# Atomic Structure

**ATOMIC STRUCTURE **

**Hints for Subjective Problems**** **

**LEVEL – I**

- b) Relate K.E. to λ with the help of an equation which relates λ of a particle to its mass and velocity.
- Total energy emitted = no. of photons emitted × energy of one photon.

**LEVEL – II**

- c) Energy required to stop ejection of electrons = K.E. of electrons.

**LEVEL – III**

- Get the value of Rydberg constant for positronium species.

- Energy of incident photon will be partially utilized in breaking the I2 molecule and residual energy in the form of K.E.
- Removal of electron involves absorption of energy and addition of electron involves release of energy.

Energy absorbed by F + Energy absorbed by Cl = Total energy absorbed

Energy released by F + Energy released by Cl = Total energy released

- Using quantization condition and the condition for the stability of the electron in the circular orbit expression for r can be derived.

**Solution to Subjective Problems**

**LEVEL – I**

- a) λ = = = 0.03313 × 10–31 m

or **3.313 ****×**** 10****–33**** m**

###### b) We are given,

Kinetic energy, mv2 = 4.55 ×10–25J

m = 9.1 ×10–31 kg

h = 6.6 ×10–34 kg m2s–1

= 6.6 ×10–34 J–sec

½ ×(9.1×10–31) v2 = 4.55 ×10–25

v2 = = 106

or v = 103 m sec–1

λ = h/mv =

**λ**** = 7.25 ****×****10****–7**** m.**

- c) mv2= 100 eV ×602 × 10–19

= 160.2 × 1019J

or

= 160.2 × 10–19 kg m2s–2

V =

= = 59.3 × 105 m/s

Now λ =

= = 0.1227 nm

or

**1.227Å**

- i) Energy of the electron

En =

= – = **– 2.43 ****×**** 10****–12**** erg**

- ii) Radius of the III orbit

rn =

= 4.75 × 10–8 cm = **4.75 Å**

iii) Frequency and wavelength

E3 = and

ΔE = E3 – E2 =

= = **19.44 ****×**** 10****–12**** erg**

iii) Frequency of radiation

ΔE = hν

Therefore v (frequency of radiation emitted)

= = = 2.95 × 1015 Hz

Wavelength of radiation

C = νλ

λ = = 1.017 × 10–5 = 1017 × 10–8 cm

= **1017Å**

3 Radius of the nth orbit is given by

rn = for hydrogen Z = 1

r1 = = 0.53 × 10–8 cm = 0.53 Å

Also, radius of the nth orbit is related to the I orbit by the relation

rn = n2r1

Therefore r2 = 4 × 0.53 = 2.12 Å

& r3 = 9 × 0.53 = 4.77 Å

Corresponding values of He+(Z = 2) are

r1 = = 0.265 Å r2 = = 1.06Å

r3 = = **2.385 Å **

- I.E. = eV per atom

= eV = 0.85 eV

or 1.36 × 10–19 J per atom

4.18 J = 1 cal

∴ I.E. = cal per atom

= 0.325 × 10–19 cal per atom

or 0.325 ×10–19 × 6.02 × 1023 cal/mole

= **19.5 kcal/mole**

- From equation En = , the energy change of a H–atom that goes from initial state of quantum number ni to a final state of quantum number nf is

ΔE = Ef – Ei =

Hence ni = 1, nf = 3, E1 = –13.6 eV

ΔE = –13.6 = **12.08 eV.**

- Z = 1 for hydrogen atom

For Paschen series, n1 = 3

For the first line, n2 = 4

∴ = RZ2

= 109673 ×12 [0.0486] = 5331 cm–1

λ = = **1.875 ****×****10****–4**** cm.**

- =

λ = = 1215× 10-8 cm = 1215Å

λ = = **6565Å**

8. The energy of each photon =

∴ λ =

λ = **6630 nm**

- a) Energy required to remove the electron from n = 2 to n2 = ∞

ΔE = 21.7 × 10–12 = 5.425 × 10–12 erg

Now, from Planck’s constant

E = hν =

λ =

= **3.67 ****×**** 10****–5**** cm**

- b) Ionisation energy for hydrogen

= –E1 (Energy of first Bohr orbit)

I.E = –E1 = – = = +13.6eV

Ionisation energy of Li2+

Z = 3

n = 2 ( for 1st exited state )

∴ I.E = –

= 30.6 eV

Ionisation energy of Be3+

Z = 4

n = 2 (for 1st exited state)

I.E =

= **54.4 eV**

- Energy of a single photon = h ν =

= = 4.42 ×10–19J …(1)

Energy emitted by the bulb = ×150 J = 12 J …(2)

Let ‘n’ photons are emitted per second

n×4.42×10–19J = 12 J

n = **27.2 ****×****10****18**

**LEVEL – II**

- a) v =

=

= 8.09 × 107 m/s

K.E. = mv2 = × 9.1 × 10–31 kg × (8.09 × 107 m/s)2

= 2.98 × 10–15J

or 1.86 × 104 eV

or **18.6 KeV**

b)

= = **4.85 ****×**** 10****–11**** m**

- i) Li2+ is a H-like (1s1) particle, whose energy levels are given by

Ea = [for Li, Z (atomic weight) = 3]

=

The energy required to excite the electron from the I to the III orbit is

E3 – E1 = = 108.8 × 1.6 × 10–19 J

Q 1 eV = 1.6 × 10–19 J

Therefore, λ =

= **113.7 ****×**** 10****–10**** m = 113.7Å**

- ii) Three spectral lines will be observed in the emission spectrum of the above excited system corresponding to the transition shown below.

3 → 1 3 → 2 2 → 1

- of spectral lines = =
**3** - a) i) Energy of photon , E = W + KE

= **1.82 + 0.73 = 2.55 eV**

- ii) The energy of an electron in nth orbit of hydrogen atom is given as

En =

E1 = –13.6 eV

E2 =

E3 = –1.51eV

E4 = –0. 85 eV

Obviously, the transition 2 ← 4 leads to the release of photon of energy

2.55 eV as

QΔE = | E2 – E4| = **2.55 eV**

- b) If λo be the threshold wavelength and φ be the work function

λo = = .

For copper = λo = = 276 nm

For sodium = λo = =540 nm

For caesium = λo = = **654 nm**

- c) Energy required to stop the ejection of electron is equal to kinetic energy i.e.,

(½ mv2)

Given K.E = 0.24 eV

λ of incident photon = 253. 7 nm

hν = hνo + K.E

Work function (hνo) = hν – K.E = – ½ mv2

= = 4.89 – 0.24 = **4.65 eV**

- λ =

= ∴ K.E. = =

= **9.65 ****×**** 10****–23****J or 6.03 ****×**** 10****–4**** eV**

- E3 for H = –2.41 ×10–12 er g

E2 for H = –5.42 ×10–12 erg

For a jump from 3rd to 2nd shell

ΔE = E3–E2 =

λ =

= = **6602.9 ****×****10****–8**** cm = 6603 Å**

6.

≥

≥ 5.2772 × 10–25 kg m sec–1

ΔVx = ≥ **5.79 ****×**** 10****5**** ms****–1**

- eVo = –φ φ = work function

φ = = 3.63 eV

K.E. = hν – φ

= = **2.57 eV**

∴ Stopping potential = 2.57 Volts

- a) E = hν

ν =

= **5.99 ****×****10****14**** sec****–1**

- b) Energy associated with photon of blue light with wave length 450 nm or

450 ×10–9 m

E = = = 4.417 ×10–19 J/ Atom

Above energy is greater than threshold energy (i.e, 3.97 ×10–19J).

**Hence electron will be ejected.**

- Number of waves made by Bohr electron in an orbit is given by

n = rn = radius of nth orbit

vn = velocity of electron in nth the orbit

= = 2.98 ≈ **3**

Number of revolutions per second made by the electron =

= = 0.0243 × 1016

= **2.43 ****×**** 10****14**** revolutions per second**

- Let the number of moles of I2 dissolved are n

Total energy = 1 kJ

× n × NA = 1000 Joule

NA ⎯→Avagadro’s no.

= 1000

On solving above equation we get

**N = 4.17 ****×**** 10****–3**** moles**

**LEVEL – III**

- p = mv

and according to de-Broglie λ =

∴

or p =

Let uncertainty in wavelength is Δλ which creates ΔP uncertainty in momentum.

∴p = ∴ p – Δp =

(as p decreases with increase in λ)

On substraction

Δp =

∴Δx =

Now since in 1 m wavelength uncertainty is = 10–6m

∴ 5 × 10–7 m wavelength uncertainty will be = 5 × 10–13 m

i.e. Δλ = 5 × 10–13m

On calculation Δx = **4 ****×****10****–2**** m**

- i) Q 1eV = 1.602 × 10–12 erg

Also ΔE = = E3 – E2 = RH.c.h.Z2

or 16.52 × 1.602 × 10–12 = 109678 × Z2 ×× 3 × 1010 × 6.626 × 10–27

∴ Z2 = 8.74 ∴ **Z ****≈**** 3**

- ii) ΔE = E4 – E3 = RH.c.h.Z2

= 109678 × 3 × 1010 × 6.626 × 10–27 × 32 × = 1.744 ×10–10 erg = **108.87 eV **

iii) = 109678 × 9

λ = **1.01 ****×**** 10****–6**** cm**

- iv) E. = mu2 = m

= = 1.962 × 10–10 erg

= **122.4 eV**

- μ = me [for hydrogen atom]

for postronium species → μ =

∴

Q I.E. ∝ R

∴ I.E. for e+ = × I.E. of hydrogen = × 13.6 = **6.8 eV**

- Number of emitted types of photons =

In initial excited state electrons are emitting three different types of photons

∴ = 3

or n = 3

i.e. when an electron is in n = 3 it can emit 3 different types of photons. Now, as according to the question initially electrons were found in two different excited state, the other excited state is obviously n = 2

Now, when sample is exposed to 2.88 eV the resulting electrons emit 10 different type of photons some having energy equal to 13.056 eV or less than 13.056 eV

or = 10

n2 – n = 20

or n = 5

When sample is exposed to 2.88 eV energy radiation electrons in n = 2 move to

n = 5 (as energy difference in n = 2 and n = 5 is 2.88 eV).

After excitation some electrons are in n = 5, some in n = 3 and some may be in n = 2 all of them when come back to lower excited state emit photons of different energies maximum value is 13.05 eV when electrons come back from n = 5 to n = 1.

Initially excited electrons are either inn = 3 or n = 2, the maximum energy difference is between n = 3 and n = 1 and minimum energy difference is n = 3 and n = 2.

∴ΔEmax= 13.6

=

= 13.6 – 1.51 = 12.09 eV

ΔEmin=

= = 3.4 – 1.5 = **1.9 eV**

- Total energy liberated during transition of electron from nth shell to first excited state (i.e., 2nd shell) = 10.20 + 17.0 = 27.20 eV = 27.20 × 1.602 × 10–12 erg

Q

∴ 27.20 × 1.602 × 10–12 = RH × Z2 × h × c … (1)

Similarly, total energy liberated during transition of electron from nth shell to second excited state (i.e., 3rd shell) = 4.25 + 5.95 = 10.20 eV

= 10.20 × 1.602 × 10–12 erg

∴ 10.20 × 1.602 × 10–12 = RH × Z2 × h × c … (2)

Dividing Eq. (1) by Eq. (2)

**n = 6 **

On substituting the value of n in Eqs. (1) or (2).

**Z= 3**

- Energy of photon = kinetic energy of the photo electron and threshold preparation.

hν1 = KE1 + hν0 …(1)

hν2 = KE2 + hν0 …(2)

Multiplying (1) by 2 and substracting equation (2) from it

2hν1 – hν2 = hν0 (Q2KE1 = KE2)

2ν1 – ν2 = ν0

ν0 = (1Å = 10–10 m)

ν0 = 1.1483 × 1015 sec–1

Also λ = = 2.6126 × 10–7 m = **2612.6Å **

- 1 g H contains = N atoms

∴ 1.8 g contains = N × 1.8 atoms

= 6.023 × 1023 × 1.8

= 10.84 × 1023 atoms

- a) ∴ of atoms in III shell =

= 292.68 × 1021 atoms

∴ No. of atoms in II shell =

= 162.6 × 1021 atoms

and No. of atoms in I shell =

= **628.72 ****×**** 10****21**** atoms **

- b) When all the atoms return to I shell, then

E′ = (E3 – E1) × 292.68 × 1021

= × 1.602 × 10–19 × 292.68 × 1021

= 5.668 × 105 joule

E′′ = (E2 – E1) × 162.6 × 1021

= × 1.602 × 10–19 × 162.6 × 1021

= 2.657 × 105 joule

E = E′ + E′′ = 5.668 × 105 + 2.657 × 105 joule

= **832.50 kJ**

- E1of atoms = – 13.6 eV

Energy give to H atom =

= 1.933 × 10–18J = 12.07 eV

∴ Energy of H atom after excitation = – 13.6 + 12.07 = – 1.53 eV

Q ∴ = n2

∴ n2 = = 9

∴ n = 3

Thus electron in H atoms is excited to 3rd shell

∴I induced λ1 =

Q E1 = – 13.6 eV; E3 = – 1.53 eV

∴ λ1 = = 1028 × 10–10 m

∴ λ = 1028 A

∴ II induced λ2 =

Q E1= – 13.6 eV; E2 = eV

∴ λ = 1216 × 10–10m = 1215Å

∴ III induced λ3 =

Q E1 = – 13.6 eV; E2 = l; E3 = eV

∴λ3 = = 6568 × 10–10 = **6558Å**

- Energy of I orbit of H like atom = 4Rh

= 4 × 2.18 × 10–18 joule

E1 for H = –2.18 × 10–18 J

Q EH like atom = E1H × Z2

∴ –4 × 2.18 × 10–18 = –2.18 × 10–18 × Z2

∴ Z = 2

i.e., Atomic no. of H like atom is 2 or it is He+ ion.

- a) For de-excitation of electron in He+ from n2 = 2 to n1 = 1

E2 – E1 =

Now E1 = 4Rh

∴ E2 = – = –Rh

∴ E2 – E1 = 3Rh = 3 × 2.18 × 10–18 J

∴ E2 – E1 =

∴ λ = = 303.89 × 10–10 = **303.89 Å**

- b) Radius (r1) of H like atom = =
**645****×****10****–9****cm** - Energy given to I2 molecule

= = = 4.417 × 10–19 J

Also energy used for breaking up to I2 molecule = = 3.984×10–19 J

∴ Energy used in imparting kinetic energy to two I atoms

= [4.417 – 3.984] × 10–19 J

∴ K.E./ iodine atoms = [(4.417 – 3.984)/2] × 10–19

= **0.216 ****×**** 10****–19**** J**

- Wave length emitted is in UV region and thus n1 = 1; For H atom

∴

= 1.097 × 107

∴ n = 2

Also the energy released is due to collision and all the kinetic energy is released in form of photon. Thus

=

or × 1.67 × 10–27 × u2 =

∴ **u = 4.43 ****×**** 10****4**** m sec****–1**

- Let x and y be the amounts of F and Cl atoms in the given mixture. Since the ionisation of atoms absorbs energy, we can write.

x(6.023 × 1023 mol–1) (27.91 × 10–22 kJ) + y (6.023 × 1023 mol–1) (20.77 × 10–22 kJ)

= 272.2 kJ …(1)

Since the addition of electron to atoms releases energy, we can write

x(6.023 × 1023 mol–1) (5.53 × 10–22 kJ) + y(6.023 × 1023 mol–1) (5.78 × 10–22 kJ)

= 68.4 kJ …(2)

Solving for x and y, we get

x = 0.054 mol and y = 0.144 mol.

Percentage of F atoms = × 100 = 27.27

Percentage of Cl atoms = 100 – 27.27 = **72.73**

- The basic expressions in Bohr model of the atom are as follows.
- a) Stability of the circular motion of the electron, i.e.

Attractive force = Centrifugal force

…(1)

Quantization of angular momentum

mvr = …(2)

Eliminating v in the above two expressions, we get

=

This gives r= n2 …(3)

- b) Now for the given problem

m = (200)me and Z = 3

Hence,

r = n2 = n2(8.856 × 10–14 m)

Equation Eq. (3) with the first Bohr orbit for the hydrogen atom, we get

n2

or n2 = 200 × 3 or n = = **24**

- c) The energy of the electron in the Bohr model of atom is

E = KE + PE = mν2 –

Using Eq. (I) , we get

E =

Substituting the expression of r from Eq. (3), we get

E =

=

= (3.899 × 10–15 J)

Hence, for the transition 1 ← 3, we get

ΔE = (3.899 × 10–15J) = 3.465 × 10–15 J

λ =

= **5.736 ****×**** 10****–11**** m = 57.36 pm**

- E of light absorbed in one photon =

Let n1 photons are absorbed, therefore,

Total energy absorbed =

Now E of light re-emitted out in one photon =

Let n2 photons re-emitted out = n2 ×

At given Eabsorbed × × Ere-emitted out

∴

∴ = **0.59**

- Energy absorbed =

=

= 5.52 × 10–11 erg

= 5.52 × 10–18 joule

Now this energy is used in overcoming forces of attraction between surface of metal and imparting velocity to electron, therefore,

Eabsorbed = E used in attractive forces + Kinetic energy of electron

∴ Kinetic energy = 5.52 × 10–18 – 7.52 × 10–19 joule

= **47.68 ****×**** 10****–19**** joule**

**Solution to Objective Problems **

**LEVEL – I**

- En ∝

= 1 :1

**∴**** (B)**

- Bohr model can not explain the spectrum of any species having multi electrons. It can explain only the spectrum of hydrogen and hydrogen like species having single electron only.

**∴**** (B)**

- The order of penetration power of s, p, d and f electrons are in order

s > p > d > f

**∴**** (A)**

- Longest wavelength in Lyman Series of hydrogen atom arises from transition between n = 2 to n = 1 whose number is given by

=

Shortest wavelength in Balmer series of He+ arises from transition between

n = α to n = 2 whose wave number is given by

= R

=

∴

**∴****(A)**

- According to Bohr theory

mvr =

∴

According to de-Broglie

λ =

∴ λ =

where r is the radius of the orbit having principal quantum number = n

Q (r ∝ n2) (here a3 = radius of 3rd orbit)

∴ λ = = 6πa1

**∴****(B)**

- ΔxΔp =

If Δx = Δp

∴ Δp2 =

(mΔv)2 =

∴ Δv =

**∴**** (A)**

- Colour in an ionic species arises when wavelength of excited photon lies in visible region. In all other cases except Cu2+(aq) energy required for excitement of electron from lower to higher level does not lie in visible region. But in Cu2+(aq) splitting of d-orbitals energy takes place under the influence of magnetic field of water and electron on excitement absorbs wavelength in visible region.

**∴**** (C)**

- Total values of m or orientation in external magnetic field or orbitals are given

by 2l + 1

**∴**** (C)**

- Orbital angular momentum =

for d, l = 2

orbital angular momentum =

=

**∴**** (A)**

- Experiment cannot be done on a single atom. Even a small weight of sample contains million of atoms.

**∴**** (B)**

- rn= , ∴ r0α n2

r3 = 9r1 = 9x

Also mvrn =

mvr3 =

mv =

λ= = 6πx,

**∴**** (B)**

- 12
*.*For one molecule

E = hν

For 1.5 moles

E = 1.5 × 6.023 ×1023 ×hν

= 1.5 × 6.023 ×1023 ×6.626 ×10–34 × 7.5 ×1014

** = 4.48 ****×****10****5**** J**

**∴**** (C)**

- The maximum number of possible sub-shells associated with a particular shell is equal to principal quantum number i.e. n ≥ 5

**∴**** (B)**

- Wave number of a spectral line

Energy =

= 6.626 ×10–34 ×3×108×5×105

**= 9.93 ****×****10****–23****kJ**

**∴**** (B)**

- = RH

= 109678

= 109678 ×

λ = **= 1215.6 Å**

**∴**** (A)**

**LEVEL – II**

- It is orbital not orbit [orbit was proposed by Bohr] which is a three dimensional area where probability of finding electron is maximum. Orbital word was coined by quantum mechanics.

**∴**** (A)**

- Energy difference between 3rd and 4th I.P.’s is very high.
- En = eV

ΔE1 → 2 = = 10.2 eV

ΔE2 → 3 = = 1.9eV

∴ On moving from lower value of n to higher value of n energy difference between successive energy levels decreases.

**∴**** (A)**

- = RH

RH =

λ =

= **486.0 nm**

**∴****(C)**

- Number of orientation of orbitals in space in presence of external magnetic field is given by 2l + 1. Since value of l for sub-shell is 3, possible orientations will be 7.

**∴**** (C)**

- Value of l are given by 0 to n – 1. For an electron in n = 3, value of l can be 0, 1 or 2.

**∴**** (D)**

- Energy difference between 3rd and 4th I.P.’s is very high.
- ΔE = hv = = = 3 × 10–17J

**∴**** (A)**

- En = , where Z = atomic number = 4

N = number of orbits

Be3+ (n = 2) (first exited state)

E1 = –13.6 eV, E2 = –13.6 × 16/4

**∴**** (A)**

- Isoelectronic species have same number of electrons.
- For a given value of l, value of m can be – l to + l. If l = 0, m can not have –1 value.

**∴**** (B)**

- Electronic configuration of ultimate and penultimate shell in ground state of these elements is as follows:

Magnetic moment μ = Bohr magneton where N is number of unpaired electrons, μ will be highest for Cr having 6 unpaired electrons.

**∴****(B)**

- Vn =

∴

∴ V3 = or

**∴**** (A)**

- … (i)

…(ii)

Dividing equation (I) by equation (ii), we get

, , ∴ Z = 2

∴ He+ **∴**** (A)** ** **

- λ = h /mv =
**= 1.060****×****10****–36****m**

**∴**** (B)**

6