NCERT Solutions Class 11 Chemistry Chapter 2 - Structure of Atom

NCERT Solutions for Class 11 Chemistry Chapter 2 (Structure of the Atom) help students understand the structure of atoms and prepare for CBSE exams, JEE Main, JEE Advanced, and NEET. This chapter explains how the study of atoms developed from early atomic models to the quantum mechanical model. These concepts also help students understand the basic principles behind technologies such as MRI machines, lasers, and semiconductor devices.

NCERT Solutions for Class 11 Chemistry Chapter 2 explain how electrons are arranged in atoms and how light interacts with matter. The solutions include answers to textbook questions and help students solve numerical problems, understand the topics, and revise for school exams, JEE, and NEET.

1.0Class 11 Chemistry Chapter 2 Structure of Atom: Key Concepts

This chapter explains the structure of an atom, the particles inside it, and how electrons behave. The key concepts include:

Discovery of Subatomic Particles: Learning how electrons, protons, and neutrons were discovered through scientific experiments.
Atomic Models: Understanding Thomson’s, Rutherford’s, and Bohr’s models of the atom and their limitations.
Electromagnetic Radiation: Learning about light and the relationship between its speed, frequency, and wavelength using $c = ν λ .$
Planck’s Quantum Theory: Understanding how energy is absorbed or released in small packets using the equation $E = h ν$ .
Photoelectric Effect: Learning how light can behave like particles called photons when it strikes a metal surface.
Bohr’s Model of Hydrogen: Understanding electron orbits, their energy levels, and the hydrogen emission spectrum.
Dual Nature of Matter: Learning how particles such as electrons can behave like waves using the de Broglie equation (\lambda = $λ = \frac{h}{m v} .$
Heisenberg’s Uncertainty Principle: Understanding why the exact position and momentum of an electron cannot be known at the same time.
Quantum Mechanical Model: Learning about atomic orbitals and the role of the wave function ((\psi)) in describing electrons.
Quantum Numbers: Understanding the four quantum numbers ((n), (l), (m_l), and (m_s)) that describe an electron’s energy, orbital, and spin.
Shapes of Atomic Orbitals: Learning about the shapes of s, p, and d orbitals and how they describe the regions where electrons are likely to be found.
Electronic Configuration: Learning how electrons are arranged in orbitals using the Aufbau principle, Pauli’s exclusion principle, and Hund’s rule.

2.0Detailed NCERT Textbook Solutions : Class 11 Chemistry Chapter 2

(i) Calculate the number of electrons which will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.

Ans.
(i) Mass of one electron $= 9.10939 \times 1 0^{- 31} kg$
∴ Number of electrons that weigh

$= 9.10939 \times 1 0^{- 31} kg = 1$

Number of electrons that will weigh 1 g

$= (1 \times 1 0^{- 3} kg) = \frac{1}{9.10939 \times 1 0 ^{- 31} kg} \times (1 \times 1 0^{- 3} kg) = 0.1098 \times 1 0^{- 3} + 31 = 0.1098 \times 1 0^{28} = 1.098 \times 1 0^{27}$

(ii) Mass of one electron $= 9.10939 \times 1 0^{- 31} kg$
Mass of one mole of electron

$= (6.022 \times 1 0^{23}) \times (9.10939 \times 1 0^{- 31} kg) = 5.48 \times 1 0^{- 7} kg$

Charge on one electron $= 1.6022 \times 1 0^{- 19}$ coulomb Charge on one mole of electron

$= (1.6022 \times 1 0^{- 19} C) (6.022 \times 1 0^{23}) = 9.65 \times 1 0^{4} C$

(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 $^{14} C$ .
(Assume that ass of a neutron $= 1.675 \times 1 0^{-}$ $^{27} kg$ )
(iii) Find (a) the total number and (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?

Ans.

(i) Number of electrons present in 1 molecule of methane $(CH_{4})$

${1 (6) + 4 (1)} = 10$

Number of electrons present in 1 mole i.e., $6. 0^{23} \times 1 0^{23}$ molecules of methane

$= 6.022 \times 1 0^{23} \times 10 = 6.022 \times 1 0^{24}$

(ii)
(a) Number of atoms of $^{14} C$ in 1 mole

$= 6.023 \times 1 0^{23}$

Since 1 atom of $^{14} C$ contains (14-6) i.e., 8 neutrons, the number of neutrons in 14 g of $^{14} C$ is $(6.023 \times 1 0^{23}) \times 8$ . Or, 14 g of $^{14} C$ contains $(6.022 \times 1 0^{23} \times 8)$ neutrons.
Number of neutrons in 7 mg

$= \frac{6.022 \times 1 0 ^{23} \times 8 \times 7 mg}{1400 mg} = 2.4092 \times 1 0^{21}$

(b) Mass of one neutron $= 1.67493 \times 1 0^{- 27} kg$
Mass of total neutrons in 7 g of $^{14} C$

$= (2.4092 \times 1 0^{21}) (1.67493 \times 1 0^{- 27} kg) = 4.0352 \times 1 0^{- 6} kg$

(iii)
(a)

$1 mole of N H_{3} = 1 (14) + 3 (1) g of N H_{3} = 17 g of N H_{3} = 6.022 \times 1 0^{23} molecules of & NH_{3}$

$T o t a l n u mb ero f p ro t o n s p rese n t in 1 m o l ec u l eo f NH_{3}$

$= 1 (7) + 3 (1) = 10$

$N u mb ero f p ro t o n s in 6.023 \times 1 0^{23} m o l ec u l eso f NH_{3} = (6.023 \times 1 0^{23}) (10)$

$= 6.023 \times 1 0^{24}$

17g of $NH_{3}$ contains $(6.023 \times 1 0^{24})$ protons.

Number of protons in 34 mg of $NH_{3}$

$= \frac{6.022 \times 1 0 ^{24} \times 34 mg}{17000 mg} = 1.2046 \times 1022$

(b) Mass of one proton = $1.67493 \times 1 0^{- 27} kg$

Total mass of protons in 34 mg of $NH_{3}$

$= (1.67493 \times 1 0^{- 27} kg) (1.2046 \times 1 0^{22}) = 2.0176 \times 1 0^{- 5} kg$

The number of protons, electrons, and neutrons in an atom is independent of temperature and pressure conditions.

Hence, the obtained values will remain unchanged if the temperature and pressure is changed.

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

$6^{13} C, 8^{16} O, 12^{24} Mg, 26^{56} Fe,_{38}^{88} Sr$

Ans. $_{6}^{13} C$

Atomic mass = 13 Atomic number = Number of protons = 6

Number of neutrons =( Atomic mass )-( Atomic number )

$= 13 - 6 = 7_{8}^{16} O$

Atomic mass =16

Atomic number = 8

Number of protons =8

Number of neutrons

=(Atomic mass)-(Atomic number)

$= 16 - 8 = 8_{12}^{24} Mg$

Atomic mass =24

Atomic number = Number of protons = 12

Number of neutrons

=(Atomic mass)-(Atomic number)

$= 24 - 12 = 12&_{26}^{56} Fe$

Atomic mass = 56

Atomic number = Number of protons = 26

Number of neutrons

=(Atomic mass)-(Atomic number)

$= 56 - 26 = 30&_{88}^{88} Sr$

Atomic mass =88

Atomic number = Number of protons = 38

Number of neutrons

=(Atomic mass)-(Atomic number)

=88-38=50

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) $Z = 4, A = 9$

Ans.

(i) $_{17}^{35} Cl$

(ii) $_{92}^{233} U$

(iii) $_{4}^{9} Be$

Yellow light emitted from a sodium lamp has a wavelength $(λ)$ of 580 nm. Calculate the frequency (v) and wave number $(\overset{v}{ˉ})$ of the yellow light.

Ans. From the expression,

$λ = \frac{c}{v}$

We get, $v = \frac{c}{λ}$

Where, $v =$ frequency of yellow light

$c$ = velocity of light in vacuum =3 $\times 1 0^{8} m / s$

$λ$ = wavelength of yellow light =580 nm

=580 $\times 1 0^{- 9} m$

Substituting the values in expression

(i) : $v = \frac{3 \times 1 0 ^{8}}{5800 \times 1 0 ^{- 9}} = 5.17 \times 1 0^{14} s^{- 1}$

Thus, frequency of yellow light emitted from the sodium lamp =5.17 $\times 1 0^{14} s^{- 1}$

Wave number of yellow light, $\overset{v}{ˉ} = \frac{1}{λ}$

$= \frac{1}{580 \times 1 0 ^{- 9}} = 1.72 \times 1 0^{6} m^{- 1}$

Find energy of each of the photons which

(i) correspond to light of frequency 3 $\times 1 0^{15} Hz .$

(ii) have wavelength of $0.50 \overset{˚}{A}$

Ans.

(i) Energy (E) of a photon is given by the expression,

$E = h ν$

Where, $h = Planck’s constant = 6.626 \times 1 0^{- 34} Js & v = frequency of light = 3 \times 1 0^{15} Hz$

Substituting the values in the given expression of E :

$E = (6.626 \times 1 0^{- 34}) (3 \times 1 0^{15}) E = 1.988 \times 1 0^{-} &^{18} J$

(ii) Energy (E) of a photon having wavelength $(λ)$

is given by the expression, $E = \frac{hc}{λ}$

$h = Planck’s constant = 6.626 \times 1 0^{- 34} Js & c = velocity of light in vacuum = 3 \times 1 0^{8} m / s$

Substituting the values in the given expression of E:

$E = \frac{( 6.626 \times 1 0 ^{- 34} )}{0.50 \times 1 0 ^{- 10}} = 3.976 \times 1 0^{- 15} J$

$∴ E = 3.98 \times 1 0^{- 15} J$

Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0 $\times 1 0^{-}^{10} s .$

Ans. Frequency (v)of light= $\frac{1}{Period}$

= $\frac{1}{2.0 \times 1 0 ^{- 10} s} = 5.0 \times 1 0^{9} s^{- 1}$

Wavelength $(λ)$ of light = $\frac{c}{v}$

Where,

c = velocity of light in vacuum =3 $\times 1 0^{8} m / s$

Substituting the value in the given expression of $λ$ :

$λ = \frac{3 \times 1 0 ^{8}}{5.0 \times 1 0 ^{9}} = 6.0 \times 1 0^{- 2} m$

Wave number $(\overset{v}{ˉ})$ of light = $\frac{1}{λ} = \frac{1}{6.0 \times 1 0 ^{- 2}}$

=1.66

$\times 1 0^{1} m^{- 1} = 16.66 m$

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

Ans. Energy (E) of a photon = $h ν$

Energy $(E_{n})$ of 'n' photons =

$nh ν \Rightarrow$ $n = \frac{E _{n} λ}{hc}$

Where, $λ$ =wavelength of light

$= 4000 pm = 4000 \times 1 0^{- 12} m & c = velocity of light in vacuum = 3 \times 1 0^{8} m / s & h = Planck’s constant = 6.626 \times 1 0^{- 34} Js$

Substituting the values in the given expression of n :

n= $\frac{( 1 ) \times ( 4000 \times 1 0 ^{- 12} )}{( 6.626 \times 1 0 ^{- 34} ) ( 3 \times 1 0 ^{8} )} = 2.012 \times 1 0^{16}$

Hence , the number of photons with a wavelength of 4000pm and energy of 1 J are $2.012 \times 1 0^{16}$

A photon of wavelength $4 \times 1 0^{- 7} m$ 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

$(1 eV = 1.6020 \times 1 0^{- 19} J)$

Ans.

(i) Energy (E) of a photon = $h ν = \frac{hc}{λ}$

Where,

h=Planck's constant = $6.626 \times 1 0^{- 34} J s$

$c = velocity of light in vacuum = 3 \times 1 0^{8} m/s$

$λ = wavelength of photon = 4 \times 1 0^{- 7} m$

Substituting the values in the given expression of E :

$E = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 3 \times 1 0 ^{8} )}{4 \times 1 0 ^{- 7}} = 4.9695 \times 1 0^{- 19} J$

Hence, the energy of the photon is 4.97 $\times 1 0^{- 19} J .$

(ii) The kinetic energy of emission $E_{k}$ is given by

$= h ν - h v_{0}$

$= (E - W) e V = (\frac{4.9695 \times 1 0 ^{- 19}}{1.6020 \times 1 0 ^{- 19}}) eV - 2.13 eV$

$= (3.1020 - 2.13) eV = 0.9720 eV$

Hence, the kinetic energy of emission is 0.97 eV.

(iii) The velocity of a photoelectron (v) can be calculated by the expression,

$\frac{1}{2} m v^{2} = h v - h v_{0} \Rightarrow v \frac{2 ( h v - h v _{0} )}{m}$

Where, $(h v - h v_{0})$ is the kinetic energy of emission in Joules and 'm' is the mass of the photoelectron. Substituting the values in the given expression of v:

$v = \frac{2 \times ( 0.9720 \times 1.6020 \times 1 0 ^{- 19} ) J}{9.10939 \times 1 0 ^{- 31} kg}$

$= 0.3418 \times 1 0^{12} m^{2} s^{- 2} v = 5.84 \times 1 0^{5} ms^{- 1}$

Hence, the velocity of the photoelectron is $5.84 \times 1 0^{5} ms^{- 1}$ .

Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in $kJ mol^{- 1}$

Ans. Energy of sodium (E)
= $\frac{N _{A} hc}{λ}$

$= \frac{( 6.023 \times 1 0 ^{23} mol ^{- 1} ) ( 6.626 \times 1 0 ^{- 34} Js ) ( 3 \times 1 0 ^{8} ms ^{- 1} )}{242 \times 1 0 ^{- 9} m}$

$= 4.947 \times 1 0^{5} J mol^{- 1}$

$494.7 \times 1 0^{3} J mol^{- 1} = 494 kJ mol^{- 1}$

A 25 watt bulb emits monochromatic yellow light of wavelength of $0.57 μ m$ Calculate the rate of emission of quanta per second.

Ans. Power of bulb, $P = 25 Watt = 25 Js^{- 1}$

Energy of one photon, $E = h v = \frac{hc}{λ}$

Substituting the values in the given expression of E:

$E = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 3 \times 1 0 ^{8} )}{( 0.57 \times 1 0 ^{- 6} )} = 34.87 \times 1 0^{- 20} J E = 34.87 \times 1 0^{- 20} J$

Rate of emission of quanta per second

$= \frac{25}{34.87 \times 1 0 ^{- 20}} = 7.169 \times 1 0^{19} s^{- 1}$

Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength $6800 \overset{˚}{A} .$ . Calculate threshold frequency $(v_{0})$ and work function $(W_{0})$ of the metal.

Ans. Threshold wavelength of radian

$(λ_{0}) = 6800 \overset{˚}{A}$ = $6800 \times 1 0^{- 10} m$

Threshold frequency $(v_{0})$ of the metal

$= \frac{c}{λ _{0}} = \frac{3 \times 1 0 ^{8} ms ^{- 1}}{6.8 \times 1 0 ^{- 7} m} = 4.41 \times 1 0^{14} s^{- 1}$

Thus, the threshold frequency $(v_{0})$ of the metal is 4.41 $\times 1 0^{14} s^{- 1} .$

Hence, work function $(W_{0})$ of the metal = $h v_{0}$

$= (6.626 \times 1 0^{- 34} Js) (4.41 \times 1 0^{14} s^{- 1}) = 2.922 \times 1 0^{- 19} J$

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 ?

Ans. The

$n_{i} = 4 to n_{f} = 2$ transition will give rise to a spectral line of the Balmer series.

The energy involved in the transition is given by the relation,

$E = 2.18 \times 1 0^{- 18} [\frac{1}{n i ^{2}} - \frac{1}{n f ^{2}}]$

Substituting the values in the given expression of E:

\begin{aligned}
E= $2.18 \times 1 0^{- 18} [\frac{1}{4 ^{2}} - \frac{1}{2 ^{2}}] = 2.18 \times 1 0^{- 18} [\frac{1 - 4}{16}]$

= $2.18 \times 1 0^{- 18} \times (- \frac{3}{16})$

E= $- (4.0875 \times 1 0^{- 19} J)$

The negative sign indicates the energy of emission.

Wavelength of light emitted $(λ) = \frac{hc}{E}$

$(since E = \frac{hc}{λ})$

Substituting the values in the given expression of $λ$ :

$λ = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 3 \times 1 0 ^{8} )}{4.0875 \times 1 0 ^{- 19}} λ = 4.8631 \times 1 0^{- 7} m = 486.3 \times 1 0^{- 9} m = 486 nm$

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 $n = 1$ =1orbit).

Ans. The expression of energy is given by,

$E_{n} = \frac{- ( 2.18 \times 1 0 ^{- 18} ) Z ^{2}}{n ^{2}}$

Where,

Z=atomic number of the atom

$n =$ principal quantum number

For ionization from $n_{1} = 5$ to $n_{1} = \infty$

$Δ E = E \infty - E 5$

$[\frac{- ( 2.18 \times 1 0 ^{- 18} J ) ( 1 ) ^{2}}{\infty ^{2}}] - [\frac{- ( 2.18 \times 1 0 ^{- 18} J ) ( 1 ) ^{2}}{5 ^{2}}]$

$= (2.18 \times 1 0^{- 18} J) (\frac{1}{( 5 ) ^{2}})$ . (Since $\frac{1}{\infty} = 0$ )

$= (0.0872 \times 1 0^{- 18} J)$

$Δ E = (8.72 \times 1 0^{- 20} J)$

Hence, the energy required for ionization from n = 5 to n = infinity is $(8.72 \times 1 0^{- 20} J)$

Energy required for

$n_{1} = 1 to n = \infty,$

$Δ E^{'} = E_{\infty} - E_{1}$

$[\frac{- ( 2.18 \times 1 0 ^{- 18} J ) ( 1 ) ^{2}}{\infty ^{2}}] - [\frac{- ( 2.18 \times 1 0 ^{- 18} J ) ( 1 ) ^{2}}{1 ^{2}}]$

$(2.18 \times 1 0^{- 18}) [1 - 0] = 2.18 \times 1 0^{- 18} J$

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

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

Ans. When the excited electron of an H atom in $\mathrm{n}=6$ drops to the ground state, the following transitions are possible:

Hence, a total number of (5+4+3+2+1) 15 lines will be obtained in the emission spectrum. The number of spectral lines produced when an electron in the

$n^{th}$ level drops down to the ground state is given by $\frac{n ( n - 1 )}{2} .$

Given, $n = 6$ Number of spectral lines

$= \frac{6 ( 6 - 1 )}{2} = 15$

(i) The energy associated with the first orbit in the hydrogen atom is $- 2.18 \times 1 0^{- 18} J$ atom 1 .What is the energy associated with the fifth orbit?

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

Ans.

(i) Energy associated with the fifth orbit of hydrogen atom is calculated as:

E 5 = \frac{- ( 2.18 \times 1 0 ^{- 18} )}{( 5 ) ^{2}} = \frac{- 2.18 \times 1 0 ^{- 18}}{25} E 5 = - 8.72 \times 1 0^{- 20} J

(ii) Radius of Bohr's $n^{th}$ orbit for hydrogen atom is given by,

$r_{n} = (0.0529 nm) n^{2}$

For,

$n = 5 r 5 = (0.0529 nm) (5)^{2} r 5 = 1.3225 nm$

Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

Ans. For the Balmer series, $n_{i} = 2$ . Thus, the expression of wavenumber $(\overset{v}{ˉ})$

is given by,

$\overset{v}{ˉ} = [\frac{1}{( 2 ) ^{2}} - \frac{1}{n _{f}^{2}}] (1.097 \times 1 0^{7} m^{- 1})$

Wave number $(\overset{v}{ˉ})$ is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, $\overline{v}$ has to be the smallest. For $\overset{v}{ˉ}$ to be minimum, $n_{f}$ should be minimum. For the Balmer series, a transition from $n_{i} = 2$ to $n_{f} = 3$

is allowed. Hence, taking $n_{f} = 3,$

we get:

$\overset{v}{ˉ} = (1.097 \times 1 0^{7}) [\frac{1}{2 ^{2}} - \frac{1}{3 ^{2}}] \overset{v}{ˉ} = (1.097 \times 1 0^{7}) [\frac{1}{4} - \frac{1}{9}] = (1.097 \times 1 0^{7}) (\frac{9 - 4}{36}) = (1.097 \times 1 0^{7}) (\frac{5}{36}) \overset{v}{ˉ} = 1.5236 \times 1 0^{6} m^{- 1}$

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 $- 2.18 \times 1 0^{11}$ ergs.

Ans. Energy (E) of the $n^{th}$ Bohr orbit of an atom is given by,

$E_{n} = \frac{- ( 2.18 \times 1 0 ^{- 18} ) Z ^{2}}{n ^{2}}$

Where,

Where, $Z =$ atomic number of the atom Ground state energy = $- 2.18 \times 1 0^{- 11}$ ergs

$= - 2.18 \times 1 0^{- 11} \times 1 0^{- 7} J = - 2.18 \times 1 0^{- 18} J$

Energy required to shift the electron from

$n = 1$ to $n = 5$ is given as:

$Δ E = E 5 - E 1$

$= \frac{- ( 2.18 \times 1 0 ^{- 18} ) ( 1 ) ^{2}}{( 5 ) ^{2}} - (- 2.18 \times 1 0^{- 18})$

$= (2.18 \times 1 0^{- 18}) [1 - \frac{1}{25}]$

$= (2.18 \times 1 0^{- 18}) (\frac{24}{25}) = 2.0928 \times 1 0^{- 18} J$

Wavelength of emitted light $= \frac{hc}{E}$

$= \frac{( 6.626 \times 1 0 ^{- 34} ) ( 3 \times 1 0 ^{8} )}{( 2.0928 \times 1 0 ^{- 18} )} = 9.498 \times 1 0^{- 8} m$

The electron energy in hydrogen atom is given by $E_{n} = (- 2.18 \times 1 0^{- 18}) / n^{2} J .$ 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?

Ans. Given,

$E_{n} = - \frac{2.18 \times 1 0 ^{- 18}}{n ^{2}} J$

Energy required for ionization from $n = 2$

is given by,

$Δ E = E \infty - E 2 = [(\frac{- 2.18 \times 1 0 ^{- 18}}{( \infty ) ^{2}}) - (\frac{- 2.18 \times 1 0 ^{- 18}}{( 2 ) ^{2}})] J$

$= [\frac{- 2.18 \times 1 0 ^{- 18}}{4} - 0] J = 0.545 \times 1 0^{- 18} J Δ E = 5.45 \times 1 0^{- 19} J$

Here, $λ$ is the longest wavelength causing the transition = $3647 \times 1 0^{- 10} m = 3647 \overset{˚}{A}$

Calculate the wavelength of an electron moving with a velocity of $2.05 \times 1 0^{7} ms^{- 1} .$

Ans. According to de Broglie's equation, $λ = \frac{h}{mv}$

Where,

$λ = wavelength of moving particle m = mass of particle v = velocity of particle h = Planck’s constant$

Substituting the values in the expression of $λ$ :

$λ = \frac{6.626 \times 1 0 ^{- 34} Js}{( 9.10939 \times 1 0 ^{- 31} kg ) ( 2.05 \times 1 0 ^{7} ms ^{- 1} )} λ = 3.548 \times 1 0^{- 11} m$

Hence, the wavelength of the electron moving with a velocity of $2.05 \times 1 0^{7} ms^{- 1}$

is $3.548 \times 1 0^{- 11} m$ .

#Please refer attached PDF for the next set of questions and solutions.

3.0NCERT Class 11 Biology – Chapter-Wise Solutions & Explanations

Get NCERT Solutions for Class 11 Chemistry with answers to textbook questions from every chapter. Learn important concepts, revise key topics, and prepare for your school exams.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Some Basic Concepts of Chemistry
Chapter 3	Classification of Elements and Periodicity in Properties
Chapter 4	Chemical Bonding and Molecular Structure
Chapter 5	Chemical Thermodynamics
Chapter 6	Equilibrium
Chapter 7	Redox Reactions
Chapter 8	Organic Chemistry – Some Basic Principles and Techniques
Chapter 9	Hydrocarbons

4.0Key Features of NCERT Solutions for Class 11 Chemistry Chapter 2

Numerical Problems: Step-by-step solutions for calculating photon energy, matter wave wavelength, and energy changes in hydrogen atoms.
Atomic Orbital Shapes: Simple explanations of the shapes, directions, and nodes of s, p, and d orbitals.
Electronic Configuration: Clear guidance on writing electronic configurations, including the special cases of Chromium (Cr) and Copper (Cu).
Hydrogen Spectrum: Using the Rydberg formula to understand spectral lines in the Lyman, Balmer, Paschen, and Brackett series.
Prepared by ALLEN Faculty: Solutions prepared by ALLEN subject experts to help students understand NCERT concepts and prepare for CBSE exams, JEE, and NEET.

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Why did Rutherford’s atomic model fail to explain atomic stability, according to NCERT Solutions Class 11 Chemistry Chapter 2?

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NCERT Solutions Class 11 Chemistry Chapter 2 - Structure of Atom

1.0Class 11 Chemistry Chapter 2 Structure of Atom: Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Chemistry Chapter 2

3.0NCERT Class 11 Biology – Chapter-Wise Solutions & Explanations

4.0Key Features of NCERT Solutions for Class 11 Chemistry Chapter 2