Atomic Structure

ATOMIC STRUCTURE

Introduction

The most special thing about this chapter is that the no. of scientists involved in developed of this concept is remarkably large. Starting with heucippus of Miletus in 440 BC it extends to 2000's. And the research is still going on. Even the great scientist EINSTEIN had his hands in developing this extraordinary concept.




Atomic No.: Total no. of protons present in nucleus.

Mass No.: Total no. of protons & Neutrons in neucleus.

Atomic No.(Z): no. of electrons (when atom is neutral).

Mass No.(A): no. of neutrons + no. of protons.
Representation





Ilustration: Calculate no. of protons, neutrons & electrons in .

Ans: Z = 35, A = 80
No. of protons = Z = 35
No. of neutrons = A - Z = 80 - 35 = 45
As atom is neutral. No. of electrons = No. of protons = 35.


Isotopes: Such atoms of same element having same atomic no. but different mass no.s are c/d isotopes.
Protiun (). Denterium () & Tritium () are isotopes.

Isobars: Atoms of different elementswhich have same mass no.s but different atomic No.
, , are isobars.

Isotones: Atomes of different elements which contain same no. of neutrons c/d isotones.
, are isotones because they have same no. of neutrons.
subatomic particle like electron at high speed varies.
m =
m₀ restmass
c velocity of light

Photoelectric effect: When radiations with certain minimum frequency (v₀) strike surface of metal electrons are ejected from surface of metal. This is c/d photoelectric effect & electrons are called photoelectrons.

Threshold frequency: It is min. frequency (v₀) below which no electrons are ejected.

Work function: Min. energy required to eject electron (h₀).
h = h₀ + mv²
Total energy = Workfunction + Kinetic energy.


Planck's Quantam Theory:

(i) Radiant energy is emitted or absorbed not continuously but discontinuously in form of small discrete packets of energy. Each packet c/d 'quanta'. In case of hight energy 'quanta' is 'photon'.

(ii) Energy of each quantum is directly proportional to frequency of radiation.

E = h
h Planck's Constant.
h = 6.626 x 10^-34 J/sec.
Body will be some whole no. quanta.
E = nh
n integer.


Illustration: Calculate KE of electrons ejected when yellow light of frequency 5.2 x 10¹⁴ s^-1 on a metal whose threshold frequency is 4 x 10¹⁴ s^-1.

Ans: h = h₀ + mv²
h - h₀ = mv² mv² = h(1.2 x 10¹⁴)
= 6.625 x 10^-34 x 1.2 x 10¹⁴
= 7.95 x 10^-20 J


Bohr's Model:

Postulate:

(1) As long as electrons occupy definite energy level, it does not radiate out energy. Emission or absorption of energy occures only when electrons jumps from one level to other.
E = En₂ - En₁ = h
If n₂ > n₁ emission epectra.
If n₂ <>1 absorption spectra.

n₁ Initial state.
n₂ Final state.

(2) Angular momentum of electron in closed is always quantized i.e. integer multiple of .
Angular momentum = n
mvr = n.


Dumb Question: Why angular momentum is mvr ?

Ans: Angular momentum = Moment of Inertia x angular velocity.
= = mvr

Note: In CGS unit, = 1
MKS unit, = 9 x 10⁹ Nm²c^-2


Calculation of radius of orbit:

Derivation:

Electrons revolves in orbit.

Centripetal force acting on electron away from centre & force of attraction towards centre.
For electron to revolve in same orbit.
[ = 1 in CGS unit] ....................... (i)
mvr = n ....................................................... (ii)
v =
Putting value of v from eq. (ii) in eq. (i)
.................................................... (iii)
r₀ = = 0.529 A⁰ Bohr radius

For H-atom, rn = n2 x r0 = n2 x 0.529 A0

For H-like atom, like He⁺,


Calculation energy of electron:

Total energy of electron (E) = K.E. + P.E.
= mv² -


Dumb Question: Why P.E. is - ?

Ans: P.E. is work done when electron moves from to r.
P.E. =


Dumb Question: Why Force is -ve ?

Ans: Force is work attractive. So, it is taken as -ve.
From eq. (i) mv² =
E = - = -


Dumb Question: What does -ve sign signify ?

Ans: -ve sign show's that electron is bound to that orbit & atom.
E = -
Substituting value of r from eq. (iii)
..................


Calculation of velocity of electron in any orbit:

Substituting value of r from (ii) in (i)
v =
For H-like atom,
v_n = x 2.188 x 10⁸ cms^-1
For H-atom, putting z = 1
v_n = cms^-1
From eq. (i)
v² =



Calculation of no. of revolutions of electron in an orbit per sec:

mvr = n v =
No. of rev./sec =

No. of revolutions per sec =
=


Calculation of no. of waves in any orbit:

No. of waves in any orbit =

= De Broglie relation.

Waver no.: It is reciprocal of wavelength.
For H-atom (wave no.) = R R Rydberg constant
R = 1.097 x 10⁷ m^-1
For Lyman Series n₁ = 1, n₂ = 2, 3, 4, .....................................
For Balmer Series n₁ = 2, n₂ = 3, 4, 5, .....................................
For Paschen Series n₁ = 3, n₂ = 4, 5, 6, .....................................
For Brackett Series n₁ = 4, n₂ = 5, 6, 7, .....................................
For Pfund Series n₁ = 5, n₂ = 6, 7, 8, .....................................
For H-like atom,
= Rz²
z - atomic No.

when n₂ in Redbergis formula is i.e. n₂ =


De Broglie Relation:

Matters have dual nature of particle & wave If assumed as wave, its energy.
E = h Plank's quantum theory .................................... (i)
If assumed as particle, its energy.
E = mc² Einstein Eq. .................................................. (ii)
Equating (1) & (2)
h = mc²
sinc =
h = mc²

=

Deg broglie pointed that this eq. can be applicable to any particle.
= =
= de broglie wavelength.


Calculation of de broglie wavelength of electron from potential applied to accelerate it:

If accelerating potential v is applied,
Energy by electron = ev (charge x potential = ....)
mv² = ev =
= =


Illustration: Calculate wavelength of spectral line in spwectra of Li²⁺ ion when transition takes place b/w two levels whose sum is 6 & difference is 2.

Ans: Let transition takes place b/w levels n₁ & n₂
n₁ + n₂ = 6 & n₂ - n₁ = 2
On solving n₂ = 4 n₂ = 2
= = Rz²
Here z = 3 (Li²⁺ ion)
= 109,677 cm^-1 x 3²
= 109,677 x 9
= 109,677 x
=
= 5.4 x 10^-6


Illustration: Calculate (i) I^st excitation energy yo electron of He⁺ atm.
(ii) Ionization energy of He⁺ atom.

atom,
E_n = -


Dumb Question: What is I^st excitation energy ?

Ans: It is amount of energy required to excite electron from h = 1 (ground state) to n = 2 (I^st excite state)
E = E₂ - E₁
z = 2
E₁ = & E₂ =
= - 1.312 x 10⁶ + 4 x (1.312 x ⁶) = 3 x 1.312 x 10² J/mol

= 3.936 x 10⁶ J/mol

(ii) Dumb Question: What is Ionization energy ?
Ans: Energy required to remove electron from n = 1 to n = i.e.
I.E. = E = - E₁ = 0 - (- 1.312 x 10⁶ x 4)
= 5.248 x 10⁶ J/mol


Dumb Question: Why = 0 ?

Ans: E_n & when n
then 0
So, = 0


Question ..... with electron (m = 9.1 x 10^-31 kg) moving with velocity 10³ m/s & find refarding potential required to stop electron ?

Ans: = = = 7.25 x 10^-7 m
Let v be retarding potential in volt.
To stop electron, K.E> of electron = opposing energy of potential.
mv² = ev
x 9.1 x 10^-31 x 10⁶ = 1.6 x 10^-19 x v
v = volt v = 2.844 x 10^-6 volt


Derivation of Bohr's Postulate of angular momentum from de boglie wavelength:

According to De Broglie electron is not only particle but has wave character.
So, circumference of orbit must be equal to integral nultiple of wavelength () for com[letely in phase.
i.e. 2r = n .................................... (i)
r radius of orbit.
But =
2r =
mvr =
It is impossible to measure simultaneously position and momentum of a small particle with absolute accuracy.
Product of uncertainity in momentum (P = mv) & position is constant & is equal to or greater than , where h Planck's constant.
x.P x.(mv)
x.v
It has significance only for microscopic particles.


Dumb Question: Why electron can not exist in nucleus ?

Ans: Diameter of atomic nucleus is order of 10^-15m.
So, max. uncertainity in its position would be 10^-15m (i.e. x = 10^-15m)
Mass of electron = 9.1 x 10^-31
By uncertainity Principle
x.(mv) = v =
v =
= 5.77 x 10¹⁰ m/s.
This value is much higher than velocity of light (3 x 10⁸ m/s). So, it is not possible.


Question: ......... with uncertainity of 0.02%. What is uncertainity in position.

Ans: v = x 500 = 0.1 m/s
x.v =
x = = 5.77 x 10^-4 m


Size of nucleus:

Size of nucleus (r) = (1.3 x 10^-13) m^1/3
r rad. of nucleus m Atomic no.


Dumb Question: Why radius of nucleus varies m^1/3 i.e. r m^1/3 ?

Ans: If nucleus isspherical,
density d =
d =
r³ =
r =
r = 1.3 x 10^-13 x m^1/3
r m^1/3


..........................


amplitude of wave
Coordinates of electron are (x, y, z), E is total energy of electron, v is its potential energy, m is mass of electron & h is planck's constant.


Significance of wave function: Square of wave function () gives probability of finding electron at that point.
Probability of finding electron = ²


Quantum No.: Set of foor no.s with help of which we can get complete information about all electrons in an atom i.e., location, energy, shap & orientation of orbital.

Principal quantam No. (n):

(i) It gives average distance of electron from nucleus.

(ii) It completely determines energy of electron in H-atom & H-like atoms.
E_n = -
For I^st principal shell (k), n = 1 which means it has lowest energy & closest to nucleus.

(iii) Max. no. of elctrons present in any principal shell is given by 2n² n no. of principal shell.


Question: ........................

Ans: is wave .......... schrodinger eq. It can be -ve or +ve but square of it always +ve & probability of finding electron at any place can not be -ve. So, it represents probability.

(i) It gives no. of subshells present in main shell.

(ii) Angular momentum of electron present in any sub shell.

(iii) Shapee of various subshells present within some principal shell.

Angular momentum of electron =
Azimuthal quantum no.
For given value of n, it have integral value ranging from 0 to n - 1.
For 1^st shell, n = 1, = 0 & have only one subshell c/d s-subshell.
For II^nd shell, n = 2, = 0 & 1 & have two subshell s-subshell ( = 0)
& P-subshell ( = 1)
For III^rd shell, n = 3, = 0, 1, 2 & havethree subshell s, p & d - subshells. For d-subshell ( = 1).
For IV^th shell, n = 4, = 0, 1, 2, 3 & have 4 subshells s, p, d & f & for f subshell
( = 0)

Note: (1) No. of subshells present in any principal shell is equal to no. of principal quantum no.

(2) Energies of different subshells are found to be in order
s < p < d < f


(1) It gives no. of orbitals present in any subshell.
For given , m has in5tegral value - to + .
For = 0, m has only one value i.e. m = 0
This means s-subshell has only one orientation in space.
For = 1, m has 3 value i.e. m = - 1, 0 & + 1
This means p-subshell has 3 orbitals. These 3 orbitals are oriented along x-axis, y & z-axis. i.e. P_x, P_y & P_z.
For = 2, m have 5 values i.e. m = - 2, -1, 0, + 1 & + 2 d-subshell has 5 orbitals.
For = 3, m has 7 value, So, has 7 orbitals.

Note: By convention, for P_z, m = 0, & for P_x & P_y m ± 1 but not fixed.
Five d-orbitals. For dz², m = 0, for dxy & dx² - y², m = ± 1.


Degenerate orbitals: Those orbitals which have same subshell having equal energy.

(4) Spin Quantum No.:

(1) To explain magnetic properties of substance.

(2) Direction of electron spin i.e. clockwise or anticlockwise.
For any valve of m, s = + 1/2 or - 1/2


Illustration: An electron is in 4f orbital. What possible values for quantum numbers n, , m & s can it have ?

Ans: Since electron is in 4f orbital, value of principal quantum no., n = 4
For f-orbital, Azimuthal quantum no. = 3. Values of magnetic quantum (m) are - to + in which zero.
= 3, m = - 3, -2, -1, 0, +1, +2, +3
For each value of m, spin quantum no., s has two values, i.e., s = + 1/2 and s = - 1/2.


Pauli Exclusion Principle:

(1) No two electrons in an atom can have same set of four quantum numbers.
(2) An orbital can have maximum two electrons & these must have opposite spins.


Illustration: Let for 3s-orbital, It has n = 3, = 0 & m = 0. Since for each value of m, there are two values, spin quantum no. i.e. + and - . 3-s orbital can have two electrons; one with quantum no. n = 3, = 0, m = 0 & s = +
& other n = 3, = 0, m = 0 & s = -
No. of orbitals in subshell = 2 + 1
Max. no. of electrons in subshell = 2(2 + 1)
No. of orbitals in n^th shell = n²
Max. no. of electrons in n^th shell = 2n²


Shape of s-orbitals

Shape of s-orbital is spherical.
For 1s, PRobability of electron is max. near nucleus & as distance .
For 2s, Prob. of electron is max. neaqr nucleus & then to zero & again & then as distance from nucleus .

Dumb Question: What is node ?

Ans: It is region where probability of finding electron is zero c/d node.


Shape of p-orbital:

Shape of p-orbital is dumb bell shape.

Note: No. of spherical/radial nodes in any orbital = n - -1
No. of planar nodes in any orbital =
In s-orbital, spherical nodes & in p, d & f orbitals planar nodes.

= a(z - b)

frequency z nuclear charge (atomic no.)


Illustration: If straight line is at any angle 45⁰ with intercept 1 on -axis. Calculate frequency when atom c No. is 50.

Ans: = az - ab
y = mx + c
Slope = mx -axis intercept = - ab
tan45⁰ = 1 = a
- ab = 1 b = - 1
= z + 1
= (50 + 1)
= (51)² = 260 S^-1


Explanation of difference of energy in subshell of multielectron atoms:

In case of H-atom, only force of integration is force of attraction b/w -vely charged electron & +vely charged nucleus. But in multielectron atom, in addition to force of attraction b/w electrons & nucleus, there are forces of repulsion among electrons.

Atom is stable b/c net forces of attraction are greater than forces of repulsions.

Most imp. repulsive forces are on electrons of outer shell by electrons of inner shells. On other hand, attractive forces on electrons increase with of nuclear charge (+ze). but these attractive forces on outershell electrons are greatly reduced by presence of inner shell electrons which produce a screening effect or shielding effect. b/w outer shell
..........
This is known as effective nuclear charge.

Note: (1) s-orbital, being spherical in shape, shields electron from nucleus more effectively than p-orbitals which in turn sheilds more effectively than d-orbitals.

(2 As effective nuclear charge depends upon actual nuclear charge. So, similar orbitalsof different atoms donot have same energy.
eg. E_2s(H) <>2s(L) <>2s(Na)


Shpaes of orbitals:

s orbitals: Its shape is spherical.

p - orbitals:

d - orbitals:

Aufbau Principle: In ........... are filled in order of their increasing energies.

In neutral, isolated atom, lower value of (n + ) for orbital, lower is its energy.

Note: If two different orbitals have same value of (n + ), orbital with lower value of n has lower energy.


Hund's Rule of Max. Multiplicites:

Electron pairing in p, d & f orbitals cannot occur until each orbital of a given subshell contains one electron each or is singly occupied.


Illustration: Consider Nitrogen atom z = 7

Rule for finding Group No.:

(i) If last shell contains 1 or 2 electrons, then group no. is 1 & 2 respectively.

(ii) If last shell contains more than 2 elctrons in last shell plus.

(iii) If electrons are present in (n n - 1)d orbital in addition to ns orbital, then gp. No. is equal to total no. of electrons present in (n - 1)d orbital & ns orbital.


Q.1. Two particles A & B are in motion. If wavelength associate with particle A is 5 x 10^-8 m. Caclulate wavelength of B if momentum is half of A.

Ans:
But P_B = P_A
B = 2 <sub>A

B</sub> = 10^-7 m


Q.2. Show that wavelength of a moving particle is related to its K.E. (E) as =

Ans: Acording de Broglie =
But mv² = E v =



Q.3. A 25 watt bulb emits mouochromatic yellow light of wavelength of 0.57 µm. Calculate rate of emission of quanta/second.

Ans: Energy emitted/sec = 25 J/s
Energy of one photon (E) = h = h
= = 3.48 x 10^-19 J
No. of photons enitted per sec =
= 7.18 x 10¹⁹


Question: Calculate energy required to ........ from n = 2 orbit. What is longest wavelength of light in cm. can used ?

Ans: E = - E₂ = 0 - = 5.45 x 10^-19 J/atom.
E = h =
= 3.647 x 10^-7


Q.5. An ion with mass no. 56 contains 30 mts +ve charge & 30.4% more neutrons than electrons. What is symbol of ion ?

Ans: Suppose no. of electrons in ion M³⁺ = x
No. of neutrons = x + x = 1.304 x
No. of electrons in neutral atom = x + 3
No. of protons = x + 3
Mass no. = No. of photons + No. of neutrons
56 = x + 3 + 1.304 x
x = 23
No. of protons = 23 + 3 = 26
Symbol of will be


Q.6. In photon of wavelength 150p, strikes an atom & an bound electron ejected out with velocity 1.5 x 10⁷ m/s. Calculate energy with which it bound to nucleus.

Ans: Energy of incident photon =
= 13.25 x 10^-16 J
= (9.11 x 10^-31) x 1.5 x 10⁷
= 1.025 x 10^-16 J
Energy with which electron was bound
= 13.25 x 10¹⁶ J - 1.025 x 10^-16 J = 12.225 x 10^-16 J


Q.7. If uncertainities in measurement of position and momentum of electron are equal in magnitude, what is uncertainity in measurement of velocity.

Ans: x x p =
As x = p (p)² =
p =

= 7.98 x 10¹² m/s which is absurd.


Q.8. A bulb emits light of wavelength 4500 A⁰. Bulb is rated as 150 watt & 8% of energy is emitted as light. How many photons are emitted by bulb per second.

Ans: Energy emitted/sec = 150 J/s
Energy emitted as light = x 150 = 12 J = E
E = nh = nh
n =
= 2.717 x 10¹⁹


Question: If H₂ is exposed to radiation of wavelength 253 ..., waht % of radiant energy will be converted into K.E. ?

Ans: Energy required to break H-H bond =
= 7.83 x 10^-19 J
Energy of photon =
= 7.83 x 10^-19 J
Energy left after dissociation = K.E. = (7.83 - 7.15) x 10^-19
% of energy used in K.E. = x 100 = 8.68 %


Q.10. Find velocity (m/s) of electron in I^st Bohr orbit of radius a₀. Find be Broglie wavelength (in m). Find angular momentum of 2p orbital of H-atom.

Ans: For H & H-like atoms,
v_n = 2.188 x 10⁶ x m/s
For H-atom, z =1 & for I^st orbit n = 1
v = 2.188 x 10⁶
Debroglie wavelength, =
= 3.33 x 10^-10 m
Orbital angular momentum =
For 2p, = 1,
Orbital angular momentum =
=


Q.1. A photon of = 4 x m strikes on metal surface, waork function of metal being 2.13 ev. Calculate (i) energy of photon (ev) (ii) K.E. of electron (iii) velocity of photoelectron.

Ans: (i) Energy of photon = h
= 4.97 x 10^-19 J
1ev = 1.6 x 10^-19 J
= 3.1 ev

(ii) K.E. of electrons = h - h₀ = 3.1 - 2.13 = 0.97 ev

(iii) mv² = 0.97 ev = 0.97 x 1.6 x 10^-19
v² = 0.341 x 10¹² v = 5.84 x 10⁵ m/s


Q.2. In H-atom, energy of electron is n^th orbit is given E_n = - ev. Show that
E_{n + 1} - E_n = ev for large values of n.

Ans: E_{(n + 1)} - E_n =

E_{n + 1} - E_n
As n , 0
E_{n + 1} - E_n =

Q.3. Wave function of 2s electron is given by . It has node at r = r₀. Find relation b/w r₀ & a₀.

Ans:
²₂₅ = 0 at r = r₀
[Dumb Question: Why ²₂₅ or prob. of finding electron is zero ?
Ans: At r = r₀, it has node where probability of finding electron is zero.]

In this expression, only factor can be zero is
2 - = 0 r₀ = 2a₀


Q.4. A dye absorbs light of = 4530 A⁰ & then fluorescence light of 50870 A⁰. 47% of absorbed energy is reemitted out as flurescence. Calculate ratio of quanta emitted out to no. of quanta absorbed.

Let n₁ photons areabsorbed. _absorbed
Total energy absorbed =
E of light re-emitted out in one photon =
Let n₂ photons are reemitted then,
Total energy reemitted out =
As given
E_absorbed x = E_{reemitted out}
x = n₂ x




Dumb Question: What is fluorescence ?

Ans: Emission of radiation from particle ?


Q.5. Waht H-like ion has wavelength difference b/w I^st lines of Balmer & Lyman series equal to 59.3 nm R_H = 109678 cm^-1

Ans: = R_Hz²
For I line of Balmer series
= R_Hz² x R_H x z²
_B =
R_Hz
_L =
_B - _L = 59.3 x 10^-7 cm.
= 59.3 x 10^-7
[7.2 - 1.333] = 59.3 x 10^-7
z² =
z = 3
H-like atom is Li²⁺.

HARD TYPE

Q.1. 1.8g H - atoms are excited to radiation. Spectra shows that 27% of atoms are in III^rd energy level & 1.5% of atoms in II^nd energy level & rest in ground state. Ionization potential of H is 13.6 ev. Calculate
(i) No. of atoms present in III, II energy level.
(ii) Total energy envolved when all atoms return to ground state.

Ans: 1g H contains = No atoms
1.8g H contains = 6.023 x 10²³ x 1.8
= 10.84 x 10²³ atoms

(a) No. of atoms in III shell =
= 292.68 x 10²¹ atoms.
= = 162.6 x 10²¹ atoms.

(b) When all atoms return ti I shell, then
E' = (E₃ - E₁) x 292.68 x 10²¹
E_n = - ev
E' = (- + 13.6) x 292.68 x 10²¹ x 1.6 x 10^-19 [ 1 ev = 1.6 x 10^-19 J]
= 5.668 x 10⁵ J
E'' = (E₂ - E₁) x 162.6 x 10²¹
= (- + 13.6) x 1.6 x 10^-19 x 162.6 x 10²¹
= 2.657 x 10⁵ J
E = E' + E'' = (5.668 + 2.657) x 10⁵ J
Total energy envolved = 832.5 KJ


Q.2. Find quantum no. 'n' corresponding to excited state of He⁺ ion if on transitiontoground state that ion emits two photons in succession with wavelengths 18.5 & 30.4 nm respectively.

Ans: Suppose electron in excited state is present in n₂ shell. First it falls from n₂ to n₁ & then from n₁ to ground state (n = 1).
(i) n₂ n_{1 1} = 108.5 nm
(ii) n₁ G.State, ₂ = 30.4 nm

= 1 - 0.75 n₁² =
n₁ = 2

Again,

= 109677 x 2²
= 0.25 - 0.21 = 0.04
n₂² = = 25 n₂ = 5


Q.3. The wave function of orbital is given by (r) = . Find thevalue of r for which probability of finding electron is max.

Ans: Probability of finding electron
²(r) =
For max. probability finding electron
&
..................
= e^-c[(-4ar)e^-2br + e^-2br(-2b)]
= e^-ce^-2br[- 4ar - 2b]
[- 4ar - 2b] = 0
4ar = - 2b

= (- a) + (- 4ar - 2b)²
= [- 4a + (4ar + 2b)²]
=[- 4a + 16a²r² + 4b² + 16abr]
Putting r = -
=[- 4a + 16a² x + 4b² + 16ab x -]
= [- 4a + ]

Max. prob. of finding electron
when r = -


Key Words

Bohr's Model of atom.
Photoelectric effect.
Emission Spectra.
Absorption Spectra.
Rydberg's Formula.
De broglie Wavelength.
Heisen berg's Uncertainity Principle.
Wave Function
Quantum No. &
Principal Quantum No.
Azimuthal Quantum No.
Amgnetic Quantum No.
Spin Quantum No.
Paulieexclusion Principle.
Nodal Plane.
Nund's Law.
Aufbaw's Principle.