Welcome to the Chapter 2 - Structure of Atom, Class 11 Chemistry NCERT Solutions page. Here, we provide detailed question answers for Chapter 2 - Structure of Atom. The page is designed to help students gain a thorough understanding of the concepts related to natural resources, their classification, and sustainable development.
Our solutions explain each answer in a simple and comprehensive way, making it easier for students to grasp key topics Structure of Atom and excel in their exams. By going through these Structure of Atom question answers, you can strengthen your foundation and improve your performance in Class 11 Chemistry. Whether you’re revising or preparing for tests, this chapter-wise guide will serve as an invaluable resource.
(i) Mass of an electron = 9.1 x 10-28 g
Or
9.1 x 10-28 g contains = 1 electron
Therefore 1g contains = 1/9.1 x 10-28 *1 = 1.098 x 1027 electrons
(ii) We know, one mole of electron = 6.022 x 1023 electron
Mass of one electron = 9.1 × 10–28 g
Or
mass of 6.022 x 1023 electron = 9.1 x 10-28 x 6.022 x 1023 = 5.48 x 10-4 g
Charge on one electron = 1.6 × 10–19 coulomb
Charge on one mole of electron = (1.6 × 10–19 C x 6.022 × 1023)
= 9.63 × 104 C
From E= hv or hc/ λ, we have λ = 242nm or 242 x 10-9m
C = 3 x 108 m/s
h = 6.62 x 10-34 Js
Now putting these values in the equation we get
E= (6.62 x 10-34 Js x 3 x 108 m/s) / (242 x 10-9m)
= 0.0821 x 10-17 J/atom
Or
E = (0.0821 x 10-17 ) / (1000 x 6.02 x 1023)
= 494 KJ/mol
(Given)Power of bulb, P = 25 Watt = 25 Js–1
We know, Energy of one photon, E = hν λ 
Substituting the values in the given expression of E:
E = 34.87 × 10–20 J
Rate of emission of quanta per second is given by R = P/ E, Where R is the rate of emission, P is the power & E is the energy
Substituting the values in the equation we get
Given λ = 6800 amgstrom or 6800x 10-10 m = 6.8 x 10-7m
C = 3 x 108 m/s
Also λ = c/v or v = c/ λ Therefore putting the values in the equation, we get
=3 x 108 / 6.8 x 10-7m = 4.41 x 1014 s-1
Hence, work function (W0) of the metal = hν0
= (6.626 × 10–34 Js) (4.41 × 1014 s–1)
= 2.922 × 10–19 J
According to Balmer formula
Wave no = RH [1/n12 – 1/n22]
Here n1 = 2 , n2 = 4, RH = 109678
Putting these values in the equation we get
Wave number =109678(1/22 - ¼2) = 109678 x 3/16
Also λ = 1/ wave number
Therefore λ = 16 / 109678 x 3 = 486 nm
The expression of energy is given by,
Where,
Z = atomic number of the atom
n = principal quantum number
For ionization from n1 = 5 to 
,
Therefore ΔE = E2- E1 = - 21.8 X10-19 (1/n22-1/n12)
= 21.8 X10-19(1/n22-1/n12)
= 21.8 X10-19 (1/52-1/∞)
= 8.72 x 10-20 J
For ionization from 1st orbit, n1= 1,
Therefore ΔE’ = 21.8x10-19(1/12-1/∞)
= 21.8x10-19 J
Now ΔE’/ ΔE = 21.8x10-19 / 8.72x10-20 = 25
Thus the energy required to remove electron from 1st orbit is 25 times than the required to electron from 5th orbit.
Number of lines produced when electron from nth shell drops to ground state = n(n-1)/2
Now according to question lines produced when electron drops from 6th shell to ground state
= 6(6-1) / 2 = 15
These are produced due to following transition
6 to 5 5 to 4 4 to 3 3to 2 2 to 1 (5 lines)
6 to 4 5 to 3 4 to 2 3 to 1 (4 lines)
6 to 3 5 to 2 4 to 1 (3 lines)
6 to 2 5 to 1 (2 lines)
6 to 1 (1 lines)
Energy of an electrons = -(2.18x 10-18)/n2
Where n= principal quantum number
Now the Energy associated with the fifth orbit of hydrogen atom is
\begin{align}E_{5}=\frac{-(2.18\times10^{-18})}{(5)^{2}}=\frac{-2.18\times10^{-18}}{25}\end{align}
E5 = –8.72 × 10–20 J
(ii) Radius of Bohr’s nth orbit for hydrogen atom is given by,
rn = (0.0529 nm) n2
For,
n = 5
r5 = (0.0529 nm) (5)2
r5 = 1.3225 nm
According to Balmer formula
ṽ=1/λ = RH[1/n12-1/n22]
For the Balmer series, ni = 2.
Thus, the expression of wavenumber(ṽ) is given by,
Wave number (ṽ) is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, ṽ has to be the smallest.
For ṽ to be minimum, nf should be minimum. For the Balmer series, a transition from ni = 2 to nf = 3 is allowed. Hence, taking nf = 3,we get:
ṽ= 1.5236 × 106 m–1
Ground state energy (E1) = – 2.18 × 10–11 ergs (given)
= –2.18 × 10–11 × 10–7 J
= – 2.18 × 10–18 J
Also the energy required to shift the electron from n = 1 to n = 5 is given as ΔE = E5 – E1
The expression for the energy of hydrogen of electron is
En = -2π2me4Z2/n2h2
Where m= mass of electrons
Z=atomic mass of atom
e = charge of electron
h = planck’s constant
Now putting the values in the equation ΔE = E5 – E1 we get
The expression for the energy of hydrogen of electron is
En = -2π2me4Z2/n2h2
Where m= mass of electrons
Z=atomic mass of atom
e = charge of electron
h = planck’s constant
When n = 1 then En = - (2.18X10-18 ) (GIVEN)
When n = 2 then En= - (2.18X10-18 )/4 = 0.5465X10-18 J
Therefore the energy required for ionization from n = 2 is 5.45 x 10-19 J
Now wavelength of light needed
= E=hv = hc/λ
Or
λ = [{(6.62x10-24)(3x108)} / 5.45] x 10-19 = 3647
(i) Molecule of CH4 (methane) contains electron = 10
Therefore 1 mole (6.022 x 1023 atoms) contains electron = 6.022 x 1024
(ii) a) 1g atom of 14C = 14g = 6.022 x 1023 atoms = 6.022 x 1024 x 8 neutrons
Thus 14g or 14000 mg have 6.022 x 1024 x 8 neutrons
Therefore 7 mg will have neutrons = 6.022 x 1024 x 8 / 14000 x 7 = 2.4088 x 1022
b) mass of 1 neutron = 1.675 x 10-27 kg
Therefore mass of 2.4088 x 1021 neutrons = 2.4088 x 1021 x 1.67 x 10-27 = 4.0347 x 10-6 kg
(iii) a) 1 mol of NH3 = 17g NH3 = 6.022 x 1023 molecules of NH3 = (6.022x1023)(7 + 3) proton = 6.022 x 1024 protons
Therefore 34 mg i.e 0.034 g NH3 = 6.022 x 1024 x 0.034/1 = 1.2044 x 1022 protons
b) mass of 1 proton = 1.6726 x 10-27 kg
Therefore mass of 1.2044 x 1022 protons = (1.6726 x 10-27)(1.2044 x 1022) kg = 2.0145 x 10-5 kg
No, the answer will not change with change in temperature & pressure.
V= 2.05 x107 (given)
According to de Broglie’s equation,
Where, λ = wavelength of moving particle
m = mass of particle (9.10939 x 10-31)
v = velocity of particle
h = Planck’s constant (6.62 x 10-34)
Substituting the values in the expression of λ:
Hence, the wavelength of the electron moving with a velocity of 2.05 × 107 ms–1 is 3.548 × 10–11 m.
We know , KE = ½ mv2
Or
v = (2KE/M)1/2
Or
v = 811.579 m/s
Now putting the values in the de Broglie’s equation
We get
Hence, the wavelength of the electron is 8.9625 × 10–7 m.
Isoelectronic species = they are the species belonging to different atoms or ions which have same number of electrons but different magnitude of nuclear charge.
Now, A positive charge denotes the loss of an electron & A negative charge denotes the gain of an electron by a species.
1) Number of electrons in sodium (Na) = 11
Therefore, Number of electrons in (Na+) = 10
2) Number of electrons in K+ = 18
3) Number of electrons in Mg2+ = 10
4) Number of electrons in Ca2+ = 18
5) Number of electrons in sulphur (S) = 16
∴ Number of electrons in S2- = 18
6) Number of electrons in argon (Ar) = 18
Hence, the following are isoelectronic species:
1) Na+ and Mg2+ (10 electrons each)
2) K+, Ca2+, S2– and Ar (18 electrons each)
Electronic configuration of an atom is defined as the representation of the position of electrons in the various energy shell & subshells.
Now A negative charge on the species indicates the gain of an electron by it & A positive charge denotes the loss of an electron
(i) (a) H– ion
The electronic configuration of H atom is 1s1.(atomic number = 1)
∴ Electronic configuration of H– = 1s2
(b) Na+ ion
The electronic configuration of Na atom is 1s2 2s2 2p6 3s1.(atomic number = 11)
∴ Electronic configuration of Na+ = 1s2 2s2 2p6 3s0 Or 1s2 2s2 2p6
(c) O2– ion
The electronic configuration of 0 atom is 1s2 2s2 2p4.(atomic number = 8)
∴ Electronic configuration of O2– ion = 1s2 2s2 p6
(d) F– ion
The electronic configuration of F atom is 1s2 2s2 2p5.(atomic number = 9)
∴ Electron configuration of F– ion = 1s2 2s2 2p6
(ii) (a) 3s1
Completing the electron configuration of the element as 1s2 2s2 2p6 3s1.
∴ Number of electrons present in the atom of the element
= 2 + 2 + 6 + 1 = 11
∴ Atomic number of the element = 11(sodium)
(b) 2p3
Completing the electron configuration of the element as 1s2 2s2 2p3.
∴ Number of electrons present in the atom of the element = 2 + 2 + 3 = 7
∴ Atomic number of the element = 7(nitrogen)
(c) 3p5
Completing the electron configuration of the element as 1s2 2s2 2p5.
∴ Number of electrons present in the atom of the element = 2 + 2 + 5 = 9
∴ Atomic number of the element = 9(fluorine)
(iii) (a) [He] 2s1
The electronic configuration of the element is [He] 2s1 = 1s2 2s1.
∴ Atomic number of the element = 3 (lithium , a p-block element)
(b) [Ne] 3s2 3p3
The electronic configuration of the element is [Ne] 3s2 3p3= 1s2 2s2 2p6 3s2 3p3.
∴ Atomic number of the element = 15(phosphorous, a p block element)
(c) [Ar] 4s2 3d1
The electronic configuration of the element is [Ar] 4s2 3d1= 1s2 2s2 2p6 3s2 3p6 4s2 3d1.
∴ Atomic number of the element = 21(scandium , a d block element)
Quantum numbers = the set of four number which gives a complete information about the electrons in an atom.
Here n = principal quantum number
As for any value ‘n’ of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n – 1).
For n = 1 (K shell) has l = 0 (one subshell)
For n = 2 (L shell) has l= 0,1 (2 subshell)
For n = 3 (M shell) has l = 0,1,2 (3 subshell)
For n = 4 (N shell) has l = 0,1,2,3 (4 subshell)
For n = 5 (O shell) has l = 0,1,2,3,4 (5subshell)
∴ For l = 4, minimum value of n = 5
When l = 2 the value of m = -2,-1, 0 ,+1,+2
Now, for the 3d orbital:
Principal quantum number (n) = 3
Azimuthal quantum number (l) = 2
Magnetic quantum number (ml) = – 2, – 1, 0, 1, 2
An atom consists of protons, electrons & neutrons.A positive atom will contain higher number of proton(H+), a negative atom will have higher number of electrons(e-) & a neutral atom will contain same number of protons & electrons.Therefore,
(i) For an atom to be neutral, the number of protons is equal to the number of electrons.
∴ Number of protons in the atom of the given element = 29
(ii) The electronic configuration of the atom is
1s2 2s2 2p6 3s2 3p6 4s1 3d10. Which is the electronic configuration of copper
Now A negative charge on the species indicates the gain of an electron by it & A positive charge denotes the loss of an electron
: Number of electrons present in hydrogen molecule (H2) = 1 + 1 = 2
∴ Number of electrons in = 2 – 1 = 1
H2: Number of electrons in H2 = 1 + 1 = 2
: Number of electrons present in oxygen molecule (O2) = 8 + 8 = 16
∴ Number of electrons in = 16 – 1 = 15
Here n= principal quantum number, l= azimuthal quantum number
(a) when n = 1, l = 0 (Given) The orbital is 1s. (can have maximum of 2 electron)
(b) For n = 3 and l = 1 The orbital is 3p. (can have maximum of 6 electrons).
(c) For n = 4 and l = 2 The orbital is 4d. (can have maximum of 10 electrons)
(d) For n = 4 and l = 3 The orbital is 4f. (can have maximum of 14 electrons)
We know atomic number(z) = no of protons = no of electrons
Also mass number(A) = no of protons + no of neutrons
136C:
Here mass number = 13
Atomic number = Number of protons = 6
Number of neutrons = (mass number) – (Atomic number)
= 13 – 6 = 7
:
Here mass number = 16
Atomic number = 8 Number of protons = 8 Number of neutrons = (mass number) – (Atomic number)
= 16 – 8 = 8
:
Here Mass number = 24
Atomic number = Number of protons = 12
Number of neutrons = (mass number) – (Atomic number)
= 24 – 12 = 12
:
Here mass number = 56
Atomic number = Number of protons = 26
Number of neutrons = (mass number ) – (Atomic number)
= 56 – 26 = 30
:
Here mass number = 88
Atomic number = Number of protons = 38
Number of neutrons = (mass number) – (Atomic number)
= 88 – 38 = 50
(a) The given set of quantum numbers is not possible because the value of the principal quantum number (n) cannot be zero.
(b) The given set of quantum numbers is possible.
(c) The given set of quantum numbers is not possible when n = 1, l is not equal to 1
(d) The given set of quantum numbers is possible.
(e) The given set of quantum numbers is not possible when n = 3 ,l = is not equal to 3
(f) The given set of quantum numbers is possible.
Total number of electrons in an atom for a value of n = 2n2
a) Total no of electrons when n=4 ,then n= 2x42 = 32 & half of them have ms = -1/2
b) When n= 3, l=0 means 3s orbital which can have 2 electrons .
The angular momentum of an electron = mvr = nh/2π --------------1
Also according to de Broglie equation λ = h/mv
Or
mv = h/v -----------2
Putting 2 in 1
h/v =nh/2π or 2λr= nλ
Since ‘2πr’ represents the circumference of the Bohr orbit (r), it is proved by equation (3) that the circumference of the Bohr orbit of the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit.
For an atom 
= 1/λ = RHZ2 (1/n12-1/n22)
For He+ spectrum Z = 2, n2=4, n1= 2
Therefore = 1/λ = RH x 4(1/22-1/42) = 3RH/4
For hydrogen spectrum = 3RH/4, Z = 1
Therefore = 1/λ = RH X 1(1/n12-1/n22)
Or
RH (1/n12-1/n22) = 3RH/4
Or
1/n12-1/n22 = 3/4
Which can be so for n1=1 & n2 = 2, i.e the transition is from n = 2 to n=1
For H like particles,
En = - 2π2mZ2e4 / n2h2
For H atom I.E = E – E1 = 0 – (- 2π2me4 / 12h2) = 2π2me4 / h2
= 2.18x 10-18J /atom (given)
For the given process, energy required = En – E1
= 0 - ( - 2π2me4 / 12h2)
= 4 x 2π2me4 / h2
= 4 x 2.18 x 10-18J
= 8.72 x 10-18 J
Diameter of carbon atom = 0.15nm = 0.15 x 10-9m = 1.5 x 10-10m
Length along which atoms are to be placed = 20cm = 20x10-2m = 2 x 10-1m
Therefore no of carbon atoms which can be placed along the length
= (2 x 10-1m) / (1.5x10-10 )
=1.33x109
Total Length = 2.4 cm
Total number of atoms along the length = 2 × 108
Therefore diameter of each atom = 2.4 / (2 × 108)
= 1.2 x 10-8 cm
And radius of the atom = (1.2 x 10-8) / 2
= 0.60 x 10-8cm
= 0.060 x 10-9m
= 0.060nm
(a) radius = 2.6/2 = 1.3Å = 1.3 x 10-10m = 130pm
(b) given Length of the arrangement
= 1.6 cm = 1.6 × 10–2 m
Diameter of zinc atom = 2.6 × 10–10 m
∴ Number of zinc atoms present in the arrangement
Charge carried by one electron = 1.6022 × 10–19 C
Therefore electrons present in particle carrying 2.5x10-16C charge =
Charge carried by the oil drop = 1.282 ×10–18C
Charge carried by one electron = 1.6022 × 10–19C
∴electrons present on the oil drop carrying 1.282x10-18C charge
1) the element with atomic number(Z) 17 & mass number (A) 35 is chlorine = 
2) the element with atomic number(Z) 92 & mass number (A)233 is uranium = 
3) the element with atomic number(Z) 4 & mass number (A) 9 is berellium = 
In 1911, Rutherford performed alpha rays scattering experiment to demonstrate the structure of atom. Heavy atoms have a heavy nucleus carrying a large amount of positive charge.Hence,some alpha particles are easily deflected back on hitting the nucleus.Also a number of alpha particles are deflected through small angles because of large positive charge on the nucleus.If light atoms are use,their nuclei will be light & moreover,they will have small positive charge on the nucleus.Hence, the number of particles deflected back & those deflecte through some angle will be negligible.
The general convention of representing an element along with its atomic mass (A) and atomic number (Z) is 
Atomic number of an element is fixed.However,mass number is not fixed as it depends upon the isotope taken.Hence it is essential to indicate mass number
Mass number = 81 i.e p + n = 81
Let us suppose number of protons = x and neutrons
= x+31.7Xx /100 = 1.317x
Therefore x + 1.317x = 81
Or
2.317x = 81
Or x = 81/2.317 = 35 T
hus number of protons = 35 ,i.e atomic number = 35
Hence the symbol is 8135 Br
Let the number of electrons in the ion carrying a negative charge be x.
Then,
Number of neutrons present
= x + 11.1% of x
= x + 0.111 x
= 1.111 x
Number of electrons in the neutral atom = (x – 1)
(When an ion carries a negative charge, it carries an extra electron)
∴ Number of protons in the neutral atom = x – 1
Therefore 37 = 1.111x + x -1
Or 2.111 x = 38
Or x = 18
Therefore no of protons = atomic no = x -1 = 18 – 1 = 17
∴The symbol of the ion is 
Let the number of electrons present in ion ,S3+ = x
∴ Number of neutrons in it = x + 30.4% of x = 1.304 x
Since the ion is tripositive,
⇒ Number of electrons in neutral atom = x + 3
∴ Number of protons in neutral atom = x + 3
Now mass number = no of protons + no of neutrons
56 = x + 3 + 1.304x
Or 2.304x = 53
Or x = 23
Therefore no of protons = atomic no = x + 3 = 23 + 3 = 26
∴ The symbol of the ion 
Cosmic rays < X-rays < radiation from microwave ovens < amber light < radiation of FM radio
E = Nhv = Nhc/λ
Where N = number of photons emitted
h = Planck’s constant
c = velocity of radiation
λ = wavelength of radiation
Substituting the values in the given expression of Energy (E):
= 3.33 × 106 J
Hence, the power of the laser is 3.33 × 106 J.
λ = 616nm or = 616x10-9m
a) Frequency , v=c/ λ = 3.8x108 / 616x10-9 = 4.87x1014 s-1
b) Velocity of the radiation = 3x108 m/s
Therefore distance travelled in 30s = 30x3x108 = 9.0x109m
c) E = hv = hc/ λ = (6.626x10-34)(3.0x108) / (616x10-9 ) = 32.27x10-20J
d) No of quanta in 2J of energy = 2/32.27 x 10-28 = 6.2x1018
Energy of 1 photon = hv= hc/ λ = (6.626x10-34)(3x108) / 600x10-9 = 3.313 x 10-19J
Total energy received = 3.15 x10-8J
Therefore no of photons received = 3.15x10-18 / 3.313x10-19 = 9.51
Frequency = 1/2.x10-9 = 0.5x109 / sec
Energy = Nhv = (2.5x105)(6.626x10-34)(0.5x109)
= 8.28x10-10J
Here λ = 580 nm(given)
We have to find out ν & 
Therefore from the equation 
we can write, 
Where, ν = frequency of yellow light
c = velocity of light in vacuum = 3 × 108 m/s
λ = wavelength of yellow light = 580 nm = 580 × 10–9 m
Substituting the values in we get
Thus, frequency of yellow light emitted from the sodium lamp
= 5.17 × 1014 s–1
Wave number = it is defined as the number of wavelengths which can be accommodated in 1cm length along the direction of propagation.
Wave number of yellow light is given by the equation 
We have λ = 580 nm(given),putting the value in the equation we get
λ1 = 589 nm = 5.89X10-9m
therefore v1=c/ λ1 = 3x108/589x10-9 = 5.093x1014/sec
λ2 = 589.6nm = 589.6x10-9m
therefore v2 = c/ λ2 = 3x108/589.6x10-9
= 5.088x1014/sec
ΔE=E2-E1 = h(v2-v1)
= (6.626x10-34)(5.093x-5.088)x1014
= 3.31x10-22J
A) Work function of caesium (WO) = hvo
Therefore vo = WO/h = 1.9 x1.602 x10-19 / 6.626x10-34
= 4.59x1014/sec
B) λo =c/vo = 3x108 / 4.59x1014 = 6.54x10-7m
C) K.E of ejected electron = h(v-vo) = hc(1/λ – 1/ λo)
=(6.626x3x10-26) (1/500x10-9 – 1/654x10-9)
=(6.626x3x10-26) / 10-9(154/500x654)
= 9.36x10-20J
k.E = 1mv2/2 = 9.36x10-20J
=9.1x10-31/2 = 9.36x10-20J
Or
v2 = 20.55x1010m2s-2
Or
v = 4.53x105ms-1
Following results are observed when sodium metal is irradiated with different wavelengths.
Calculate (a) threshold wavelength and, (b) Planck’s constant.
| λ (nm) | 500 | 450 |
400 |
| v × 10–5 (cm s–1) | 2.55 | 4.35 | 5.35 |
(a) Let us suppose threshold wavelength to be λD nm (= λ0 x 10-9 m), the kinetic energy of the radiation is given as:
h(v-v0)=1/2 mv2
Or
hc(1/λ - 1/λD) = 1/2 mv2
Threshold wavelength (λD)= 540 nm
Substituting this value in equation iii,we get
= h x (3x108)/10-9 [1/400-1/540]
= [(9.11x10-31)(5.20x106)2] / 2
=6.66x10-34Js
From the principle of conservation of energy, the energy of an incident photon (E) is equal to the sum of the work function (W0) of radiation and its kinetic energy (K.E) i.e.,
Energy of incident radiation(E)= hc/λ = (6.626x10-34)(3x108)/(256.7x 10-9)
= 7.74x10-19 J
Since the potential applied gives the kinetic energy to the radiation ,therefore K.E of the electron = 0.35Ev
Therefore work function = 4.83 – 0.35 = 4.48 eV
Energy of incident photon (E) is given by,
= 10.2480 × 10–17 J
= 1.025 × 10–16 J
Energy with which the electron was bound to the nucleus
= 13.25 × 10–16 J – 1.025 × 10–16 J
= 12.225 × 10–16 J
= 12.225 × 10–16 J/1.602 x10-19 eV
= 7.63x103eV
V = c/λ = 3x108/1285x10-9
= 3.29 x 1015(1/32 – 1/n2)
= 1/n2 = 1/9 - 3x108/1285x10-9 x (1 / 3.29 x 1015)
= 0.111- 0.071 = 0.04 or 1/25
Or
n2 = 25,
n= 5
The spectrum lies in the infra-red region.
The radius of the nth orbit of hydrogen-like particles = 0.529n2/Z Å
Now r1 = 1.3225 nm or 1322.5 pm = 52.9n12
And
r2 = 211.6pm = 52.9n22/Z
Therefore r1/r2 = 1322.5 / 211.6 = n12/n22
or n12/n22 = 6.25
or n1/n2 = 2.5
therefore n2 = 2 , n1 = 5.
Thus the transition is from 5th orbit to 2nd orbit. It belongs Balmer series
Wave number 
for the transition is given by,
1.097 × 107 m–1 (1/22-1/52)
=1.097 x 107m-1 (21/100)
= 2.303 × 106 m–1
Wavelength (λ) associated with the emission transition is given by,
= 0.434 ×10–6 m
λ = 434 nm
the region is visible region
From de Broglie’s equation,
λ = h/mv
Substituting the values in this equation,we get
6.626x 10-34 / (9.11x10-31)(1.6x106)
= 4.55 x 10-10m
= 455pm
Mass of neutron = 1.675 x10-27 kg
We know λ = h/mv
or
v = h/mλ
Substituting the value ,we get
6.626x10-34 / (1.675x10-27)(800x10-12)
= 4.94x104 m/sec
According to de Broglie’s equation,
λ = h/mv
Where,
λ = wavelength associated with the electron
h = Planck’s constant
m = mass of electron
v = velocity of electron
Substituting the values in the expression of λ:
λ = 6.626x10-34 / (9.11x10-31)(2.19x106)
=3.32x10-10
= 332 pm
Planck’s quantum theory = Based on the assumption that all atoms on the surface of the heated solid vibrate at the frequency, Planck developed a model that came to be known as Planck’s equation. Through experiments of frequencies and temperature, Planck was able to generate a constant, Planck’s constant
h = 6.62607 x 10-34 J s
Using this constant he was able to restate his theory: energy was directly proportional to frequency. He wrote his equation as
E=hν
where E is energy, h is Planck’s constant, and v is frequency.
(i) Energy (E) of a photon is given by the expression,
E=hν
Where, h = Planck’s constant = 6.626 × 10–34 Js
ν = frequency of light = 3 × 1015 Hz
Substituting the values in the given expression E = hv we get
E = (6.626 × 10–34) (3 × 1015)
E = 1.988 × 10–18 J
(ii) Energy (E) of a photon having wavelength (λ) is given by the expression,
where , h = Planck’s constant = 6.626 × 10–34 Js
c = velocity of light in vacuum = 3 × 108 m/s
Substituting the values in the given expression of E:
λ = h/mv
= 6.626x10-34 / (0.1)(4.37x105)
= 1.515x10-28m
From Heisenberg’s uncertainty principle,
Where,
Δx = uncertainty in position of the electron
Δp = uncertainty in momentum of the electron
Δx = 0.002nm = 2x10-12m(given)
Therefore, Substituting the values in the expression of Δp:
Δp = h/4π Δx or
= 2.637 × 10–23 Jsm–1
Δp = 2.637 × 10–23 kgms–1 (1 J = 1 kgms2s–1)
Actual momentum = h/4πx 0.05nm
= 6.626x10-34/4 x3.14 x 5 x10-11
= 1.055 x 10-24 kg m/sec
Quantum number – they are the index numbers which gives the complete address of an electron
Here n =principal quantum number
l = azimuthal quantum number
ms = Spin quantum number
For n = 4 and l = 2, the orbital occupied is 4d.
For n = 3 and l = 2, the orbital occupied is 3d.
For n = 4 and l = 1, the orbital occupied is 4p.
Hence, the six electrons i.e., 1, 2, 3, 4, 5, and 6 are present in the 4d, 3d, 4p, 3d, 3p, and 4p orbitals respectively.
Therefore, the increasing order of energies is 5(3p) < 2(3d) = 4(3d) < 3(4p) = 6(4p) < 1 (4d).
4p electrons, being farthest from the nucleus,experience the lowest effective nuclear charge.
Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of a multi-electron atom. The closer the orbital, the greater is the nuclear charge experienced by the electron (s) in it.
(i) 2s is closer to the nucleus than 3s.Hence 2s will experience larger effective nuclear charge.
(ii) 4d will experience greater nuclear charge than 4f since 4d is closer to the nucleus than 4f.
(iii) 3p will experience greater nuclear charge since it is closer to the nucleus than 3f because 3p is closer to nucleus than 3f.
Nuclear charge is defined as the net positive charge experienced by an electron in a multi-electron atom.
Silicon has greater nuclear charge(+14) than aluminium (+13).
Hence the unpaired 3p electron in case of silicon will experience more effective nuclear charge.
(a) Phosphorus (P): 1s2 2s2 2p6 3s2 3p3
No of unpaired electron = 3
(b) Silicon (Si): 1s2 2s2 2p6 3s2 3p2
No of unpaired electron = 2 (since p orbital can have maximum 6 electron )
(c) Chromium (Cr): 1s2 2s2 2p6 3s2 3p6 4s1 3d5
No of unpaired electrons = 6 (since 1 electron is to be added to 4s & 5 electron to be added to 3d orbital.
(d) Iron (Fe): 1s2 2s2 2p6 3s2 3p6 4s2 3d6
No of unpaired electrons = 4
(e) Krypton (Kr): 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6
Since all orbitals are fully occupied, there are no unpaired electrons in krypton.
(a) n = 4 (Given)
For a given value of ‘n’, ‘l’ can have values from zero to (n – 1).
∴ l = 0, 1, 2, 3
Thus, four sub-shells are associated with n = 4, which are s, p, d and f.
(b) Number of orbitals in the nth shell = n2
For n = 4
Number of orbitals = 16
Each orbital has one electron with ms = -1/2
Hence there will be 16 electrons with ms = -1/2
Given period = 2.0 x 10-10s
We know, Frequency (ν) of light 
,therefore putting the values .we get
We also know
Where,c = velocity of light in vacuum = 3×108 m/s
Substituting the value in the given expression of λ, we get
Wave number 
of light 
Energy (E) of a photon = hc/λ
Where,λ = wavelength of light = 4000 pm = 4000 ×10–12 m = 4x 10-9 m
c = velocity of light in vacuum = 3 × 108 m/s
h = Planck’s constant = 6.626 × 10–34 Js
thefore the energy of photon (E)
= 6.626 × 10–34 Js X 3 × 108 m/s / 4x 10-9 m = 4.965 x 10-16 J
Now 4.965 x 10-16 J is the energy of = 1 photon
Therefore 1 J will be the energy of = 1/4.965 x 10-16 J = 2.014 x 1015 photons
(1)We know λ = 4 × 10–7 m(given)
C = 3 x 108
From the equation E= hv or hc/ λ
Where, h = Planck’s constant = 6.626 × 10–34 Js
c = velocity of light in vacuum = 3 × 108 m/s
λ = wavelength of photon = 4 × 10–7 m
Substituting the values in the given expression of E:
Hence, the energy of the photon is 4.97 × 10–19 J.
(ii) The kinetic energy of emission Ek is given by
= (3.1020 – 2.13) eV
= 0.9720 eV
Hence, the kinetic energy of emission is 0.97 eV.
(iii) The velocity of a photoelectron (ν) can be calculated by the expression,
Where, (hv-hv0) 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 = 5.84 × 105 ms–1
Hence, the velocity of the photoelectron is 5.84 × 105 ms–1.
can be written, whereas symbols 
are not acceptable. Answer briefly.