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Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE PDF Download

Q.97. The binding energy of a valence electron in a Li atom in the states 2S and 2P is equal to 5.39 and 3.54 eV respectively. Find the Rydberg corrections for S and P terms of the atom.

Ans. From ihe Rydbeig formula we write
Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

we use Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEEThen for n = 2 state 

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

α1 = -0.41 

for p state

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

α= -0.039 

 

Q.98. Find the Rydberg correction for the 3P term of a Na atom whose first excitation potential is 2.10 V and whose valence electron in the normal 3S state has the binding energy 5.14 eV. 

 Ans. The energy of the 3p state must be - Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE where - E0 is the energy of the 3S state.

Then

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

so  Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

 

Q.99. Find the binding energy of a valence electron in the ground state of a Li atom if the wavelength of the first line of the sharp series is known to be equal to λ1 = 813 nm and the short-wave cutoff wavelength of that series to λ2  = 350 nm.

Ans. For the first line of the sharp series Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE in a Li atom

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

For the short wave cut-off wave-length of the same series

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

From these two equations we get on subtraction

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Thus in the ground state, the binding energy of the electron is

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

 

Q.100. Determine the wavelengths of spectral lines appearing on transition of excited Li atoms from the state 3S down to the ground state 2S. The Rydberg corrections for the S and P terms are —0.41 and —0.04. 

Ans. The energy of the 3 S state is

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

The energy of a 2 S state is

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

The energy of a 2 P state is

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

We see that

E (2S ) <E (2P)< E ( 3 S)

The transitions are 3S → 2P and 2P →2S.

Direct 3S → 2S transition is forbidden by selection rules. The wavelengths are determined by

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Substitution gives

                 λ = 0.816 μm ( 3S → 2P )

and           λ = 0.674 μm ( 2P → 2S )

 

Q.101. The wavelengths of the yellow doublet components of the resonance Na line caused by the transition 3P → 3S are equal to 589.00 and 589.56 nm. Find the splitting of the 3P term in eV units.

Ans. The splitting of the Na lines is due to the fine structure splitting of 3 p lines (The 3 s state is nearly single except for possible hyperfine effects.) The splitting of the 3 p level then equals the energy difference

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEEIrodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Here Δλ = wavelength difference & λ = average wavelength. Substitution gives

ΔE = 2.0 meV

 

Q.102. The first line of the sharp series of atomic cesium is a doublet with wavelengths 1358.8 and 1469.5 nm. Find the frequency intervals (in rad/s units) between the components of the sequent lines of that series. 

Ans. The sharp series arise from the transitions ns → mp . The s lines are unsplit so the splitting is due entirely to the p level. The frequency difference between sequent lines is Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE and is the same for all lines of the sharp series. It is

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Evaluation gives

1.645 x 1014 rad/s

 

Q.103. Write the spectral designations of the terms of the hydrogen atom whose electron is in the state with principal quantum number n = 3. 

Ans. We shall ignore hyperfine interaction. The state with principal quantum number n = 3 has orbital angular momentum quantum number

/ = 0, 1,2

The levels with these terms are 3 5 , 3 P, 3 D. The total angular momentum is obtained by combining spin and angular momentum. For a single electron this leads to

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

We then get the final designations

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

  

Q.104. How many and which values of the quantum number J can an atom possess in the state with quantum numbers S and L equal respectively to
 (a) 2 and 3;
 (b) 3 and 3;
 (c) 5/2 and 2?

Ans. The rule is that if Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE then J takes the values

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

in step of 1. Thus :

(a) The values are 1, 2, 3, 4, 5

(b) The values are 0, 1, 2, 3, 4, 5, 6

(c) The values are  Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

 

Q.105. Find the possible values of total angular momenta of atoms in the states 4P and 6D. 

Ans. For the state 4 p, L = 1, S =  Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE  ( since 2s + 1 = 4 ) . For the state 5 d , Z , = 2 , s = 2 . 

The possible values of  J are

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

The value of the magnitude of angular momentum is Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE Substitution gives the values 

4 P :

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEEIrodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

and Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

 

Q.106. Find the greatest possible total angular momentum and the corresponding spectral designation of the term
 (a) of a Na atom whose valence electron possesses the principal quantum number n = 4;
 (b) of an atom with electronic configuration 1s22p3d. 

Ans. (a) For the Na atoms the valence electron has principal quantem number n = 4, and the possible values of orbital angular momentum are l = 0 , 1, 2 , 3 so lmax = 3 . The state is 2F, maximum value of J is  Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Thus the state with maximum angular momentum will be

For this state    Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

(b) For the atom with electronic configuration 1 s2 2 p 3 d . There are two inequivalent valence electrons. The total orbital angular moments will be 1, 2, 3 so we pick l = 3. The total spin angular momentum will be s = 0, 1 so we pick up s = 1. Finally 7 will be 2, 3 , 4 so we pick up 4. Thus maximum angular momentum state is

For this state Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

 

Q.107. It is known that in F and D states the number of possible values of the quantum number J is the same and equal to five. Find the spin angular momentum in these states. 

Ans. For the f state L = 3 , For the d state L = 2. Now if the state has spin s the possible angular momentum are

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

The number of J angular momentum values i s 2 S + 1  if L ≥ S and 2 L + 1 if L < S. Since the number of states is 5 , we must have S ≥ L = 2 for D state while S ≤ 3 and 2 S + 1 = 5 in ply S = 2 for F state. Thus for the F state total spin angular momentum

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

while for D state Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

 

Q.108. An atom is in the state whose multiplicity is three and the total angular momentum is Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEEWhat can the corresponding quantum number L be equal to? 

Ans. Multiplicity is 2S + l so S = 1 .

Total angular m om entum is  Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE so J = 4. Then

L must equal 3, 4, 5 

in order that J = 4 may be included in

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

 

Q.109. Find the possible multiplicities x of the terms of the types (a) xD2; (b) xHP3/2; (c) xF1

Ans. (a) Here J = 2, L = 2. Then S = 0, 1 , 2, 3, 4

and the multiplicities( 2S - 1 ) are 

1, 3, 5, 7, 9.

(b) Here J = 3 / 2 , L = 1 Then 

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE and the multiplicities are 6, 4, 2

(c) Here J = 1, L = 3. Then S = 2 , 3 , 4 

and the multiplicities are 5, 7, 9

 

Q.110. A certain atom has three electrons (s, p, and d), in addition to filled shells, and is in a state with the greatest possible total mechanical moment for a given configuration. In the corresponding vector model of the atom find the angle between the spin momentum and the total angular momentum of the given atom. 

Ans. The total angular momentum is greatest when L, S are both greatest and add to form J. Now for a triplet of s, p> d electrons Maximum spinIrodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEEcorresponding to

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Maximum oibital angular momentum     →     L = 3

corresponding to Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Maximum total angular momentum Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

corresponding to  Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

In vector model Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE or in magnitude squared

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Thus Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Substitution gives  Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

 

Q.111. An atom possessing the total angular momentum Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE is in the state with spin quantum number S = 1. In the corresponding vector model the angle between the spin momentum and the total angular momentum is θ = 73.2°. Write the spectral symbol for the term of that state. 

Ans. Total angular momentum  Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEEmeans J = 2. It is gives that S = 1. 

This means that L = 1,2, or 3 . From vector model relation

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Thus E = 2 and the spectral symbol of the state is

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

 

Q.112. Write the spectral symbols for the terms of a two-electron system consisting of one p electron and one d electron. 

Ans. In a system containing a p electron and a d electron

S = 0,1

L = 1,2,3

For S = 0 we have the terms

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

For S = 1 we have the terms

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

 

Q.113. A system comprises an atom in 2P3/2  state and a d electron. Find the possible spectral terms of that system.

Ans. The atom has 

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

The electron has Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE so the total angular momentum quantum number must be

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

In L - S compling we get S = 0, 1 . L = 1 ,2 , 3 and the terms that can be formed are the same as written in the problem above. The possible values of angular momentum are consistant

with the addition Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

The latter gives us                J = 0, 1, 2, 3 ; 1, 2, 3, 4

All these values are reached above.

 

Q.114. Find out which of the following transitions are forbidden by the selection rules: Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEEIrodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Ans. Selection rules are ΔS = 0

 Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Thus Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE is allowed

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE not allowed

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE  is not allowed  (ΔL = 2)

Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE  is allowed

 

Q.115. Determine the overall degeneracy of a 3D state of a Li atom. What is the physical meaning of that value? 

Ans. For a 3 d state of a Li atom,  Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE because there is only one electron and L = 2.
The total degeneracy is

g = (2L + 1) (2S + 1) = 5 x 2 = 10.

The states are Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE and  Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEEand we check that

g = 4 + 6 =Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

The document Irodov Solutions: Properties of Atoms. Spectra- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.
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FAQs on Irodov Solutions: Properties of Atoms. Spectra- 1 - I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

1. What are the properties of atoms?
Ans. Atoms have several important properties. They are the basic building blocks of matter, and they are incredibly small, with a size on the order of 10^-10 meters. Atoms are composed of protons, neutrons, and electrons. Protons are positively charged particles, neutrons are neutral particles, and electrons are negatively charged particles that orbit the nucleus of the atom. Atoms also have a unique atomic number, which corresponds to the number of protons in the nucleus, and an atomic mass, which is the combined mass of the protons and neutrons.
2. What is the significance of atomic spectra?
Ans. Atomic spectra refer to the unique patterns of light emitted or absorbed by atoms. These spectra are often used to identify the elements present in a sample. Each element has its own characteristic set of spectral lines, which are produced when electrons in the atom move between different energy levels. By analyzing the pattern of spectral lines, scientists can determine the composition of a substance and gain insights into the energy levels and electronic structure of atoms.
3. How are atomic spectra produced?
Ans. Atomic spectra are produced when electrons in atoms undergo transitions between different energy levels. When an electron absorbs energy, it moves to a higher energy level, and when it releases energy, it moves to a lower energy level. These transitions result in the emission or absorption of light at specific wavelengths. The emitted or absorbed light corresponds to the energy difference between the two energy levels involved in the transition, and this energy difference determines the color or frequency of the light.
4. What is the relationship between atomic spectra and electron configuration?
Ans. The electron configuration of an atom refers to the arrangement of electrons in its energy levels or orbitals. The electron configuration determines the energy levels that electrons can occupy and the transitions they can undergo. As electrons move between different energy levels, they emit or absorb light, leading to the production of atomic spectra. Therefore, the electron configuration of an atom directly influences the pattern of spectral lines observed for that particular element.
5. How are atomic spectra used in practical applications?
Ans. Atomic spectra have numerous practical applications. They are extensively used in astronomy to identify the composition of stars and other celestial objects. Spectral analysis is also employed in chemistry to determine the presence of specific elements in compounds or samples. Additionally, atomic spectra are utilized in various fields such as forensic science, environmental monitoring, and materials science. By studying the unique spectral fingerprints of different elements, scientists can gain valuable insights into the properties and behavior of matter in a wide range of applications.
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