2. Structure of Atom

Class 11 Chemistry Chapter 2 Structure of Atom

Chapter 2 : Structure of Atom

Class 11 Chemistry Chapter 2 Structure of Atom Internal Problems and Answers :

Problem 2.1 : Calculate the number of protons, neutrons and electrons in .

Answer : Number of mass (A) = 80

Number of protons or atomic number(Z) = 35

Number of neutrons = Mass number – Atomic number = 80 - 35 = 45

Number of electrons = Number of protons = 35

Problem 2.2 : The number of electrons, protons and neutrons in a species are equal to 18, 16 and 16 respectively. Assign the proper symbol to the species.

Answer : The atomic number or the number of protons (Z)= 16.

The element is sulphur (S).

Atomic mass number (A) = number of protons + number of neutrons = 16 + 16 = 32

Species is not neutral as the number of protons is not equal to electrons. It is anion (negatively charged) with charge equal to excess electrons = 18 – 16 = 2. The symbol is .

Problem 2.3 : The Vividh Bharati station of All India Radio, Delhi, broadcasts on a frequency of 1,368 kHz (kilo hertz). Calculate the wavelength of the electromagnetic radiation emitted by transmitter. Which part of the electromagnetic spectrum does it belong to?

Solution : Here, m/s , ,

We know that ,

Therefore, the wavelength of the electromagnetic radiation emitted by transmitter is 219.3 m .

This is a characteristic radiowave wavelength .

Problem 2.4 : The wavelength range of the visible spectrum extends from violet (400 nm) to red (750 nm). Express these wavelengths in frequencies (Hz). (1nm = m)

Solution : For frequency of red light : Here, m/s , ,

We know that ,

For frequency of red light : Here, m/s , ,

We have,

The range of visible spectrum is from to Hz in terms of frequency units.

Problem 2.5 : Calculate (a) wavenumber and (b) frequency of yellow radiation having wavelength 5800 Å .

Solution: (a) Here, m ,

We know that,

The wave number ()

(b) Here, m/s , and m

We know that , The wavelength

Problem 2.6 : Calculate energy of one mole of photons of radiation whose frequency is Hz.

Solution : Here, Hz and Js

We know that ,

The energy of a quantum of radiation

Energy of one mole of photons

J/mol

= 199.51 kJ/mol

Problem 2.7 : A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb.

Solution: Here , Js , m/s , m and power of the bulb = 100 watt = 100 J

Energy of one photon

The number of photons emitted

Problem 2.8 : When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy of J . What is the minimum energy needed to remove an electron from sodium? What is the maximum wavelength that will cause a photoelectron to be emitted?

Solution: Here , Js , m/s , m

Energy of one photon

The energy of one mole of photons J

The minimum energy needed to remove one mole of electrons from sodium J

The minimum energy for one electron J

We know that ,

m m nm

Problem 2.9 : The threshold frequency for a metal is . Calculate the kinetic energy of an electron emitted when radiation of frequency hits the metal.

Solution : Here, Js , and

We know that , Kinetic energy

Therefore , the kinetic energy of an electron emitted is J .

Problem 2.10 : What are the frequency and wavelength of a photon emitted during a transition from state to the state in the hydrogen atom?

Solution: Here, and

We know that ,

It is an emission energy .

The frequency of the photons

The wavelength of a photon

Problem 2.11 : Calculate the energy associated with the first orbit of . What is the radius of this orbit?

Solution: We know that ,

For : ,

The radius of the orbit is given by equation,

Problem 2.12 : What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of 10 m ?

Solution : Here, Js , kg and m/s

We know that,

Therefore, the wavelength of a ball is m .

Problem 2.13 : The mass of an electron is kg. If its K.E. is J, calculate its wavelength.

Solution : Here, kg , K.E. J

We know that ,

m/s

We have ,

Therefore, the wavelength is 897 nm.

Problem 2.14 : Calculate the mass of a photon with wavelength 3.6 Å.

Solution : Here, , Js , m/s

We know that,

Therefore, the mass of a photon is kg .

Problem 2.15 : A microscope using suitable photons is employed to locate an electron in an atom within a distance of 0.1 Å. What is the uncertainty involved in the measurement of its velocity?

Solution: Here, m , Js , kg

We know that ,

m/s

Therefore, the velocity is m/s .

Problem 2.16 : A golf ball has a mass of 40g, and a speed of 45 m/s. If the speed can be measured within accuracy of 2%, calculate the uncertainty in the position.

Solution: Here, g kg , Js , m/s

We know that ,

Therefore, the uncertainty in the position is m .

Problem 2.17 : What is the total number of orbitals associated with the principal quantum number n = 3 ?

Answer : Number of Orbitals

For n = 3, the number of orbitals would be:

Number of Orbitals

So, when n = 3, there are 9 orbitals associated with that principal quantum number. These orbitals are distributed among the subshells within the third energy level (n = 3).

There are three types of subshells at this energy level: s, p, and d, each with a specific number of orbitals:

(i) 1 s orbital

(ii) 3 p orbitals

(iii) 5 d orbitals

In total, there are 9 orbitals, which can be distributed as 1s, 3p, and 5d orbitals in the third energy level.

Problem 2.18 : Using s, p, d, f notations, describe the orbital with the following quantum numbers (a) n = 2, l = 1, (b) n = 4, l = 0, (c) n = 5, l = 3, (d) n = 3, l = 2

Solution:

(a) n = 2, l = 1:

For n = 2 and l = 1, we have a p orbital. So, the notation is 2p.

(b) n = 4, l = 0:

For n = 4 and l = 0, we have an s orbital. So, the notation is 4s.

For n = 5 and l = 3, we have an f orbital. So, the notation is 5f.

(d) n = 3, l = 2:

For n = 3 and l = 2, we have a d orbital. So, the notation is 3d

Class 11 Chemistry Chapter 2 Structure of Atom Exercise Questions and Answers :

2.1 (i) Calculate the number of electrons which will together weigh one gram.

(ii) Calculate the mass and charge of one mole of electrons.

Solution: (i) The mass of an electron gram .

Number of electrons electron electron electron

(ii) The mass of one mole of electrons = the molar mass of electrons mol-1 .

The mass of an electron gram .

The mass of one mole of electrons gram gram

Again, The charge of a single electron coulombs

The charge of one mole of electrons C

2.2 (i) Calculate the total number of electrons present in one mole of methane.

(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of 14C. (Assume that mass of a neutron = kg ).

(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of at STP. Will the answer change if the temperature and pressure are changed ?

Solution: (i) The total number of electrons in one molecule of methane

Total electrons in one mole of

(ii) Given, Mass of one neutron

Mass of isotope = 14 atomic mass units (u), and

(a) Total number of neutrons in 7 mg of : Here,

Number of atoms

Total neutrons

(b) Total mass of neutrons in 7 mg of :

The mass of a single neutron is

(iii) Total number and mass of protons in 34 mg of :

Given, Molar mass of

Mass of one proton

(a) Total number of protons in 34 mg of : Here,

Total protons

(b) Total mass of protons in 34 mg of :

The mass of one proton is

Yes, the volume of gases like will change if the temperature and pressure are altered, according to the Ideal Gas Law (). However, the total number and mass of protons in are determined by the number of molecules of ,which is not directly dependent on temperature and pressure.

Therefore, the number and mass of protons will remain constant, as these are based on the composition of molecules and not on the conditions of temperature and pressure.

2.3 How many neutrons and protons are there in the following nuclei ?

Solution : For :

Atomic number (Z) = 6

Mass number (A) = 13

Neutrons = Mass number (A) - Atomic number (Z)

Neutrons in = 13 – 6 = 7

Protons = Atomic number (Z) = 6

For :

Atomic number (Z) = 8

Mass number (A) = 16

Neutrons = Mass number (A) - Atomic number (Z)

Neutrons in = 16 – 8 = 8

Protons = Atomic number (Z) = 8

For :

Atomic number (Z) = 12

Mass number (A) = 24

Neutrons = Mass number (A) – Atomic number (Z)

Neutrons in = 24 – 12 = 12

Protons = Atomic number (Z) = 12

For:

Atomic number (Z) = 26

Mass number (A) = 56

Neutrons = Mass number (A) – Atomic number (Z)

Neutrons in = 56 – 26 = 30

Protons = Atomic number (Z) = 26

For :

Atomic number (Z) = 38

Mass number (A) = 88

Neutrons = Mass number (A) – Atomic number (Z)

Neutrons in = 88 – 38 = 50

Protons = Atomic number (Z) = 38

2.4 Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)

(i) Z = 17, A = 35. (ii) Z = 92, A = 233. (iii) Z = 4, A = 9.

Solution : (i) For Z = 17 , A = 35 :

The complete symbol is (Chlorine)

(ii) For Z = 92, A = 233 :

The complete symbol is (Uranium) .

(iii) For Z = 4, A = 9 :

The complete symbol is (Beryllium) .

2.5 Yellow light emitted from a sodium lamp has a wavelength () of 580 nm. Calculate the frequency () and wavenumber () of the yellow light.

Solution : Here, nm m m/s

We know that, Speed of light

The wavenumber

The wavenumber is

2.6 Find energy of each of the photons which

(i) correspond to light of frequency Hz.

(ii) have wavelength of .

Solution : Here, Js , m/s

(i) For light with a frequency Hz

We know that ,

(ii) For light with a wavelength m

We know that ,

2.7 Calculate the wavelength, frequency and wavenumber of a light wave whose period is s.

Solution : For a wave with a period s

We have,

Again, The wavelength

The wavenumber

2.8 What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?

Solution : Here, m or m , m/s and Js

We know that ,

Therefore the numbers of Photons

2.9 A photon of wavelength m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon (eV),

(ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron () .

Solution : Given, Wavelength of photon,

Work function of the metal,

(i) Here, Js ,

We have,

So, the energy of the photon is 3.1 eV.

(ii) We have,

So, the kinetic energy of the emitted photoelectron is 0.97 eV.

(iii) Here,

We have,

So, the velocity of the photoelectron is .

2.10 Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ .

Solution : Here, ; ;

We have,

Again,

The ionization energy of sodium is 495 kJ/mol.

2.11 A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57μm. Calculate the rate of emission of quanta per second.

Solution : Here, Power of the bulb

Wavelength of light ; ;

We have,

Rate of emission

The rate of emission of quanta (photons) per second is photons per second.

2.12 Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å . Calculate threshold frequency () and work function ( ) of the metal.

Solution : Here, Wavelength of radiation ; ;

We have,

Since the energy of a photon at the threshold frequency is equal to the work function () of the metal:

We have,

2.13 What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?

Solution : Here, ;

We have,

[Since ]

The wavelength of the light emitted when the electron transitions from in a hydrogen atom is 486 nm.

2.14 How much energy is required to ionise a H atom if the electron occupies n = 5 orbit?

Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n = 1 orbit).

Solution : The energy required to ionize an electron from a particular orbit can be calculated using the formula for the energy levels of the hydrogen atom:

For , the energy is:

Ionization energy

For , the energy is:

Ionization energy

Thus, the ionization energy for the electron in the orbit is much smaller than the ionization energy for the electron in the orbit. The further the electron is from the nucleus (higher value), the lower the ionization energy required because the electron is less tightly bound to the nucleus.

2.15 What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?

Solution : When an excited electron in a hydrogen atom falls from to the ground state (), it can emit a series of photons corresponding to the energy difference between various energy levels.

The maximum number of emission lines corresponds to the total number of possible transitions the electron can make as it drops from higher levels to lower ones.

An electron in the level can drop to any of the lower levels: .

The possible transitions are as follows:

After reaching any of the lower levels, the electron can make further transitions to levels closer to the ground state:

The total number of transitions can be calculated by summing all the possible transitions between levels as the electron falls:

From (5 transitions)

From (4 transitions)

From (3 transitions)

From (2 transitions)

From (1 transition)

Thus, the total number of transitions (or emission lines)

Therefore, the maximum number of emission lines that can be observed when an excited electron in a hydrogen atom drops from to the ground state is 15.

2.16 (i) The energy associated with the first orbit in the hydrogen atom is J . What is the energy associated with the fifth orbit?

(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.

Solution : The energy of an electron in a hydrogen atom in the n-th orbit is given by the formula:

For orbit:

(ii) The radius of the n-th orbit in Bohr's model of the hydrogen atom is given by the formula:

Where,

For orbit :

2.17 Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.

Solution :

Solution : The wave number () for an electronic transition in hydrogen is given by the Rydberg equation:

Where,

(final state in the Balmer series),

(initial state for the longest wavelength transition).

The wave number for the longest wavelength transition in the Balmer series of hydrogen is

2.18 What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is ergs.

Solution : The energy of an electron in a Bohr orbit is given by the formula:

Energy in the first Bohr orbit ( ):

Energy in the fifth Bohr orbit ( ):

The energy required to move the electron from the first orbit to the fifth orbit is the difference in energy between these two orbits:

So, the energy required to shift the electron from the first Bohr orbit to the fifth Bohr orbit is

We have,

2.19 The electron energy in hydrogen atom is given by J . Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?

Solution : The electron energy in hydrogen atom is given by J

For orbit :

To remove the electron completely from the orbit, we need to bring the electron to the orbit, which has energy (since the electron is free at infinity).

Thus, the energy required to remove the electron from the orbit is:

So, the energy required to remove the electron from the orbit is .

The relationship between energy and wavelength given by the equation:

Where, ;

We have,

2.20 Calculate the wavelength of an electron moving with a velocity of m .

Solution : Here, ; ;

We have,

The wavelength of the electron moving with a velocity of is or 35.4 pm (picometers).

2.21 The mass of an electron is kg. If its K.E. is J, calculate its wavelength.

Solution : Here, ; ;

The momentum (p) of the electron is related to its kinetic energy (KE) by the equation:

We have,

The wavelength of the electron is or 2.83 μm (micrometers).

2.22 Which of the following are isoelectronic species i.e., those having the same number of electrons?

Solution : Na⁺: Sodium (Na) has an atomic number of 11. When it loses one electron to become Na⁺, it has 10 electrons.

K⁺: Potassium (K) has an atomic number of 19. When it loses one electron to become K⁺, it has 18 electrons.

Mg²⁺: Magnesium (Mg) has an atomic number of 12. When it loses two electrons to become Mg²⁺, it has 10 electrons.

Ca²⁺: Calcium (Ca) has an atomic number of 20. When it loses two electrons to become Ca²⁺, it has 18 electrons.

S²⁻: Sulphur (S) has an atomic number of 16. When it gains two electrons to become S²⁻, it has 18 electrons.

Ar: Argon (Ar) has an atomic number of 18. It has 18 electrons in its neutral state.

Isoelectronic species (having the same number of electrons):

Na⁺, Mg²⁺, and Ar each have 10 electrons.

K⁺, Ca²⁺, and S²⁻ each have 18 electrons.

2.23 (i) Write the electronic configurations of the following ions:

(a) (b) (c) (d)

(ii) What are the atomic numbers of elements whose outermost electrons are represented by

(a) (b) and (c) ?

(iii) Which atoms are indicated by the following configurations ?

(a) [He] (b) [Ne] (c) [Ar]

Solution : (i) Electronic Configurations:

(a) (b) (c) (d)

(ii) (a)

Electron configuration :

Atomic Numbers (Sodium)

(b)

Electron configuration :

Atomic Numbers (Nitrogen)

(c)

Electron configuration :

Atomic Numbers (Chlorine)

(iii) (a) [He]

Electron configuration :

Atomic Numbers . The element is lithium (Li).

(b) [Ne]

Electron configuration :

Atomic Numbers

The element is Phosphorus (P).

Electron configuration :

Atomic Numbers

The element is scandium (Sc).

2.24 What is the lowest value of that allows orbitals to exist ?

Solution : For g orbitals,

The value of must always be less than (the principal quantum number). Therefore, to have g orbitals (), the principal quantum number must be at least 5 (since can take values from 0 to ).

Thus, is the lowest value of that allows for the existence of orbitals.

2.25 An electron is in one of the 3d orbitals. Give the possible values of and for this electron.

Solution : Since the electron is in a orbital, the principal quantum number is 3.

For orbitals, the azimuthal quantum number is always 2 (since corresponds to orbitals, corresponds to orbitals, and corresponds to orbitals).

The magnetic quantum number takes integer values from , including zero.

For , the possible values of are:

2.26 An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.

Solution : (i) The atom contains 29 electrons.

In a neutral atom, the number of electrons is equal to the number of protons. So, the number of protons in this atom is also 29.

Thus, the number of protons is 29.

Since the number of protons is 29, the element is copper (Cu), which has the atomic number 29.

(ii) The electronic configuration of copper (Cu) is:

Electron configuration of : or .

2.27 Give the number of electrons in the species

Solution : is a cation formed by removing one electron from .

Thus, the number of electrons in is: 2 electrons −1 electron = 1

So, has 1 electron.

The number of electrons in is: 1 electrons + 1 electron = 2

So, has 2 electrons.

O₂ (oxygen molecule) consists of two oxygen atoms, each with 8 electrons.

Total of oxygen molecule

is a cation formed by removing one electron from .

So, the number of electrons in is: 16 electrons − 1 electron = 15 electrons

Thus, has 15 electrons.

2.28 (i) An atomic orbital has n = 3. What are the possible values of and ?

(ii) List the quantum numbers ( and ) of electrons for orbital.

(iii) Which of the following orbitals are possible?

1p, 2s, 2p and 3f

Solution : (i) An atomic orbital has .

For an orbital with a principal quantum number :

The azimuthal quantum number can take values from .

So, for , the possible values of are:

Thus, the possible values of and for are:

(ii) For a orbital, the principal quantum number and the azimuthal quantum number

For ( orbital), the possible values of the magnetic quantum number are:

Thus, the quantum numbers for the electrons in a orbital are:

(iii) 1p: The value of (principal quantum number) means that the maximum value of is

(for a orbital) is not possible when .

So, 1p is not possible.

2s: For , the possible values of are 0 ( orbital) and 1 ( orbital).

corresponds to an orbital.

So, is possible.

2p: For , the possible values of are 0 ( orbital) and 1 ( orbital).

corresponds to a orbital.

So, is possible.

3f: For , the possible values of are 0 ( orbital), 1 ( orbital), and 2 ( orbital).

corresponds to an f orbital, but is not possible for because the maximum value of is .

So, 3f is not possible.

2.29 Using notations, describe the orbital with the following quantum numbers.

(a) ; (b) (c) ; (d) .

Solution : (a)

: The principal quantum number indicates the first energy level.

: The azimuthal quantum number corresponds to an orbital.

Thus, the orbital is a orbital.

(b)

: The principal quantum number indicates the third energy level.

: The azimuthal quantum number corresponds to a orbital.

Thus, the orbital is a orbital.

(c)

: The principal quantum number indicates the fourth energy level.

: The azimuthal quantum number corresponds to a orbital.

Thus, the orbital is a orbital.

(d)

: The principal quantum number indicates the fourth energy level.

: The azimuthal quantum number corresponds to an orbital.

Thus, the orbital is a orbital.

2.30 Explain, giving reasons, which of the following sets of quantum numbers are not possible.

(a)

(b)

(c)

(d)

(e)

(f)

Solution : (a)

The principal quantum number must be a positive integer, i.e., .

Since is not valid, this set is not possible.

(b)

For , must be between 0 and , so is valid.

For , can only be 0, so is valid.

The spin quantum number is a valid value.

This set is possible.

(c)

For , must be between 0 and , somust be 0, is not allowed for .

This set is not possible.

(d)

For , can be 0 or 1, so is valid.

For, can be -1, 0, or +1, so is valid.

The spin quantum number is valid.

This set is possible.

(e)

For , must be between 0 and . So, are valid choices, is not valid for .

This set is not possible.

(f)

For , can be , so is valid.

The spin quantum number is valid.

This set is possible.

2.31 How many electrons in an atom may have the following quantum numbers?

(a)

(b)

Solution : (a)

Total possible electrons in shell:

The possible values of (Azimuthal Quantum Number) range from 0 to , i.e., .

The number of orbitals in each subshell is given by .

	Subshell	Orbitals ()	Electrons with
0	4s	1	1
1	4p	3	3
2	4d	5	5
3	4f	7	7

Total number of electrons with is

(b)

, the electron is in the 3rd energy level.

, this means the electron is in the 3s subshell.

The number of orbitals in the 3s subshell

Each orbital can hold two electrons (one with , one with ).

Total number of electrons in the 3s subshell is 2 electrons.

2.32 Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.

Solution : In Bohr's model, the electron in the hydrogen atom moves in a circular orbit around the nucleus with a quantized angular momentum. The quantization condition Bohr proposed is

Where, is the mass of the electron, is the velocity of the electron, is the radius of the orbit, is a positive integer (the principal quantum number) and is Planck's constant.

The de Broglie wavelength associated with a particle with momentum is given by:

Where, is the momentum of the electron.

Thus, the de Broglie wavelength of the electron moving in a circular orbit is:

From Bohr's quantization condition, we have

This shows that the circumference of the Bohr orbit () is an integral multiple () of the de Broglie wavelength ().

2.33 What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of spectrum ?

Solution : The wavelength of a spectral line in a hydrogen-like ion () is given by the Rydberg formula:

Where, Z=2 for ; ,

We have,

For hydrogen ( ), the Rydberg formula becomes:

Since,

If , then

Thus, the transition in hydrogen that corresponds to the same wavelength is

2.34 Calculate the energy required for the process

The ionization energy for the H atom in the ground state is J

Solution : The energy required for the process:

The ionization energy of a hydrogen atom in the ground state is given as:

For a hydrogen-like ion with atomic number , the ionization energy is given by

For ( ), the ionization energy is

2.35 If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.

Solution : The diameter of a carbon atom is given as 0.15 nm.

The length of the scale is 20 cm.

The number of carbon atoms (N) that can fit along the scale is given by:

2.36 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.

Solution : Given, Number of carbon atoms arranged side by side .

Total length of the arrangement

Since the carbon atoms are arranged side by side, the total length of the arrangement is equal to the number of atoms multiplied by the distance between the centers of two adjacent atoms.

The radius of a carbon atom is .

2.37 The diameter of zinc atom is 2.6 Å. Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.

Solution : (a) Radius of zinc atom in picometers (pm):

The radius

(b) The diameter of one zinc atom

The number of atoms (N) that fit into 0.016 m is:

atoms

2.38 A certain particle carries C of static electric charge. Calculate the number of electrons present in it.

Solution : Given, Charge on the particle = C , Charge of one electron

The number of electrons () is

2.39 In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is C, Calculate the number of electrons present on it .

Solution : Given, The charge of a single electron .

The static electric charge on the oil drop is

The number of electrons

So, the number of electrons present on the oil drop is 8 electrons.

2.40 In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results ?

Answer : If a thin foil of light atoms like aluminum were used in Rutherford’s experiment instead of heavy atoms like gold or platinum, the scattering of α-particles would be less pronounced. Lighter atoms have fewer protons in their nucleus, resulting in a weaker Coulomb force acting on the α-particles. As a result, α-particles would experience less deflection and fewer instances of large-angle scattering compared to heavy atom foils.

2.41 Symbols and can be written, whereas symbols and are not acceptable. Answer briefly.

Answer : is the correct notation, where 35 is the atomic number (indicating bromine) and 79 is the mass number.

is incorrect because the atomic number cannot be greater than the mass number. The atomic number should always be less than or equal to the mass number.

2.42 An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.

Solution : Let the number of protons be .

Neutrons are 31.7% more than protons, so the number of neutrons,

The mass number

A/Q,

Thus, the number of neutrons

The element with 35 protons is Bromine (Br).

The atomic symbol is .

2.43 An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion.

Solution : Let the number of electrons in the ion be .

Since the ion has a −1 charge, the number of protons .

The mass number , where is the number of neutrons.

Neutrons are more than electrons: .

A/Q,

Thus, , and .

The element with 17 protons is Chlorine (). The ion is .

2.44 An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.

Solution : Mass number .

The ion has a charge of +3, meaning the number of protons is 3 more than the number of electrons.

If the number of electrons is , then the number of protons will be:

The number of neutrons is 30.4% more than the number of electrons, so the number of neutrons can be expressed as:

Now, we know the mass number

Now, the number of protons

The element with atomic number 26 is Iron ().

Therefore, the symbol for this ion is .

So, the ion is Iron-56 with a +3 charge:

2.45 Arrange the following type of radiations in increasing order of frequency: (a) radiation from microwave oven (b) amber light from traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) X-rays.

Answer: The increasing order of frequency for the given types of radiation is:

(c) Radiation from FM radio < (b) Amber light from traffic signal < (a) Radiation from microwave oven < (e) X-rays < (d) Cosmic rays from outer space.

This order is based on the electromagnetic spectrum, where FM radio has the lowest frequency and cosmic rays have the highest.

2.46 Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is , calculate the power of this laser.

Solution : Here, ; ; ;

The energy of a photon is given by:

Now, the total energy emitted per second is:

The power is simply the energy emitted per second,

Thus, the power of the nitrogen laser is 3.3 MW (megawatts).

2.47 Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, Calculate (a) the frequency of emission, (b) distance traveled by this radiation in 30 s (c) energy of quantum and (d) number of quanta present if it produces 2 J of energy.

Solution : (a) Here, ;

We have,

Thus, the frequency of emission is .

(b) Here, ;

The distance traveled by the radiation is given by:

Thus, the distance traveled by the radiation in 30 seconds is .

The energy of a single photon (quantum) is given by:

Thus, the energy of one quantum (photon) is .

(d) Given, ;

The number of quanta can be calculated using the formula:

Thus, the number of quanta present is photons.

2.48 In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of J from the radiations of 600 nm, calculate the number of photons received by the detector.

Solution : Here, ; ; ;

The energy of a single photon is given by the formula:

Thus, the energy of a single photon is

The number of photons

So, approximately 10 photons were received by the detector.

2.49 Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is , calculate the energy of the source.

Solution : The frequency of the radiation

The energy of a photon is given by the equation,

Where, is the number of photons emitted (), is the energy of a single photon, is Planck's constant (6.626×10-34 J·s) , is the frequency of the radiation.

We have,

2.50 The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.

Solution : Here, ; ; ;

Frequency () is related to the wavelength () by the equation:

For :

The energy difference () is:

2.51 The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

Solution : Given, Work function ,

Incident wavelength ;

(a) The threshold wavelength threshold is the wavelength of light that corresponds to the work function, i.e., the minimum energy required to eject an electron.

Using the equation for the photoelectric effect:

So, the threshold wavelength is 653 nm.

(b) The threshold frequency is given by:

So, the threshold frequency is

Now, the kinetic energy of the ejected photoelectron is:

The velocity of the ejected photoelectron can be found using the kinetic energy formula:

where is the mass of the electron

We have,

So, the velocity of the ejected photoelectron is .

2.52 Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant.

(nm)	500	450	400
(cm)	2.55	4.35	5.35

Solution :

2.53 The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.

Solution :

2.54 If the photon of the wavelength 150 pm strikes an atom and one of this inner bound electrons is ejected out with a velocity of ms-1 , calculate the energy with which it is bound to the nucleus.

Solution :

2.55 Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represeted as . Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.

Solution :

2.56 Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.

Solution :

2.57 Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is m , calculate de Broglie wavelength associated with this electron.

Solution :

2.58 Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, Calculate the characteristic velocity associated with the neutron.

Solution :

2.59 If the velocity of the electron in Bohr’s first orbit is m , Calculate the de Broglie wavelength associated with it.

Solution :

2.60 The velocity associated with a proton moving in a potential difference of 1000 V is m . If the hockey ball of mass 0.1 kg is moving with this velocity , Calculate the wavelength associated with this velocity.

Solution :

2.61 If the position of the electron is measured within an accuracy of + 0.002 nm, Calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is nm, is there any problem in defining this value.

Solution :

2.62 The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists:

Solution :

2.63 The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electron in 4p orbital. Which of these electron experiences the lowest effective nuclear charge ?

Solution:

2.64 Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p.

Solution:

2.65 The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus ?

Solution:

2.66 Indicate the number of unpaired electrons in : (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.

Solution:

2.67 (a) How many subshells are associated with n = 4 ? (b) How many electrons will be present in the subshells having value of –1/2 for n = 4 ?

Solution: (a) For a given principal quantum number , the possible values of the angular momentum quantum number are:

Thus, for , the possible values of are:

Each value of corresponds to a different subshell:

corresponds to the s-subshell

corresponds to the p-subshell

corresponds to thed-subshell

corresponds to the f-subshell

Therefore, the number of subshells associated with is 4.

(b) For , we already know there are 4 subshells: , corresponding to respectively.

Each subshell has a number of orbitals, and each orbital can accommodate 2 electrons with opposite spins ( ).

For the s-subshell (): 1 orbital → 2 electrons (one with and one with )

For the p-subshell ( 3 orbitals → 6 electrons (3 orbitals × 2 electrons each)

For the d-subshell (): 5 orbitals → 10 electrons (5 orbitals × 2 electrons each)

For the f-subshell (): 7 orbitals → 14 electrons (7 orbitals × 2 electrons each)

Since each orbital can hold one electron with , we calculate the number of electrons in each subshell with :

s-subshell: 1 electron with

p-subshell: 3 electrons with (one in each of the 3 orbitals)

d-subshell: 5 electrons with (one in each of the 5 orbitals)

f-subshell: 7 electrons with (one in each of the 7 orbitals)

Total electrons with in :

electrons

So, the number of electrons with in the subshells for is 16 electrons.