the target per second =
energy incident per second on the target = n.eV = = 5 x 10² J/S
He at produced = 500 x 0.999 = 4.99.5 J/S

MEDIUM
1. A hydrogen like atom (atomic number Z) is in a higher axcited state of quantum number n. The excited atom can make a transition to the second excited state by successively emitting two photos of energies 4.25 eV and 5.95 eV respectively. Determine the values of n and Z.
Solution:- From the given conditions
E_n - E₂ = (10.2 + 17)eV = 27.2 eV ........................................ (i)
E_n - E₃ = (4.25 + 5.95)eV = 10.2 eV ...................................... (ii)
from (i) and (ii) we get
E₃ - E₂ = 17.0
or Z² - 13.6 = 17.0
=> Z = 3
from equation (i) Z²(13.6) = 27.2
Substituting Z = 3 we get, n = 6
Z = 3 and n = 6


2. A Hydrogen like atom of atomic Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition of quantum state n, a photon energy 40.8 eV is emitted. Find n, Z and the ground state energy (in eV) of this atom. Also calculate the maximum energy (in eV) that can be emitted by this atom during de-excitation. Ground state energy of hydrogen atom is - 13.6 eV.
Solution :- Let the ground state in eV be E₁
then E_2n - E₁ = 204 eV => - E₁ = 204 eV and E_2n- E_n = 40.8 .................... (i)
= 40.8 eV, ;E₁ = 40.8 eV ................................... (ii)
from (ii) E₁ = - n²(40.8) eV
E₁ = - (2)²(40.8) eV => E₁ = - 217.6 eV
E₁ = - 13.6 Z²,
z² = => Z = 4
E_min = E_2n - E_{2n - 1} =
[here n = 2]
E_min = 10.58 eV

3. A moving hydrogen atom makes a head on collision with a stationary hydrogen atom. Before collision both atoms are in ground state and after collision they move together. What is the minimum value of the kinetic energy of the moving hydrogen atom, such that one of the atoms reaches one of the excitation state.
Solution:- Let K be the kinetic energy of the moving hydrogen atom and K', the kinetic energy of combined mass after collision.




from conservation of linear momentum,
P = P' or
or, K = 2K'........................................................................................ (i)
from conservation of energy, K = K' + E .........................................(ii)
Solving Eqn. (i) and (ii), we get E = K/2
Now, minimum value of E for hydrogen atom is 10.2 eV
or

The minimum kinetic energy of moving hydrogen is 20.4 eV

HARD:-
1. An imaginary particle has a charge equal to that of an electron and mass 100 times the man of the electron. If moves in a circular orbit around a nucleus of charge + 4e. Take the mass of the nucleus to be infinite. Assuming the Bohr's model is applicable to this system.
(a) Derive an expression for the radius of nth Bohr orbit.
(b) Find the wavelength of the radiation emitted when the particle jumps from fourth orbit to the second orbit.
Solution:- (a) We have ....................... (i) [Centripetal force]
The quantization of angular momentum gives,
m_pvr_n = .................................... (ii)
Solving (i) & (ii), we get
r = ..................................... (iii)
Substituting m_p = 100 m and Z = 4
(where m = mass of electron)
We get, r_n =

(b) As we know, E₁^H = - 13.60 eV
and E_n
for the given particle, E₄ = x 100 = - 1300 eV
E₂ = x 100 = - 5440 eV
E = E₄ - E₂ s= 4080 eV
= 3.0

2. The energy levels of hypothetical are electron atom are given by
E_n = - eV
where n = 1, 2, 3 .......................
(a) Compute the four lower energy level diagram.
(b) What is the excitation potential of the stage n = 2 ?
(c) What wavelength () can be emitted when then atoms in the ground state are bombarded by electrons that have been accelerated through a potential difference of 16.2 V.
Solution:- (a) E₁ = = - 18.0 eV
E₂ = = - 4.5 eV
E₃ = = - 2.0 eV
and E₄ = = - 1.125 eV

The energy level diagram is shown below.

Diagram

(b) The excitation potential of stage n = 2 is 18.0 - 4.5 = 13.5 Volts.

(c) Energy of the electron accelerated by a potential diffrence of 16.2 V is 16.2 eV with this energy the electron can write the atom from n = 1 to n = 3 as
E₄ - E₁ = - 1.125 - (- 18.0) = 16.875 eV > 16.2 eV
and E₃ - E₁ = - 2.0 - (- 18.0) = 16.0 eV < 16.2 eV
Now, ₃₂ = = 4950 Ans.
₃₁ = = 773 Ans.
and ₂₁ = = 917