# Atomic Structure 1 Hints and Solutions

###### Atomic Structure

** **

# Hints to Subjective Problems

* *

#### Level – I

- Longest wavelength is associated with least energy.
- Total energy emitted = number of photons emitted × energy of 1 photon
- At the closest distance of approach, the alpha particle will have the same kinetic and potential energies
- Period of revolution = T =

#### Level – II

- a) KE for a H – like atom is same as the magnitude of total energy
- b) l µ 1/m if the velocity is same.
- Use quantization condition for a Bohr orbit in order to evaluate the n values.
- The energy of the incident light should always be greater than the work function for photo electric emission to take place.

#### Level – III

- Identify the wavelengths corresponding to the transitions; which one is shorter or larger. Using the appropriate evaluate the work function and hence stopping potential.
- Evaluate the uncertainty in position with the help of radius of first orbit of He
^{+}ion - d) If the energy of the incoming electron is less than DE then it cannot interact with the atom and if the energy is more then transition is made from one level to another and the remaining energy will be in the form of K.E.
- One Rydberg = IP of H atom.
- Energy of incident photon will be partially utilized in breaking the I
_{2}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.

** **

# Solutions to Subjective Problems

#### Level – I

For He; Z = 2; For Paschen series n_{1} = 3

For longest wavelength n_{2} = 4

= 109678×4×

= 109678 ×4×

**l**** = 4689 A ^{o}**

- wave number of first line of Balmer, =

\wave length of first line of Balmer =

wave number of ultimate line of Balmer, = =

\wave length of ultimate line of Balmer =

** ****\****Ratio = **

- Energy corresponding to 60 W bulb is

E = 60 J/sec = 60 ×3600 J/hr

For 10 hours the energy is

E = 60×3600×10

Energy corresponding to the photon =

= 3.37×10^{-19} J

\ No. of photons emitted =

** = 6.404 ×10 ^{24}**

- From equation E
_{n}= , the energy change of a H–atom that goes from initial state of quantum number n_{i}to a final state of quantum number n_{f}is

DE = E_{f} – E_{i} =

Hence n_{i} = 1, n_{f} = 3, E_{1} = –13.6 eV

DE = –13.6 = **12.08 eV**.

- (a) From equation r
_{n}= n^{2}a_{o}

n = » **435**

(b) E_{n} =

(Such an atom would obviously be extremely fragile and can be easily ionized).

** **

- At the distance of closest approach of an alpha particle (r)

P.E. = K.E.

v^{2} =

=

**v= 6.35 ×10 ^{6} m/sec**

= =

1.54 =

**V = 63.24 volts**

- 8. Dx × Dp ³

(Dp)^{2} = (Dx)^{2} = (since Dp = Dx)

Dp = =

mDv = Dp =

Dv =** **

m = 9.1 ×10^{-31} Kg

\ **D****v = 7.98 × 10 ^{12} m/sec**

- n= = = 5.45 ´10
^{17}s^{–1}

½ mv^{2} = hn–hn_{o}

= (6.63´10^{–27} ergs–sec) (5.45´10^{17}s^{–1}) –2.62 ´10^{–9 }ergs

** K.E. = 9.93 ****´****10 ^{–10} ergs**

** **

- Let the number of photons required = n

n

n = = = 27.6 **= 28 photons **

- Energy corresponding to the incident photon

E = 6.626 × 10^{-34} × 3.2 × 10^{16}

= 2.12 ×10^{-17} J

K.E of photoelectron = th energy of photon

= ×2.12×10^{-17}

= 1.590 × 10^{-17} J

hn = hn_{0} + KE

2.12 × 10^{-17} = hv_{0} + 1.590 ×10^{-17}

hv_{0} = 5.297 × 10^{-18 }J

n_{0} =

**= 7.995 10 ^{15} Hz**

- Time taken for one complete revolution is given by T =

T a n^{3}

, n^{2} = 4 Þ n = 2

\ the ratio of time taken = 1^{3} : 2^{3}

**= 1:8**

- For nth orbit mvr =

= mv =

Again from de-Beoglie’s equation we have

\

**l** **=**

- Energy of a single photon = h n =

= = 4.42 ´10^{–19}J …(1)

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

Let ‘n’ photons are emitted per second

n´4.42´10^{–19}J = 12 J

** n = 27.2 ****´****10 ^{18} **

- According to Rydberg equation

= RZ^{2}

for lowest frequency in Lyman series n_{2} = 2 and n_{1} = 1

\ = 1.09678 ´10^{7} ´ (Z = 1 for hydrogen)=

l= 1.2156´10^{–7} metre

** ****l**** = 1215.6 Å**

Frequency = n = =

** = 2.4679****´****10 ^{15}H_{z}**

Energy E = hn = 6.626´10^{–34} ´2.4679´10^{15}

= 16.352´10^{–19} J = eV **= 10.2 eV**

Now energy for corresponding spectral line of Li^{2+} species

= Z^{2}E_{H} (for Li^{2+}, Z =3)

= (3)^{2} ´10.2 **= 91.80 eV**

** **

** **

**Level – II**

- Energy corresponding to the photon =

= 3.822 ×10^{-19} J

the power output = 2.79 Watt/m^{2}

= 2.79 J/sec/m^{2}

\no. of photons emitted per sq. meter sec =

**= 7.30 ×10 ^{18}**

- i) KE of an electron in the orbits of hydrogen and hydrogen like atoms

=

= 3.4 eV

- ii) =

= 6.654 × 10^{-10} m

**= 6.654 A ^{o}**

** **

- a) Energy of photon , E = W + KE

= 1.82 + 0.73 **= 2.55 eV**

- b) The energy of an electron in n
^{th}orbit of hydrogen atom is given as

E_{n} =

E_{1} = –13.6 eV

E_{2} =

E_{3} = –1.51eV

E_{4} = –0. 85 eV

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

2.55 eV as

QDE = | E_{2} – E_{4}| = 2.55 eV

- If l
_{o}be the threshold wavelength and f be the work function

l_{o}= = .

For copper = l_{o} = = **276 nm**

For sodium = l_{o} = =**540 nm**

For caesium = l_{o} = = **654 nm**

- a) =

**= 1.46 × 10 ^{-9} m**

- b)

when velocity is same l µ

- a) Energy corresponding to 8205.8 A
^{o}=

= 2.422 × 10^{-19} J

= 1.572 eV

1.512 eV = E_{1H} × (1)^{2} ×

1.512 eV =

E_{1 H} = 13.608 eV

\ Ionisation energy of hydrogen atom = **13.6 eV**

- b)

**= 916 A ^{o}**

- Angular momentum of the electron = 5.2728 × 10
^{-34}kgm^{2}/sec

= 5.272 × 10^{-34}

n_{1} = 4.99 » 5

Angular momentum of electron after transition = 3.1636 × 10^{-34} Kgm^{2}/sec

= 3.1636 × 10^{-34}

n_{2} = 2.999 » 3

\ the electron makes trasition from . 5 ® 3

n = = c × R_{h} × Z^{2}

n = 3 × 10^{10} × 109678 × 3^{2} ×

**= 2.1058 × 10 ^{15} Hz**

- Energy corresponding to n = 2 =

= 5.425 × 10^{-12} erg

5.425 × 10^{-12}

**l**** = 3.66 ****× 10 ^{-5} cms**

n_{2} = 4

next level is n = 5

velocity of electron in 5^{th} level of He^{+} =

= = 87.39 ×10^{6} cm/sec

velocity of the electron in 5^{th} level : velocity of light **= 1:344**

- Lowest energy in u.v. region is

E = hc R_{H}

_{ }6.626 ×10^{-34} × 3 ×10^{8} × 1.09 × 10^{7} ×

= 1.625 × 10^{-18} J

This is energy corresponding to 1 atom of H

For 1 g atom the energy is

E = 1.625 × 10^{-18} × 6.023×10^{23} J

**= 978.753 KJ mole ^{-1} **

- eV
_{o}= –f f = work function

f = **= 3.63 eV**

K.E. = hn – f

= = 2.57 eV

\ Stopping potential = **2.57 Volts **

** **

- a) E = hn

n = **= 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 ^{–19}J). **

** Hence electron will be ejected.**

- Kinetic energy of electron = hn – hn
_{critical}

= –

KE = **= 1.22 eV**

(Since 1240 nm = 1eV)

- 14
*.*Energy required to stop the ejection of electron is equal to kinetic energy i.e., (½ mv^{2})

Given K.E = 0.24 eV

l of incident photon = 253. 7 nm

hn = hn_{o} + K.E

Work function (hn_{o}) = hn – K.E = – ½ mv^{2}

= = 4.89 – 0.24

** = 4.65 eV**

- Let the number of moles of I
_{2}dissociated are n

Total energy = 1 kJ

= 1000 Joule

N_{A} ® Avogadro’s no.

On solving above equation we get

**n = 4.17 ****´****10 ^{–3} moles **

level – iiI

- a) Q 1eV = 1.602 ´ 10
^{–12}erg

Also DE = = E_{3} – E_{2} = R_{H}.c.h.Z^{2}

or 16.52 ´ 1.602 ´ 10^{–12} = 109678 × Z^{2} ×´ 3 ´ 10^{10} ´ 6.626 ´ 10^{–27}

\ Z^{2} = 8.74 \ **Z ****»**** 3**

- b) DE = E
_{4}– E_{3}= R_{H}.c.h.Z^{2}

= 109678 ´ 3 ´ 10^{10} ´ 6.626 ´ 10^{–27} ´ 3^{2} ´ = 1.744 ´10^{–10} erg

** = 108.87 eV **

- c) = 109678 ´ 9

** ****l**** = 1.01 ****´**** 10 ^{–6} cm**

- d) E. = mu
^{2 }= m

= **= 1.962 ****´**** 10 ^{–10} erg**

** = 122.4 eV**

- Let wavelength for the transition 5®4 and be that of 4 ® 3 transition

= = 4.502 × 10^{-7} m

= 450.2 mm

= 2.089 × 10^{-7} m = 208.9 nm

shorter wavelength is , longer wavelength is

stopping potential for the photo electrons ejected by the light of wavelength

l_{2} is 3.95 Volts

Energy corresponding to l_{2 }is

= 5.95 eV

since stopping potential = 3.95 V;

K.E. = 3.95 eV

E = f + KE_{1}

5.95 = f + 3.95

f = **2.00 eV**

Energy corresponding to l_{1} is

= 2.75 eV

E_{1} = f + K.E_{2}

2.75 = 2.00 + KE_{2}

KE_{2} = 0.75 eV

\stopping potential = **0.75 V**

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

Q

\ 27.20 ´ 1.602 ´ 10^{–12} = R_{H} ´ Z^{2} ´ h ´ c … (1)

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

= 10.20 ´ 1.602 ´ 10^{–12} erg

\ 10.20 ´ 1.602 ´ 10^{–12} = R_{H} ´ Z^{2} ´ h ´ c … (2)

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

** n = 6 **

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

Z= 3

- Radius of 1
^{st}orbit of the He^{+}ion =

= 0.265 = 0.265 × 10^{-10} m

Uncertainty in the location = 1 % of the radius of the 1^{st} orbit

= 0.265 × 10^{-12} m

Dp × Dx ³

Dv ³

³

³ 2.18 × 10^{8} m/sec

velocity of the electron

v = m/sec

**= 4.368 ×10 ^{6} m/sec** (since Z = 2, n = 1)

- a) I.P. of the atom is

I.P. = E_{¥} – E_{1}

= 0 – (-15.6) **= 15.6 eV**

- b) Series limit at n = 2 is

E = E_{¥} – E_{2}

= 0-(-5.3) = 5.3 eV

= 5.3 eV

= 5.3 eV

**l**** = 233.9 nm**

- c) D E for the transition n = 3 ® n = 1 is

D E = E_{3} – E_{1}

= -3.08 – (-15.6)

= 12.52 eV

= 2.005 ×10^{-18} J

= 2.005 ×10^{-18}

** = 1.009 × 10 ^{7} m^{-1}**

- d) The difference in energy between n = 1 and n = 2 is 10.3 eV
- i) Since the incoming electron has 6 eV energy, it will not be able to do anything hence it will not interact with the atom.
- ii) In case it has 11 eV energy, it will impact 10.3 eV energy to this atom and (11 eV – 10.3 eV) 0.7 eV energy is carried by this electron. Due to transfer of 10.3 eV of energy it will jump from n = 1 to n = 2 level.

- 1 g H contains = N atoms

\ 1.8 g contains = N ´ 1.8 atoms

= 6.023 ´ 10^{23} ´ 1.8

= 10.84 ´ 10^{23} atoms

(a) \ No. of atoms in III shell =

** = 292.68 ****´**** 10 ^{21} atoms **

\ No. of atoms in II shell =

** = 162.6 ****´**** 10 ^{21} atoms **

and No. of atoms in I shell =

** = 628.72 ****´**** 10 ^{21} atoms **

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

E¢ = (E_{3} – E_{1}) ´ 292.68 ´ 10^{21}

= ´ 1.602 ´ 10^{–19} ´ 292.68 ´ 10^{21}

= 5.668 ´ 10^{5} joule

E¢¢ = (E_{2} – E_{1}) ´ 162.6 ´ 10^{21}

= ´ 1.602 ´ 10^{–19} ´ 162.6 ´ 10^{21}

= 2.657 ´ 10^{5} joule

E = E¢ + E¢¢ = 5.668 ´ 10^{5} + 2.657 ´ 10^{5} joule

= **832.50 kJ**

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

hn_{1} = KE_{1} + hn_{0} …(1)

hn_{2} = KE_{2} + hn_{0} …(2)

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

2hn_{1} – hn_{2} = hn_{0} (Q2KE_{1} = KE_{2})

2n_{1} – n_{2} = n_{0}

n_{0} = (1Å = 10^{–10} m)

n_{0} = 1.1483 ´ 10^{15} sec^{–1}

Also l = = 2.6126 ´ 10^{–7} m

** = 2612.6****Å**

- Energy of I orbit of H like atom = 4R
_{h}

= 4 ´ 2.18 ´ 10^{–18} joule

E_{1} for H = –2.18 ´ 10^{–18} J

Q E_{H like atom} = E_{1H} ´ Z^{2}

\ –4 ´ 2.18 ´ 10^{–18} = –2.18 ´ 10^{–18} ´ Z^{2}

\ 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 n_{2}= 2 to n_{1}= 1

E_{2} – E_{1} =

Now E_{1} = 4R_{h}

\ E_{2} = – = –R_{h}

\ E_{2} – E_{1} = 3R_{h} = 3 ´ 2.18 ´ 10^{–18} J

\ E_{2} – E_{1} =

\ l = = 303.89 ´ 10^{–10 }m

** = 303.89 Å**

- b) Radius (r
_{1}) of H like atom =**= 2.645****´****10**^{–9}cm

- Energy given to I
_{2}molecule

^{ }= = = 4.417 ´ 10^{–19} J

Also energy used for breaking up to I_{2} 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 n
_{1}= 1; For H atom

\

= 1.097 ´ 10^{7}

\ 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} ´ u^{2} =

\ 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 ´ 10^{23} mol^{–1}) (27.91 ´ 10^{–22} kJ) + y (6.023 ´ 10^{23} 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 ´ 10^{23} mol^{–1}) (5.53 ´ 10^{–22} kJ) + y(6.023 ´ 10^{23} 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 = **7****2.73**

- 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= n^{2} …(3)

- b) Now for the given problem

m = (200)m_{e} and Z = 3

Hence,

r = n^{2}

= n^{2}(8.856 ´ 10^{–14} m)

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

n^{2}

or n^{2} = 200 ´ 3 or n = = **24**

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

E = KE + PE = mn^{2} –

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

DE = (3.899 ´ 10^{–15}J) = 3.465 ´ 10^{–15} J

l =

= 5.736 ´ 10^{–11} m = **5****7.36 pm**

- O
_{2}¾® O_{N}+

O_{2} ¾® O_{N} + O_{N}

E = 498 ´10^{3} J / mole

= per molecule = 8.268 ´10^{–19} J

Energy required for excitation = 1.967 eV

= 3.146 ´10^{–19}J

Total energy required for photochemical dissociation of O_{2}

= 8.268 ´10^{–19} + 3.146 ´10^{–19 } = 11.414 ´ 10^{–19} J

= 11.414 ´10^{–19} J

l =

= 1.7415 ´10^{–7} m **= 1741.5 Å** ** **

- 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,

E_{absorbed} = 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**

- Energy of light absorbed by one photon =

Let n_{1} photons are absorbed, therefore,

Total energy absorbed =

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

Let n_{2} photons are re-emitted then,

Total energy re-emitted out = n_{2} ´

As given E_{absorbed} ´ = E_{re-emitted out}

´ n_{1} ´_{ = }n_{2 }´

_{ }\ = ´ =

\ = **1****.25**

** Solutions to Objective Problems**

# Level – I

- Since average mass is M + 0.5

= M + 0.5

**\**** (B)**

- For the last electron of Ga the configuration is 4p
^{1}. The added electron enters the vacant p orbital making configuration 4p^{2}

For 4p electron n = 4 and l = 1

**\**** (A)**

- We know that r
_{n}= 0.529 Å, = 0.529 (3^{2}–2^{2}) Å

(For hydrogen Z = 1, n = 1)

= 0 .529 ´5´10^{–8}cm

= **2.645 ****´****10 ^{–8 }cm**

** ****\**** (A)**

- Total spin (s) of an atom is given by

s = n (Where n is number of unpaired electrons) = ´ 4 = **2**

** ****\****(C)**

- Orbital angular momentum =

for d, l = 2

orbital angular momentum =

=

** ****\**** (A)**

**Hint:**Isosters have same number of electrons and atoms

**Hint:**7+3×8+1 = 32

* **\** (B)*

- 12
*.*For one molecule

E = hn

For 1.5 moles

E = 1.5 ´ 6.023 ´10^{23} ´hn

= 1.5 ´ 6.023 ´10^{23} ´6.626 ´10^{–34} ´ 7.5 ´10^{14 }** = 4.48 ****´****10 ^{5} J**

**\**** (C)**

n*Hint:*_{1}= 2 and n_{2}= 3, 4, 5, …… corresponds to Balmer series- Wave number of a spectral line

Energy =

= 6.626 ´10^{–34} ´3´10^{8}´5´10^{5}

** = 9.93 ****´****10 ^{–23}kJ**

** ****\**** (B)**

* *

- = R
_{H}

= 109678

= 109678 ´

l = **= 1215.6 Å**

**\**** (A)**

**Hint:**r µ**Hint:**E_{n}of H- like species =

# Level – II

**Hint:**For lowest wave length, the energy difference between two energy states should be highest.

- Energy difference between 3
^{rd}and 4^{th}I.P.’s is very high.

- Number of quanta absorbed =

=

= ** = 1.065 ****´****10 ^{22}**

** ****\**** (A)**

- = R
_{H}

R_{H} =

l =

= **486.0 nm**

** ****\****(C)**

- One of the possible excited state of He
^{+}ion should have energy

E_{n} = eV

\ E_{n} = = **-6.04 eV**

- Energy difference between 3
^{rd}and 4^{th}I.P.’s is very high.

- The ion has configuration [Ar] 4s
^{0}3d^{6}

Before losing the electrons it would have been [Ar] 4s^{2}3d^{7}

\ the ion is Co^{3+}

**\**** (C) **

- E
_{n}of H- like species =

D E for H like species = D E for H × Z^{2}

For Be^{3+} Z = 4

**D**** E = 10.2 ×16 = 163.2 **

** ****\**** (D) **

**Hint:**Isoelectronic species have same number of electrons.

**Hint:**The order of energies of orbitals is

2s < 2p < 3p < 4s < 3d

The transition from higher to lower levels result in emission of light.

- … (i)

** ** …(ii)

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

, , \ Z = 2

\ He^{+} **\**** (A) **

- l = h /mv

= **= 1.060 ****´**** 10 ^{–36} m **

** ****\**** (B)**

- = R

= (1.09678 ´ 10^{7}) ´ 2^{2} ´ , = 4.3871 ´ 10^{7}

or l = **2****.279 ****´**** 10 ^{–8} m**

**\****(A)**

** **

- l = =

10^{–10} =

On solving we get

**V = 0.0826 Volt **

** ****\****(C)**

- hc=

** =** **1.6203 ×10 ^{6} m^{-1}**

**\**** (B) **