# NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom

**NCERT solutions for class 11 chemistry chapter 2 Structure of Atom- **In this chapter, we will study about the discovery of fundamental particles like electron, proton and neutrons. NCERT solutions for class 11 chemistry chapter 2 structure of atom deals with the questions based on various advancements being made to explain different atomic models which gave the basic concept of quantization of energy and gave a modern structure of atoms. In this chapter, there are 67 question in the exercise. The NCERT solutions for class 11 chemistry chapter 2 structure of atom are prepared by our subject experts which provide all the answers. These NCERT solutions will help students to clear their doubts in other subjects and other classes as well. By referring to the NCERT solutions for class 11 , students can understand all the important concepts and practice questions well enough before their examination

The rich diversity of chemical behavior of various elements can be discovered to the differences in the internal structure of atoms of these elements. This chapter starts with some experimental observations made by scientists towards the end of 19th and beginning of the 20th century. According to these observations, atoms are made of sub-atomic particles, such that protons, electrons and neutrons. This concept is very different from Dalton's atomic theory. NCERT solutions for class 11 chemistry chapter 2 structure of atom has covered all the topic wise answers as well as the exercise answers.

After completing NCERT solutions for class 11 chemistry chapter 2 Structure of Atom, students will be able to know about the discovery of proton, electron and neutron; describe Rutherford, Thomson and Bohr atomic models; understand nature of Planck's quantum theory and electromagnetic radiation; explain the photoelectric effect; state the de Broglie relation and Heisenberg uncertainty principle. This chapter also explains atomic orbitals in terms of quantum numbers, Pauli exclusion principle, Aufbau principle and Hund's rule of maximum multiplicity and at the end of chapter students will learn how to write the electronic configurations of atoms. Please scroll down to get free NCERT solutions for class 11 chemistry chapter 2 Structure of Atom.

**Topics of NCERT class 11 chemistry chapter 2 Structure of Atom **

2.1 Discovery of Sub-atomic Particles

2.2 Atomic Models

2.3 Developments Leading to the Bohr’s Model of Atom

2.4 Bohr’s Model for Hydrogen Atom

2.5 Towards Quantum Mechanical Model of the Atom

2.6 Quantum Mechanical Model of Atom

## **NCERT solutions for class 11 chemistry chapter 2 Structure of Atom - Solved Exercise Questions **

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

** Answer ** :

As the mass of one electron we know is .

Therefore,

.

** Question ** ** 2.1 (ii) ** Calculate the mass and charge of one mole of electrons.

** Answer ** :

As the mass of one electron is equal to

Therefore, Mass of 1 mole or electrons .

Charge on one electron is .

Therefore, the charge on 1 mole of electrons will be:

.

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

** Answer ** :

1 molecule of methane contains electrons.

Therefore, 1 mole of methane will contain:

** electrons. **

** Question ** ** 2.2 (ii) ** Find (a) the total number of neutrons in 7 mg of ^{ 14 } C. (Assume that mass of a neutron ).

** Answer ** :

As 1 atom of ^{ 14 } C contains .

and the number of atoms in ^{ 14 } C in 1 mole is atoms.

Therefore, the number of neutrons in 14g of ^{ 14 } C in 1 mole .

The number of neutrons in :

.

** Question ** ** 2.2 (ii) ** Find (b) the total mass of neutrons in 7 mg of ^{ 14 } C. (Assume that mass of a neutron = 1.675 × 10 kg).

** Answer ** :

As the mass of one neutron is .

Then the mass of total neutrons in 7grams of ^{ 14 } C:

.

** Question ** ** 2.2 (iii) ** Find (a) the total number of protons in 34 mg of NH _{ 3 } at STP.

Will the answer change if the temperature and pressure are changed ?

** Answer ** :

1 mole of ammonia .

and 1 atom of contains .

Therefore, the number of protons in 1 mole of .

Number of protons in 3mg of :

** No, there will be no effect of temperature and pressure. **

** Question ** ** 2.2 (iii) ** Find (b) the total mass of protons in 34 mg of NH _{ 3 } at STP.

Will the answer change if the temperature and pressure are changed?

** Answer ** :

As the mass of one proton is

Therefore, the mass of will be:

** No, there will be no effect of temperature and pressure. **

** Question ** ** 2.3(i) ** How many neutrons and protons are there in the following nuclei?

** Answer ** :

Given the nucleus of carbon:

Atomic number ** (Z) ** = 6

Mass number ** (A) ** = 13

Number of protons ** (Z) ** = 6

Number of neutrons ** (A-Z) ** = 13-6 = 7

** Question ** ** 2.3(ii) ** How many neutrons and protons are there in the following nuclei?

** Answer ** :

Given the nucleus of oxygen:

Atomic number ** (Z) ** = 8

Mass number ** (A) ** = 16

Number of protons ** (Z) ** = 8

Number of neutrons ** (A-Z) ** = 16-8 = 8

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

Edit Q

** Question ** ** 2.3(iii) ** How many neutrons and protons are there in the following nuclei?

** Answer ** :

Given the nucleus of Magesium:

Atomic number ** (Z) ** = 12

Mass number ** (A) ** = 24

Number of protons ** (Z) ** = 12

Number of neutrons ** (A-Z) ** = 24-12 = 12

** Question ** ** 2.3(iv) ** How many neutrons and protons are there in the following nuclei?

** Answer ** :

Given the nucleus of Iron:

Atomic number ** (Z) ** = 26

Mass number ** (A) ** = 56

Number of protons ** (Z) ** = 26

Number of neutrons ** (A-Z) ** = 56-26 = 30

=

** Question ** ** 2.3(v) ** How many neutrons and protons are there in the following nuclei?

** Answer ** :

Given the nucleus of Strontium:

Atomic number ** (Z) ** = 38

Mass number ** (A) ** = 88

Number of protons ** (Z) ** = 38

Number of neutrons ** (A-Z) ** = 88-38 = 50

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

** Answer ** :

For the given atomic number ** Z=17 ** and mass number ** A=35; **

Atom is .

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

** Answer ** :

For the given atomic number ** Z=92 ** and mass number ** A=233; **

Atom is .

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

** Answer ** :

For the given atomic number ** Z=4 ** and mass number ** A=9; **

Atom is .

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

** Answer ** :

Given the wavelength of the yellow light emitted from a sodium, lamp is .

And the frequency will be:

Therefore the wavenumber,

** Question ** ** 2.6 ** Find energy of each of the photons which

(i) correspond to light of frequency .

** Answer ** :

If a photon has a frequency of .

Then, the energy of each of the photons will be:

** Question ** ** 2.6 ** Find energy of each of the photons which

(ii) have wavelength of 0.50 Å.

** Answer ** :

For the wavelength .

The energy of each of the photons will be:

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

** Answer ** :

Given frequency, wavelength, and the wave number of a light wave:

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

** Answer ** :

Given the wavelength of light .

and Energy is 1J of energy:

Therefore, the number of photons of light with a wavelength of 4000 pm that provides 1J of energy is:

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

** Answer ** :

The photon is having a wavelength of strikes on a metal surface, where the work function of the metal being is .

So, Energy of the photon:

** Answer ** :

The photon is having a wavelength of strikes on a metal surface, where the work function of the metal being is .

The kinetic energy of the emission will be:

** Answer ** :

The photon is having a wavelength of strikes on a metal surface, where the work function of the metal being is .

From the previous part, we have the Kinetic Energy (K.E.):

** Answer ** :

Given the wavelength of the electromagnetic radiation is which is just sufficient to ionize the sodium atom.

So, the ionization energy required will be:

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

** Answer ** :

Given that the light is monochromatic yellow of wavelength .

Hence the energy emitted by the bulb will be:

Therefore, the number of photons emitted per second:

** Answer ** :

Given the wavelength of radiation is .

But the electrons are emitted with zero velocity from a metal surface when it is exposed to radiation. That means ** the kinetic energy will be zero ** .

So, the ** Threshold frequency ** will be:

and the ** Work function ** will be:

** Answer ** :

When an electron in a hydrogen atom undergoes a transition from an energy level with to an energy level , there will be an emission of energy whose wavelength can be found by:

and wavelength will be equal to:

** Answer ** :

The energy which is required to ionize an H atom if the electron occupies n=5 orbit is:

For ionization from 5th orbit,

Therefore,

For ionization from 1st orbit,

Therefore,

Hence, 25 times less energy is required to ionize an electron in the 5th orbital of the hydrogen atom as compared to that in the ground state.

** Question ** ** 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?

** Answer ** :

The number of lines produced when an electron from shell drops to the ground state:

.

According to the question, the maximum number of emission lines when the excited electron of an H atom in drops to the ground state will be:

These are produced because of the following transitions:

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

** Answer ** :

The energy associated with the first orbit in the hydrogen atom is

The energy of an electron in shell is given by:

So, the energy associated with the ** fifth orbit ** would be:

** Question ** ** 2.16 (ii) ** Calculate the radius of Bohr’s fifth orbit for hydrogen atom.

** Answer ** :

The radius of Bohr's orbit for the hydrogen atom is given by,

So, for we have

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

** Answer ** :

Balmer formula:

As we can note from the formula that the wavenumber is inversely proportional to the wavelength.

Hence, for the longest wavelength transition in the Balmer series of atomic hydrogen wavenumber has to be the smallest or should be minimum i.e., .

For the Balmer series,

Thus, the expression of wavenumber is given by,

** Question ** ** 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 .

** Answer ** :

The ground state energy:

The energy required to shift the electron from the ** first ** Bohr orbit to the ** fifth ** Bohr orbit is:

And the expression for the energy of an electron is given by:

where ** m ** is mass of an electron, ** Z ** is the atomic mass of an atom, ** e ** is a charge of an electron, and ** h ** is the Planck's constant.

Now, substituting the values in the equation, we get

Hence, the wavelength of the emitted light will be:

** Question ** ** 2.19 ** The electron energy in hydrogen atom is given by . 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?

** Answer ** :

The expression for the energy of an electron in hydrogen is:

Where ** m ** is mass of electrons, ** Z ** is the atomic mass of an atom, ** e ** is the charge of an electron, and ** h ** is the Planck's constant.

and electron energy in the hydrogen atom is given by,

The electron energy in orbit is:

Therefore, the energy required for the ionization from is

Now, the longest wavelength of light that can be used to cause this transition will be:

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

** Answer ** :

The wavelength of an electron is given by the ** de Broglie's ** equation:

Where,

is the wavelength of moving particle,

** m ** is the mass of the particle, i.e.,

** v ** is the velocity of the particle,i.e., ( ** Given) **

and ** h ** is the Planck's constant value, i.e.,

Now, substituting the values in the equation, we get

** Hence, the wavelength of the electron moving with a velocity of is . **

** Question ** ** 2.21 ** The mass of an electron is If its K.E. is , calculate its wavelength.

** Answer ** :

The wavelength of an electron can be found by de Broglie's equation:

Given the K.E. of electron which is equal to .

Hence we get,

Hence the wavelength is given by,

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

** Answer ** :

Calculating the number of electrons for each species.

has 11 electrons then, will have electrons.

has 19 electrons then, will have electrons.

has 12 electrons then, will have electrons.

has 20 electrons then, will have electrons.

has 16 electrons then, will have electrons.

has 18 electrons.

Hence, the following are isoelectronic species:

having 10 electrons each.

having 18 electrons each.

** Question ** : ** 2.23 ** (i) Write the electronic configurations of the following ion:

(a)

** Answer ** :

The electronic configuration of is :

Now, the electronic configuration of will be .

** Question ** ** 2.23 ** (i) Write the electronic configurations of the following ion :

(b)

** Answer ** :

The electronic configuration of having is :

Now, the electronic configuration of will be .

** Question ** ** 2.23 ** (i) Write the electronic configurations of the following ion :

(c)

** Answer ** :

The electronic configuration of having is :

Now, the electronic configuration of will be .

** Question ** ** 2.23 ** (ii) What are the atomic numbers of elements whose outermost electrons are represented by

** Answer ** :

With given outermost electrons ,

The complete electronic configuration is

Hence the number of electrons present in the atom of the element is:

.

Therefore, the atomic number of the element is which is ** Sodium (Na) ** .

** Question ** ** 2.23 ** (ii) What are the atomic numbers of elements whose outermost electrons are represented by

** Answer ** :

With given outermost electrons ,

The complete electronic configuration is

Hence the number of electrons present in the atom of the element is:

.

Therefore, the atomic number of the element is 7 which is ** NItrogen (N) ** .

** Question ** ** 2.23 ** (ii) What are the atomic numbers of elements whose outermost electrons are represented by

** Answer ** :

With given outermost electrons ,

The complete electronic configuration is

Hence the number of electrons present in the atom of the element is:

.

Therefore, the atomic number of the element is 17 which is ** Chlorine (Cl) ** .

** Question ** ** 2.23 ** (iii) Which atoms are indicated by the following configurations?

** Answer ** :

The electronic configuration of the element is or

Therefore, the atomic number of the element is ** 3 ** which is ** Lithium ** , a p-block element.

** Question ** ** 2.23 ** (iii) Which atoms are indicated by the following configurations?

** Answer ** :

The electronic configuration of the element is or

Therefore, the atomic number of the element is ** 15 ** which is ** Phosphorus ** , a p-block element.

** Question ** ** 2.23 ** (iii) Which atoms are indicated by the following configurations?

** Answer ** :

The electronic configuration of the element is or

Therefore, the atomic number of the element is ** 21 ** which is ** Scandium ** , a d-block element.

** Question ** ** 2.24 ** What is the lowest value of * n * that allows g orbitals to exist?

** Answer ** :

For g-orbital, the value of Azimuthal quantum number (l) will be 4.

As for any value 'n' of the principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n-1).

** Therefore, for l =4, the minimum value of n should be 5. **

** Question ** ** 2.25 ** An electron is in one of the 3 * d * orbitals. Give the possible values of * n * , * l * and * m _{ l } * for this electron.

** Answer ** :

For d-orbital, the value of Azimuthal quantum number (l) = 2.

When l =2, the values of m are:

Now, for the 3d orbital:

The value of Principal quantum number,

Azimuthal quantum number,

Magnetic quantum number, =

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

** Answer ** :

Given an atom of an element contains 29 electrons and 35 neutrons.

Now, for an atom to be neutral, the number of protons is equal to the number of electrons.

Therefore, the ** number of protons ** in the atom of the given element will be ** 29 ** .

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

** Answer ** :

Given an atom of an element contains 29 electrons and 35 neutrons.

The electronic configuration of the atom will be:

which is the electronic configuration of ** copper ** .

** Question ** ** 2.27 ** Give the number of electrons in the species , and

** Answer ** :

The number of electrons in molecule is .

The number of electrons in molecule will be one less than the number of electrons in molecule. i.e, .

The number of electrons present in molecule will be one less than the number of electrons present in molecule. i.e.,

** Question ** ** 2.28 ** (i) An atomic orbital has * n = 3 * . What are the possible values of * l * and * m _{ l } * ?

** Answer ** :

For a given value of the principal quantum number , the azimuthal quantum number can have values from to .

Therefore, for given atomic orbital ,

The value of can take values from to , i.e., .

And for a given value of , the Magnetic quantum number can have values.

When value of then, ,

or then,

or then,

or then,

** Question ** ** 2.28 ** (ii) List the quantum numbers ( * m _{ l } * and

*l*) of electrons for 3d orbital.

** Answer ** :

For 3d-orbital, the values of Principal quantum number is and Azimuthal quantum number .

Therefore, for ,

, magnetic quantum number can have values.

i.e.,

** Question ** ** 2.28 ** (iii) Which of the following orbitals are possible? * 1p * , * 2s * , * 2p * and * 3f *

** Answer ** :

is ** NOT ** possible because for , the value of is zero.

is possible because, when . .

is possible because when . .

is ** NOT ** possible because for , the value of .

** Question ** ** 2.29 ** Using s, p, d notations, describe the orbital with the following quantum numbers.

** Answer ** :

Here, is principal quantum number and is azimuthal quantum number.

Then the orbital with given quantum numbers is which can have a maximum of 2 electrons.

** Question ** ** 2.29 ** Using s, p, d notations, describe the orbital with the following quantum numbers.

** Answer ** :

Here, is principal quantum number and is azimuthal quantum number.

Then the orbital with given quantum numbers is which can have a maximum of 6 electrons.

** Question ** ** 2.29 ** Using s, p, d notations, describe the orbital with the following quantum numbers.

** Answer ** :

Here, is principal quantum number and is azimuthal quantum number.

Then the orbital with given quantum numbers is which can have a maximum of 10 electrons.

** Question ** ** 2.29 ** Using s, p, d notations, describe the orbital with the following quantum numbers.

** Answer ** :

Here, is principal quantum number and is azimuthal quantum number.

Then the orbital with given quantum numbers is which can have a maximum of 14 electrons.

** Question ** ** 2.30 (a) ** Explain, giving reasons, which of the following sets of quantum numbers are not possible.

** Answer ** :

Given quantum numbers :

** NOT ** possible, because n cannot be equal to zero.

** Question ** ** 2.30 (b) ** Explain, giving reasons, which of the following sets of quantum numbers are not possible.

** Answer ** :

Given quantum numbers :

It is possible and it is ** 1s orbital. **

** Question ** ** 2.30 (c) ** Explain, giving reasons, which of the following sets of quantum numbers are not possible.

** Answer ** :

Given quantum numbers :

It is ** NOT ** possible because when .

** Question ** ** 2.30 (d) ** Explain, giving reasons, which of the following sets of quantum numbers are not possible.

** Answer ** :

Given quantum numbers :

It is possible and it is ** 2p orbital. **

** Question ** ** 2.30 (e) ** Explain, giving reasons, which of the following sets of quantum numbers are not possible.

** Answer ** :

Given quantum numbers :

It is ** NOT ** possible because when , .

** Question ** ** 2.30 (f) ** Explain, giving reasons, which of the following sets of quantum numbers are not possible.

** Answer ** :

Given quantum numbers :

It is possible and it is ** 3p ** orbital.

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

** Answer ** :

The total number of electrons in an atom for a value of n is given by:

Therefore, the total no. of electrons when ,

and half of them i.e. 16 will have .

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

** Answer ** :

When then it is orbital which can have ** 2 electrons ** .

** Answer ** :

According to the Bohr's postulate of angular momentum,

which can be written as: ** .....................................(1) **

Then according to de Broglie's equation for wavelength,

** ......................................(2) **

Now, substituting the values of equation (2) in equation (1) we get,

Thus, the circumference of the Bohr's orbit for the hydrogen atom is an integral multiple of de Broglie's wavelength associated with the electron revolving around the orbit.

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

** Answer ** :

For the transition of H-like particles,

For transition spectrum,

, , and

Therefore,

Then for the hydrogen spectrum,

and

Therefore,

The values of can be found by the hit and trial method in the above equation.

So, we get and , i.e., the transition is from to .

** Question ** ** 2.34 ** Calculate the energy required for the process

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

** Answer ** :

For the hydrogen-like particles,

For H-atom, Ionization energy:

For the given process, the energy required will be:

** Question ** ** 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.

** Answer ** :

Given the diameter of a carbon atom which is

Then the number of carbon atoms which can be placed side by side in a straight line across the length of the scale of length long will be:

** Question ** ** 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.

** Answer ** :

The arrangement length is given which is .

and the number of atoms of carbon which are arranged in this length is given .

Let the radius of carbon atom be then,

or

** Hence the radius of carbon atom is . **

** Question ** ** 2.37 ** The diameter of zinc atom is 2.6 Å. Calculate (a) radius of zinc atom in pm

** Answer ** :

If the diameter of zinc atom is then, its ** radius ** would be:

** Answer ** :

The number of atoms present in a length of will be:

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

** Answer ** :

As the charge carried by one electron is

Therefore, the number of electrons present in particle carrying charge will be:

** Question ** ** 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 , calculate the number of electrons present on it.

** Answer ** :

Given charge on the oil drop is

and the charge carried by one electron is

Therefore, the number of electrons present on the oil drop carrying charge is:

** Question ** ** 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 ** :

The thin foil of heavy atoms, like gold, platinum, etc. have a nucleus carrying a large amount of positive charge. Therefore, some particles will easily get deflected back.

These particles also deflect through small angles because of the large number of a positive charge.

Hence if we use light atoms, their nuclei will have a small positive charge, hence the number of particles getting deflected even through small angles will be negligible.

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

** Answer ** :

The general way to represent an element along with its atomic mass (A) and atomic number (Z) is . Here the atomic number of an element is fixed. However, its mass number is not fixed as it depends upon the isotope taken.

Hence, it is essential to indicate the mass number.

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

** Answer ** :

Let the number of protons of an atom be .

Then the number of neutrons will be,

and the mass number is (Given).

Mass number = number of neutrons + number of protons.

Therefore,

Thus, there are numbers of protons which is also its atomic number.

** Hence, the symbol for the element is . **

** Question ** ** 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.

** Answer ** :

Given an ion has mass number 37 and possesses one unit of negative charge.

Let the number of electrons be then,

the number of neutrons will be:

The number of electrons in the neutral atom (ion possesses one unit of negative charge).

Therefore, the number of protons will be .

** Mass number = number of protons + number of neutrons **

therefore,

Therefore, the number of protons is equal to the atomic number.

** Hence, the symbol for an ion will be: **

** Question ** ** 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.

** Answer ** :

Given an ion has mass number 56 and possesses three units of negative charge.

Let the number of electrons be then,

the number of neutrons will be:

The number of electrons in the neutral atom (ion possesses three units of positive charge).

Therefore, the number of protons will be .

** Mass number = number of protons + number of neutrons **

therefore,

Therefore, the number of protons is equal to the atomic number.

** Hence, the symbol for an ion will be: **

** Answer ** :

The increasing order of frequency of radiations will be:

Radiation from FM radio ** < ** amber light from traffic signal ** < ** radiation from microwave oven ** < ** X-rays ** < ** cosmic rays from outer space.

** Answer ** :

The energy emitted by the nitrogen laser is:

** Question ** ** 2.47 ** Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission,

** Answer: **

The wavelength of neon gas is or

Hence the frequency of this radiation will be:

** Answer ** :

The velocity of neon gas radiation is .

Therefore, the distance travelled in will be:

** Question ** ** 2.47 ** Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (c) energy of quantum

** Answer ** :

The energy of quantum will be:

** Answer ** :

If it produces 2J of energy then, the number of quanta present in it will be .

Therefore,

or ,

Where and from the previous part.

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

** Answer ** :

Let the number of photons received by the detector be .

Then, the total energy it receives from the radiation of will be:

Or

Where,

,

,

and

Substituting the values in the equation above, we get

** number of photons is received by the detector. **

** Question ** ** 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.

** Answer ** :

Given the duration of a radiation source and the number of pulse source is , then its frequency will be:

and the energy of the source for the given frequency will be:

** Answer ** :

We have two wavelengths of and .

Calculating the frequency for each:

Therefore, the energy difference between two excited states will be:

** Question ** ** 2.51 ** The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength

** Answer ** :

Given the work function for the Caesium atom is .

i.e., Or

As

** Therefore, the threshold wavelength is 6.53× 10 ^{ -7 } m. **

** Question ** ** 2.51 ** The work function for caesium atom is 1.9 eV. Calculate (b) the threshold frequency of the radiation.

** Answer ** :

To find threshold frequency:

Where

h = Planck’s constant

= threshold frequency

** Answer ** :

** Finding the kinetic energy of the ejected electrons: **

K.E of the ejected photoelectron:

** Finding the Velocity of the ejected electrons: **

Where,

= mass of electron

= velocity of electron

Therefore, the velocity is given by,

** Question ** ** 2.52 ** Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength

** Answer ** :

Let us assume the threshold wavelength to be and the kinetic energy of the radiation is given as:

** .................................(1) **

Similarly, we can also write,

** ...................................(2) **

** ...................................(3) **

Now, dividing the equations (3) with (1),

Therefore, the wavelength is .

** Question ** ** 2.52 ** Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (b) Planck’s constant.

** Answer ** :

We have the threshold wavelength.

Then substituting this value in any of the equation ( look in the previous part), we get

Taking the mass of an electron to be .

approximately.

** Question ** ** 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.

Answer:

Given work function of the metal, and the Wavelength,

From the Law of conservation of energy, the energy of an incident photon E is equal to the sum of the work function W of radiation and its kinetic energy K.E i.e.,

The energy of incident radiation

Or

Since the potential applied gives the kinetic energy to the radiation, therefore K.E of the electron

** Therefore, Work Function **

** Question ** ** 2.54: ** If the photon of the wavelength 150 pm strikes an atom and one of it's inner bound electrons is ejected out with a velocity of , calculate the energy with which it is bound to the nucleus.

** Answer ** :

Given the wavelength of a photon which strikes an atom is .

Then the energy associated with this photon will be:

Given the velocity of ejected inner bounded electron: .

Then, the energy associated with this electron will be, Kinetic energy.

Hence finding

Where

m = mass of electron, v = velocity of electron

Hence the energy with which the electrons are bounded to the nucleus is:

** Question ** ** 2.55 ** Emission transitions in the Paschen series end at orbit * n = 3 * and start from orbit n and can be represented as

Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.

** Answer ** :

Given transition in the Paschen series end at orbit and starts from orbit n: ** .........................(1) **

** .........................(2) **

Equating both (1) and (2) equations: we get

Therefore, the radiation corresponding to 1285 nm lies in the ** infrared region ** .

** Answer ** :

The radius of orbit of H-like particles is given by:

Or

Here, starting radius,

Ending radius,

Therefore,

If and , then the transition is from orbit to orbit.

** Therefore, it belongs to the Balmer Series ** .

Frequency is given by:

Wavelength :

.

** Therefore, it lies in the visible range. **

** Question ** ** 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 , calculate de Broglie wavelength associated with this electron.

** Answer ** :

According to de- Broglie's equation for the wavelength.

Given the velocity of electron

and mass of electron

So, the wavelength will be:

or

** Answer ** :

Given the wavelength of neutron:

and the mass of neutron

So, According to the de-Broglie's equation,

Substituting the values in above equation:

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

** Answer ** :

Given the velocity of the electron in Bohr's first orbit is .

And we know the mass of electron which is

Hence the de-Broglie's wavelength associated with the electron will be:

** Answer ** :

Given a proton is moving with velocity .

And if the hockey ball of mass is also moving with the same velocity, then

According to de-Broglie's equation we have,

** Answer ** :

We have given the uncertainty in position, i.e., .

According to Heisenberg's Uncertainty Principle:

Where,

is uncertainty in the position of the electron.

is uncertainty in the momentum of the electron.

Then,

Or

The actual momentum of the electron:

** Therefore, it cannot be defined because the actual magnitude of the momentum is smaller than the uncertainty. **

** Question ** ** 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:

** Answer ** :

Quantum number provides the entire information about an electron of a particular atom.

Principal quantum number

Azimuthal quantum number

Magnetic quantum number

Spin quantum number .

The orbitals occupied by the electrons are:

** 1. ** 4d-orbital

** 2. ** 3d-orbital

** 3. ** 4p-orbital

** 4. ** 3d-orbital

** 5. ** 3p-orbital

** 6. ** 4p-orbital

For the same orbitals, electrons will have the same energy and higher the value of value higher is the energy.

** Therefore, the increasing order of energies: **

** Answer ** :

As the ** p-orbital ** is farthest from the nucleus hence the electrons in ** (4p)subshell ** experiences the lowest effective nuclear charge.

** Question ** ** 2.64 ** Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (i) * 2s * and * 3s *

** Answer ** :

Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of an atom exerted by the nucleus of the atom.

Closer orbitals experience more nuclear charge than outer orbitals.

Therefore, ** (i) 2s and 3s **

The 2s orbital is more closer to the nucleus than 3s orbital hence 2s will experience a larger effective nuclear charge compared to 3s.

** Question ** ** 2.64 ** Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (ii) * 4d * and * 4f *

** Answer ** :

Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of an atom exerted by the nucleus of the atom.

Closer orbitals experience more nuclear charge than outer orbitals.

Therefore, ** (ii) 4d and 4f **

The 4d orbital is more closer to the nucleus than 4f orbital hence 4d will experience a larger effective nuclear charge compared to 4f.

** Question ** ** 2.64 ** Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (iii) * 3d * and * 3p *

** Answer ** :

Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of an atom exerted by the nucleus of the atom.

Closer orbitals experience more nuclear charge than outer orbitals.

Therefore, ** (ii) 3d and 3p **

The 3p orbital is more closer to the nucleus than 3d orbital hence 3p will experience a larger effective nuclear charge compared to 3d.

** Question ** ** 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?

** Answer ** :

Silicon has a greater nuclear charge than aluminium .

Hence, the effective nuclear charge exerted on the unpaired 3p electron of silicon would be greater as compared to that of aluminium.

** Question ** ** 2.66 ** Indicate the number of unpaired electrons in:

** Answer ** :

The electronic configuration of P: .

Hence, the number of unpaired electrons are ** 3 ** in 3p orbital.

** Question ** ** 2.66 ** Indicate the number of unpaired electrons in :

** Answer ** :

The electronic configuration of Si : .

Hence, the number of unpaired electrons are ** 2 ** in orbital.

** Question ** ** 2.66 ** Indicate the number of unpaired electrons in :

** Answer ** :

The electronic configuration of Cr : .

Hence, the number of unpaired electrons are ** 6 ** (1 in 4s and 5 in 3d) ** . **

** Question ** ** 2.66 ** Indicate the number of unpaired electrons in :

** Answer ** :

The electronic configuration of Fe : .

Hence, the number of unpaired electrons are ** 4 ** (in 3d ) ** . **

** Question ** ** 2.66 ** Indicate the number of unpaired electrons in :

** Answer ** :

As Krypton (Kr) is a noble gas whose atomic number is 36 and have all orbitals filled.

** Hence, there are no unpaired electrons in Kr element. **

** Question ** ** 2.67 ** (a) How many subshells are associated with n = 4?

** Answer ** :

For a given value of , can have values from to .

Therefore, for ,

can have values from to .

i.e., .

Thus, four subshells are associated with , which are

** Question ** ** 2.67 (b) ** How many electrons will be present in the subshells having * m _{ s } * value of

*–1/2*for

*n = 4*?

** Answer ** :

The number of orbitals in the shell is equal to .

So, for , there are orbitals present.

As each orbital has one electron with spin .

Hence, there will be electrons with .

** NCERT solutions for class 11 chemistry **

**NCERT solutions for class 11 subject wise**

** **

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** Benefits of NCERT solutions for class 11 chemistry chapter 2 Structure of Atom **

- The solutions are written in a comprehensive manner in the NCERT solutions for class 11 chemistry chapter 2 Structure of Atom will help you writing answers in your exam.
- Revision will be easy because the detailed solutions will help you to remember the concepts and get you good marks.
- Homework problems will be easier for you, all you need to do is check the detailed NCERT solutions for class 11 chemistry chapter 2 Structure of Atom and you are ready to go.

If you have a doubt or question that is not available here or in any of the chapters, contact us. You will get all the answers that will help you score well in your exams.

## Frequently Asked Question (FAQs) - NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom

**Question: **Where can I find complete solutions of NCERT class 11 Chemistry

**Answer: **

Refer to this link: https://school.careers360.com/ncert/ncert-solutions-class-11-chemistry

**Question: **How are the ncert solutions beneficial for board exam

**Answer: **

Most of the questions are asked directly from NCERT, hence it is must to do ncert solutions

**Question: **Whether the chapter Structure of Atom is helpful in higher studies?

**Answer: **

Yes, it is helpful to a large extent

**Question: **Write is the official website of NCERT

**Answer: **

official website of NCERT: http://www.ncert.nic.in/

**Question: **What is the weightage of the chapter Structure of Atom for CBSE board exam

**Answer: **

It holds weightage of 6 marks

**Question: **What is the weightage of NCERT chapter structure of Atom for neet exam

**Answer: **

It holds weightage of 2%

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