MP Board Class 11th Chemistry Solutions Chapter 2 Structure of Atom

In this article, we will share MP Board Class 11th Chemistry Solutions Chapter 2 Structure of Atom Pdf, These solutions are solved by subject experts from the latest edition books.

MP Board Class 11th Chemistry Solutions Chapter 2 Structure of Atom

MP Board Class 11 Chemistry Structure of Atom Textbook Questions and Answers

Question 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) Mass of one electron = 9.11 × 10-31 kg
one gram = 10-3 kg
∴ Number of electrons in one gram = \(\frac{10^{-3} \mathrm{~kg}}{9.11 \times 10^{-31} \mathrm{~kg}}\) = 1.098 ×1027

(ii) Mass of one electron =9.11 × 10-31 kg
∴ Mass of one mole of electrons
= (9.11 × 10-31) × (6.022 × 1023)
= 5.486 × 10-7 kg
Charge on one electron = 1.602 × 10-19 C
∴ Charge on one mole of electrons
= (1.602 × 10-19) × (6.022 × 1023)
= 9.65 × 104C.

Question 2.
(i) Calculate the total number of electrons present in one mole of methane.
(ii) Find (a) the total number and (6) the total mass of neutrons in 7 mg of 14C (Assume that the mass of neutron = 1.675 × 10-27 kg)
(iii) Find (a) the total number and (6) the total mass of protons in 34 mg of NH3 at S.T.P.
Will the answer change if temperature and pressure are changed?
Solution.
(i) 1 molecule of methane (CH4) contains electrons = 6 + 4 = 10
∴ 6.022 × 1023 molecules will contain electrons = 6.022 × 1023 × 10 = 6.022 × 1024

(ii)
(a) 1 g atom of 14C = 14 g = 6.022 × 1023 atoms One atom of 14C contains neutrons = 14 – 6 = 8
Thus, 14 g or 14000 mg contain = 8 × 6.022 × 1023 neutrons
∴ 7 mg will contain neutrons = \(\frac{8 \times 6.022 \times 10^{23}}{14000}\) × 7
= 2.4088 × 1021
(b) Mass of 1 neutron = 1.675 × 10-27 kg
∴ Mass of 2.4088 × 1021 neutrons
= (2.4088 × 1021) (1.675 × 10-27 kg)
= 4.0347 × 10-6 kg

(iii) (a) 1 mol of NH3 = 17 g NH3 = 6.022 × 1023 molecules of NH3
The number of protons in 17 g of NH3 = (6.022 × 1023) × (7 + 3) protons = 6.022 × 1024 protons
The number of protons is 34 mg or 0.034 g of NH3
= \(\frac{6.022 \times 10^{24}}{17 \mathrm{~g}}\) × 0.034 g
= 1.2044 × 1022
(b) Mass of one proton = 1.673 × 10-27 kg
∴ Mass of 1.2044 × 1022 protons
= (1.673 × 1027) × (1.2044 × 1022) kg = 2.015 × 10-5 kg.
There is no effect of temperature and pressure.

Question 3.
How many neutrons and protons are there in the following nuclei?
6C13, 😯16, 12Mg24, 26Fe56, 38Sr88.
Solution.
MP Board Class 11th Chemistry Solutions Chapter 2 Structure of Atom 1

Question 4.
Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)
(i) Z = 17, A = 35 (if) Z = 92, A = 233 (iii) Z = 4, A = 9.
Solution.
(i) 17Cl35
(ii) 92U233
(iii) 4Be9.

Question 5.
Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (v) and wave number \(\bar{V} \) of the yellow light.
Solution.
Wavelength of the radiation λ = 580 nm
= 5.80 × 10-9 m = 5.80 × 10-7m
Velocity of radiation, c = 3 × 108 m/s
c = vλ

Frequency v = \(\frac{c}{\lambda}\) = \(\frac{3 \times 10^{8} \mathrm{~ms}^{-1}}{5.80 \times 10^{-7} \mathrm{~m}}\) = 5.17 × 1014 s -1
Wave number \(\bar{V} \) = \(\frac{1}{\lambda}\) = \(\frac{1}{5.80 \times 10^{-7} \mathrm{~m}}\)
= 1.72 × 106 m,sup>-1.

MP Board Class 11th Chemistry Solutions Chapter 2 Structure of Atom

Question 6.
Find energy of each of the photons which
(i) correspond to light of frequency 3 x 1015 Hz. (ii) have wavelength of 0.50 A.
Solution.
(i) E = hv
= 6.63 × 10-34 Js × 3 × 1015 s-1
= 1.989 × 10-18 J

(ii) E = hv = \(\frac{h c}{\lambda}\)
= \(\frac{6.63 \times 10^{-34} \mathrm{Js} \times 3 \times 10^{8} \mathrm{~ms}^{-1}}{0.50 \times 10^{-10} \mathrm{~m}} \)
= 3.98 × 10-15 J.

Question 7.
Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0 × 10-10 s.
Solution.
Frequency (v) of wave is inverse of period
∴ v = \(\frac{1}{2.0 \times 10^{-10} \mathrm{~s}}\)
= 5 × 109 s-1
c = vλ
∴ λ = \(\frac{c}{v}\) = \(\frac{3 \times 10^{8} \mathrm{~ms}^{-1}}{5 \times 10^{9} \mathrm{~s}^{-1}}\)
= 6 × 10-2 m
Wave number \(\overline{\mathrm{v}}\) = \(\frac{1}{\lambda}\)
= \(\frac{1}{6 \times 10^{2} \mathrm{~m}} \)
= 16.66 m-1.

Question 8.
What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?
Solution.
Suppose N photons of the light with wavelength 4000 pm can provide 1 J of energy.
Energy of N photons = Nhv
1J = Nhv = N\(\frac{h c}{\lambda}\)

N = \(\frac{1 \mathrm{~J} \times \lambda}{h c}\)
= \(\frac{1 \mathrm{~J} \times 4000 \times 10^{-12} \mathrm{~m}}{6.63 \times 10^{-34} \mathrm{Js} \times 3 \times 10^{8} \mathrm{~ms}^{-1}}\)
= 2.01 × 1016.

Question 9.
A photon of wavelength 4 x 10-7 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 = 1.602 × 10-19 J).
Solution,.
(i) Energy of the photon = hv = \(\frac{h c}{\lambda}\)
= \(\frac{6.63}{4} \cdot \frac{10^{-34} \mathrm{Js} \times 3 \times 10^{8} \mathrm{~ms}^{-1}}{4 \times 10^{-7} \mathrm{~m}}\)
= 4.97 × 10-19 J.

(ii) Energy of photon = K.E. + Work function
Kinetic energy = Energy of photon – Work function
= 4.97 × 10-19 J – 2.13 × 1.602 × 1019 J
= 1.56 × 10-19 J.
= 0.97 eV

(iii) \(\frac{1}{2}\) mv2 = 1.56 × 10-19 J
v2 = \(\frac{2 \times 1.5\left(3 \times 10^{-19}\right.}{m}\)
= \(\frac{2 \times 1.56 \times 10^{-19}}{9.1 \times 10^{-31}}\)
v2 = 0.34 × 1012
v = 5.85 × 105 ms-1.

Question 10.
Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol-1.
Solution.
Energy of one photon of the radiation having wavelength 242 nm is just sufficient to ionise the sodium atom.
∴ Ionisation energy of sodium = Energy of one photon of radiation of wavelength
Energy of photon = hv
= \(\frac{h c}{\lambda}\)
= \(\frac{6.63 \times 10^{-34} \mathrm{Js} \times 3 \times 10^{8} \mathrm{~ms}^{-1}}{242 \times 10^{-9} \mathrm{~m}} \)
= 8.22 × 10-19 J/photon
= 8.22 × 1022 kJ/photon
∴ Ionisation energy per mole
= 8.22 × 10-22 × 6.022 × 1023 kJ/mol
= 495 kJ/mol.

Question 11.
A 25-watt bulb emits monochromatic yellow light of wavelength of 0.57 pm. Calculate the rate of emission of quanta per second.
Solution.
Energy emitted by the bulb = 25 watt = 25 Js-1
Energy of one photon, E = hv = h \(\frac{c}{\lambda}\)
Here, λ = 0.57 μm = 0.57 × 10-6 m
Putting c = 3 × 108 ms-1 h = 6.63 × 10-34 Js, we get

E = \(\frac{\left(6.63 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^{8} \mathrm{~ms}^{-1}\right)}{0.57 \times 10^{-6} \mathrm{~m}}\)
= 3.49 × 10-19 J
∴ Number of photons emitted per sec = \(\frac{25 \mathrm{Js}^{-1}}{3.49 \times 10^{-19} \mathrm{~J}}\)
= 7.16 × 1019.

Question 12.
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (v0) and work function (W0) of the metal.
Solution.
Energy of photon = \(\frac{h c}{\lambda}\)
= \(\frac{6.63 \times 10^{-34} \mathrm{Js} \times 3 \times 10^{8} \mathrm{~ms}^{-1}}{6800 \times 10^{-10} \mathrm{~m}}\)
= 2.93 × 10-19 J
Energy of photon = K.E. + W0 = 0 + W0 = WQ
∴ W0 = 2.93 × 10-19 J
Threshold frequency v0 = \(\frac{c}{\lambda_{0}}\)
= \(\frac{3 \times 10^{8} \mathrm{~ms}^{-1}}{6800 \times 10^{-10} \mathrm{~m}}\)
= 4.41 × 1014 s-1.

MP Board Class 11th Chemistry Solutions Chapter 2 Structure of Atom

Question 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.
According to Rydberg’s formula, \(\bar{v}\left(\mathrm{~cm}^{-1}\right)\) = 109,677 \(\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)\)
In the present case n2 = 4 and n1= 2
\(\overline{\mathrm{V}}\) = 109,677 \(\left(\frac{1}{(2)^{2}}-\frac{1}{(4)^{2}}\right)\)
= 109,677 x \(\frac{3}{16}\) = 20564 cm-1
λ = \(\frac{1}{\bar{v}}\) = \(\frac{1}{20564}\)cm
= 4.86 × 10-5 cm = 4.86 × 10-5 × 107 nm.
= 486 nm.

Question 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.
E1 = – 2.18 × 10-18 J/atom
E5 = –\(\frac{2.18 \times 10^{-18}}{(5)^{2}}\) J/atom
= 8.72 × 10-20 J/atom

Energy required to remove the electron from 5th orbit
= E-E5
= 0 – (- 8.72 × 10-18 J)
= 8.72 × 10-20 J

Energy required to remove the electron from n = 1 orbit
= E-E1
= 0- (- 2.18 × 10-18J)
= 2.18 × 10-18 J
\(\frac{\mathrm{IE}_{5}}{\mathrm{IE}_{1}}\) = \(\frac{8.72 \times 10^{-20} \mathrm{~J}}{2.18 \times 10^{-18} \mathrm{~J}}\)
= 4 × 10-2.

Question 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.
Number of lines produced when electron from nth shell drops to ground state
= \(\frac{n(n-1)}{2}\) = \(\frac{6(6-1)}{2}\) = 15.

Question 16.
(i) The energy associated with the first orbit in the hydrogen atom is – 2.18 × 10-18 J atom-1. What is the energy associated with the fifth orbit?
(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.
Solution. En= –\(\frac{2.18 \times 10^{-18}}{n^{2}}\) J
∴ E5 =-\(\frac{2.18 \times 10^{-18}}{5^{2}}\)
= – 8.72 x 10-20 J.

(iii) For H-atom, rn = 0.529 × n2 Å
∴ r5 = 0.529 × 52 = 13.225 Å
= 1.3225 nm.

Question 17.
Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.
Solution.
The longest wavelength corresponds to minimum energy (ΔE) transition.
For Balmer series this transition is from n2 = 3 to n1 = 2.
\(\overline{\mathbf{V}}\) = 109677\(\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]\) cm-1
= 109677 \(\left[\frac{1}{(2)^{2}}-\frac{1}{(3)^{2}}\right]\) cm-1
= 15233 cm-1 = 1.5233 × 104 cm -1
= 1.5233 × 106 m-1.

Question 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 – 2.18 x 10-11 ergs.
Solution.
E1 = – 2.18 × 10-11 ergs
E5 = \(\frac{\mathrm{E}_{1}}{(5)^{2}}\) = \(\frac{-2.18 \times 10^{-11}}{(5)^{2}}\) = –\( \frac{2.18 \times 10^{-11}}{25}\)

ΔE = E5 – E1
= \(\left[-\frac{2.18 \times 10^{-11}}{25}-\left(-2.18 \times 10^{-11}\right)\right]\) ergs
= \(\left(2.18 \times 10^{-11}-\frac{2.18 \times 10^{-11}}{25}\right)\) ergs
= 2.09 × 10-11 ergs
Thus, the energy required to shift the electron from the first orbit to the fifth orbit in hydrogen atom is 2.09 × 10-11 ergs. The same amount of energy would he released in the form of a photon of radiation, the wavelength of which is given by the relation.

ΔE = \(\frac{h c}{\lambda}\)
or λ = \(\frac{h c}{\Delta \mathrm{E}}\)
= \(\frac{6.63 \times 10^{-34} \mathrm{Js} \times 3 \times 10^{8} \mathrm{~ms}^{-1}}{2.09 \times 10^{-11} \times 10^{-7} \mathrm{~J}}\)
= 9.52 × 10-8 m
= 9.52 × 10-8 × 1010 Å
= 952 Å.

Question 19.
The electron energy in hydrogen atom is given by En = (- 2.18 × 10-18)/n2 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.
E2= – \(\frac{2.18 \times 10^{-18}}{(2)^{2}}\) J
= -5.45 × 10-19
E= 0

The energy required to remove the electron completely from n = 2 orbit,
ΔE = E – E2.
= 0 -(-5.45 × 10-19) J
= 5.45 × 10-19 J.

ΔE = \(\frac{h c}{\lambda}\)
or λ = \(\frac{h c}{\Delta \mathrm{E}}\)
= \(\frac{6.63 \times 10^{-34} \mathrm{Js} \times 3 \times 10^{8} \mathrm{~ms}^{-1}}{5.45 \times 10^{-19} \mathrm{~J}} \)
= 3.65 × 10-7 m
= 3.65 × 10-7 cm.

Question 20.
Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 ms-1.
Solution.
λ = –\(\frac{h}{m}\)
= \(\frac{6.63 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-1}}{9.1 \times 10^{-31} \mathrm{~kg} \times 2.05 \times 10^{7} \mathrm{~ms}^{-1}}\)
= 3.55 × 10-11 m.

MP Board Class 11th Chemistry Solutions Chapter 2 Structure of Atom

Question 21.
The mass of an electron is 9.1 × 10-31 kg. If its K.E. is 3.0 × 10-25 J, calculate its wavelength.
Solution.
Mass of the electron m = 9.1 × 10-31 kg
K.E. = \(\frac{1}{2}\) mv2 = 3 × 10-25 J
\(\frac{1}{2}\) × 9.1 × 10-31× v2 = 3 × 10-25J
υ = \(\sqrt{\frac{2 \times 3 \times 10^{-25}}{9.1 \times 10^{-31}}}\) = 8.1 × 102 ms-1

h = 6.63 × 10-34 Js = 6.63 × 10-34 Kg ms-1
or λ = \(\frac{h}{m v}\)
= \(\frac{6.63 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-1}}{9.1 \times 10^{-31} \mathrm{~kg} \times 8.1 \times 10^{2} \mathrm{~ms}^{-1}}\)
= 8.99 × 10-7 m.

Question 22.
Which of the following are isoelectronic species i.e., those having the same number of electrons ?
Na+, K+, Mg2+, Ca2+, S2-, Ar.
Solution.
Na+ has (11 – 1) =10 electrons
K+has (19-1) = 18 electrons
Mg2+ has (12-2) = 10 electrons
Ca+ has (20 – 2) = 18 electrons
Ar has 18 electrons
Thus, Na+, Mg2+ and Ar are isoelectronic with 10 electrons K+ and Ca2+ are also isoelectronic with 18 electrons.

Question 23.
(i) Write the electronic configurations of the following ions:
(a) H
(b) Na+
(c) O2
(d) F
(ii) What are the atomic numbers of elements whose outermost electrons are represented by
(a) 3S1
(b) 2p3 and
(c) 3p5?
(iii) Which atoms are indicated by the following configurations?
(a) [He] 2s1
(b) [Ne] 3s2 3p3
(c) [Ar] 4s2 3d1.
Solution.
(i) (a) H : 1s2
(b) Na+ : 1s2 2s2 2p6
(c) O2- : 1s2 2s2 2p6.
(d) F : 1s2 2s2 2p6

(ii)
(a) 1s2 2s2 2p6 3s1
This element contains 11 electrons and hence its atomic number is 11
(b) 1s2 2s2 2p3 Atomic number = 7
(c) 1s2, 2s2 2p6 3s2 3p5 Atomic number = 17
(iii) (a) Lithium (Z = 3) (b) phosphorus (Z = 15) (c) Scandium (Z = 21).

Question 24.
What is the lowest value of n that allows g orbitals to exist?
Solution.
For g-subshell, l=4. The highest value of l for a particular energy shell is equal to n – 1. Therefore, for l to be equal to 4, the minimum value of n is equal to 5.

Question 25.
An electron is in one of the 3d orbitals. Give the possible values of n, 1 and m, for this electron.
Solution.
For 3d orbital, n = 3, l = 2 and mi can have value + 2,+ 1, 0,-1 or- 2.

Question 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) Number of protons = Number of electrons = 29.
(ii) Electronic configuration of the element is
1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s1.

Question 27.
Give the number of electrons in the species H2+, H2 and O2+
Solution.
H2+ : (2 -1) = 1 electron
H2 : (1 + 1) = 2 electrons
O2+b : (16 – 1) = 15 electrons.

Question 28.
(i) An atomic orbital has n = 3. What are the possible values of 1 and m,?
(ii) List the quantum numbers (mt and 1) of electrons for 3d orbital.
(iii) Which of the following orbitals are possible?
Ip, 2s, 2p and 3f.
Solution.
(i) When n = 3, l = 0, 1, 2.
When l – 0, m[ = 0. When / = 1, ml = – 1, 0, + 1. .
When Z = 2, m = – 2, – 1, 0, + 1, + 2
(ii) For 3d-orbital, n = 3,1 = 2. For l = 2, ml = — 2, — 1, 0, + 1, + 2.

(iii) 1p is not possible because when n = 1,1 cannot be 1. For p-subshell 1=1. 2s is possible because when n = 2,1 can be 0. For s-subshell 1 = 0.
2p is possible because when n = 2,1 can be 1.
3f is not possible because when n = 3,1 cannot be 3. For f-subshell 1 = 3.

MP Board Class 11th Chemistry Solutions Chapter 2 Structure of Atom

Question 29.
Using s, p, d notations, describe the orbital with the following quantum numbers:
(a) n = 1,1 = 0
(b) n = 3,1 = 1
(c) n = 4,1 = 2
(d) n = 4,1 = 3.
Solution.
(a) 1s
(b) 3p
(c) 4d
(d) 4f.

Question 30.
Explain, giving reasons, which of the following sets of quantum numbers are not possible.
MP Board Class 11th Chemistry Solutions Chapter 2 Structure of Atom 2
Solution.
(a) is not possible because n cannot be zero.
(c) is not possible because the value of l cannot be equal to value of n.
(e) is not possible because the value of l cannot be equal to the value of n.

Question 31.
How many electrons in an atom may have the following quantum numbers?
(a) n = 4, ms = – 1/2
(b) n = 3,1 = 0
Solution.
(a) Total number of electrons in n = 4 is 2n2, i.e., 2 × 42 = 32. Half of them,
i.e., 16 electrons have ms = –\(\frac{1}{2}\)
(b) n = 3,1 = 0 means 3s orbital which can have 2 electrons.

Question 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.
According to Bohr’s model,
mvr = n \(\frac{h}{2 \pi}\)
2πr = n \(\frac{h}{m v}\)
But,\(\frac{h}{m v}\) = λ
∴ 2πr = nλ.

Question 33.
What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum?
solution.
For He+, \(\frac{1}{\lambda}\) = \(\mathrm{R}_{\mathrm{H}} \times(2)^{2}\left[\frac{1}{(2)^{2}}-\frac{1}{(4)^{2}}\right]\)
This may be simplified to the form
\(\frac{1}{\lambda}\) = RH\(\left[\frac{(2)^{2}}{(2)^{2}}-\frac{(2)^{2}}{(4)^{2}}\right]\)
RH = \(\left[\frac{1}{(1)^{2}}-\frac{1}{(2)^{2}}\right]\)
Thus, for hydrogen atom n = 2 to n = 1 transition would give radiation of the same wavelength as n = 4 to n = 2 transition for He+.

Question 34.
Calculate the energy required for the process
He+ (g) → He2+(g) + e
The ionization energy for the H atom in the ground state is 2.18 × 10-18 J atomM-1
Solution.
E1 for H atom = – I.E. = – 2.18 × 10-18 J
E1 for He+ = E1(H) × Z2 = – 2.18× 10-18 ×22 J
= – 8.72 × 1018 J
The given process represents ionization of He+.
The energy required for ionization of He+
= – E1(He+)
= 8.72 × 10-18J.

Question 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.
Diameter of carbon atom, d = 0.15 nm
= 0.15 × 10-9 m
Length of the scale, / = 20 cm = 20 × 10-2 m

Number of carbon atoms which can be placed side by side across length of the scale = \(\frac{l}{d}\)
= \(\frac{20 \times 10^{-2} m}{0.15 \times 10^{-9} m}\)
= 1.33 × 109.

Question 36.
2 × 108 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.
Number of carbon atoms, n = 2 × 108
Length of the arrangement, l = 2.4 cm = 2.4 × 107 nm
Diameter of carbon atom = \(\frac{l}{n} \)
Radius of carbon atom = \(\frac{l}{2 n} \)
= \(\frac{2.4 \times 10^{7} \mathrm{~nm}}{2 \times 2 \times 10^{8}}\)
= 0.06 nm.

MP Board Class 11th Chemistry Solutions Chapter 2 Structure of Atom

Question 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 atoms are arranged side by side lengthwise.
Solution.
(a) Radius = \(\frac{2.6 AA}{2} \)  = 1.3 Å = 1.3 × 10-10 m

= 1.3 × 10-10 × 1012 pm = 130 pm

(b) Given length = 1.6 cm = 1.6 × 10-2 m
Diameter of one atom = 2.6 Å = 2.6 ×10-10 m
∴ Number of atoms present along the length
= \(\frac{1.6 \times 10^{-2} \mathrm{~m}}{2.6 \times 10^{-10} \mathrm{~m}}\)
= 6.2 × 107.

Question 38.
A certain particle carries 2.5 × 10-16 C of static electric charge. Calculate the number of electrons present in it.
Solution.
Charge carried by one electron = 1.6022 × 10-16 C
∴Electrons present in particle carrying 2.5 × 10-16
C charge = \(\frac{2.5 \times 10^{-16} \mathrm{C}}{1.6022 \times 10^{-19} \mathrm{C}}\)
= 1560.

Question 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 – 1.282 x 10“18 C, calculate the number of electrons present on it.
Solution.
Charge on one electron = – 1.6022 × 10-19 C
The number of electrons present = \(\frac{-1282 \times 10^{-18} \mathrm{C}}{-1.6022 \times 10^{-19} \mathrm{C}}\)
= 8.

Question 40.
In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc., have been used to be bombarded by the K-particles. If the thin foil of light atoms like aluminium etc., is used, what difference would be observed from the above results?
Solution.
The nucleus of lighter atoms carries smaller positive charge and has smaller volume. As a result, the number of a-particles deflected through small angles and through large angles would be less. Moreover, angle of deflection, in case of a-particles undergoing small angle deflections would be smaller.

Question 41.
Symbols 7935Br and 79Br can be written whereas symbols 7935Br and 35Br are not acceptable. Answer briefly.
Solution.
Atomic number of an element is fixed. However, mass number is not fixed as it depends upon the isotope. Hence, it is essential to indicate mass number.

Question 42.
An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.
Solution.
Number of protons + Number of neutrons = mass number = 81
Let number of protons in the atom be = x
∴ Number of neutrons in the atom = \(\frac{x \times 131.7}{100}\) = 1.317x
x + 1.317x = 81
2.317x = 81
x = 34.96 = 35 (nearest whole number)
∴ Number of protons = 35
Atomic number of the element = 35
The element with atomic number 35 is Br
∴ Atomic symbol is 7935 Br.

Question 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.
Since the ion carries one unit negative charge, it means that in the ion the number of electrons is one more than the number of protons.
Total number of electrons and neutrons in the ion = 37 + 1 = 38
Let the number of electrons in the ion be = x

Number of electrons in the ion = \(\frac{x \times 1111}{100}\) = 1.111 x
∴ Total number of electrons and neutrons in the ion = x + 1.111 x = 2.111 x
2.111x = 38
x = \(\frac{38}{2.111}\) = 18
∴ Number of electrons in the ion = 18
Number of protons in the ion = 18 – 1 = 17
Thus, atomic number of the element is 17 which corresponds to chlorine.
∴ Symbol of the ion is 1735Cl.

Question 44.
An ion with mass number 56 contains 3 units of positive charge and 30.40% more neutrons than electrons. Assign the symbol to this ion.
Solution.
Since the ion contains three units positive charge it indicates that the number of electrons in the ion is three less than the number of protons.
Number of protons + Number of neutrons = Mass number = 56 Number of electrons + Number of neutrons = 56 – 3 = 53

Let the number of electrons in the ion be = x
∴ The number of neutrons in the ion = \(\frac{x \times 130.4}{100}\) = 1.304x
But the sum of electrons and neutrons in the ion = 53 .
x + 1.304x = 53
2.304x = 53
x = \(\frac{53}{2.304}\)
2.304 = 23

Thus, the number of electrons in the ion = 23
The number of protons in the ion = 23 + 3 = 26
The element with atomic number 26 is iron (Fe)
∴ Symbol of the ion is 2656 Fe3+.

Question 45.
Arrange the following types 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.
Solution.
Cosmic rays < X-rays < amber light < microwave < FM.

Question 46.
Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number
of photons emitted is 5.6 x 1024 per second, calculate the power of this laser.
Solution. E = N hv
= N h\(\frac{c}{\lambda}\)
= \(\frac{\left(5.6 \times 10^{24}\right)\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3.0 \times 10^{3} \mathrm{~ms}^{-1}\right)}{\left(337.1 \times 10^{-9} \mathrm{~m}\right)}\)
= 3.3 × 106 J s-1.

Question 47.
Neon gas is generally used in the signboards. If it emits strongly at 616 nm, calculate
(a) the frequency of emission,
(b) distance travelled 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) λ = 616 nm = 616 × 10-9 m, c = 3 × 108 ms-1
v = \(\frac{c}{\lambda}\) =\(\frac{3 \times 10^{8} \mathrm{~ms}^{-1}}{616 \times 10^{-9} \mathrm{~m}}\)
= 4.87 × 1014 s-1

(b) Distance travelled by the radiation in 30 s = c × t
= 3 ×108 ms-1 ×30 s = 9 × 109 m
(c) Energy of quantum, E = hv = 6.626 × 10-34 Js × 4.87 × 1014 s-1 = 3.23 × 10-19 J

(d) Number of quanta = \(\frac{\text { Total energy }}{\text { Energy per quantum }}\)
= \(\frac{2 \mathrm{~J}}{3.23 \times 10^{-19} \mathrm{~J}}\)
= 6.19 × 1018.

MP Board Class 11th Chemistry Solutions Chapter 2 Structure of Atom

Question 48.
In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 × 10-18 J from the radiations of 600 nm, calculate the number of photons received by the detector.
Solution.
Wavelength, A = 600 nm = 600 × 10-9 m
Energy per photon = \(\frac{h c}{\lambda}\)
= \(\frac{6.626 \times 10^{-34} \mathrm{Js} \times 3.00 \times 10^{8} \mathrm{~ms}^{-1}}{600 \times 10^{-9} \mathrm{~m}}\)
= 3.31 × 10-19 J

Number of photons = \(\frac{\text { Total energy received by the detector }}{\text { Energy per photon }}\)
= \(\frac{3.15 \times 10^{-18} \mathrm{~J}}{3.31 \times 10^{-19} \mathrm{~J}}\)
= 9.52 = 10 photons.

Question 49.
Lifetimes of the molecules in the excited state are often measured by using pulsed radiation source of duration nearly in the nanosecond range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 × 1015, calculate the energy of the source.
Solution.
Frequency = \(\frac{1}{2 \times 10^{-9} \mathrm{~s}}\)
= 0.5 × 109 s-1
Energy = N hv
= 2.5 × 1015 × 6.626 x 10-34 Js × 0.5 × 109 s-1
= 8.28 × 10-10 J.

Question 50.
The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and the energy difference between the two excited states.
Solution.
λ1 = 589 nm = 589 × 10-9 m.
∴ v1 = \(\frac{c}{\lambda_{1}}\) = \(\frac{3.0 \times 10^{8} \mathrm{~ms}^{-1}}{589 \times 10^{-9} \mathrm{~m}}\)
= 5.093 × 1014 s-1
λ1 = 589.6 nm = 589.6 × 10-9m
∴ v2 = \(\frac{c}{\lambda_{2}}\) = \(\frac{3.0 \times 10^{8} \mathrm{~ms}^{-1}}{589.6 \times 10^{-9} \mathrm{~m}}\)
= 5.088 × 1014 s-1

ΔE = E2 – E1 = h(v2 – v1)
= 6.626 × 10-34 Js × (5.093 – 5.088) × 1014 s-1
= 3.31 × 10-22 J.

Question 51.
The work function for caesium atom is 1.9 eV. Calculate
(a) the threshold wavelength and
(b) the threshold frequency of the radiation,
(c) If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.
Solution.
Work function, W0 = 1.9 eV = 1.9 × 1.6 × 10-19 J = 3.04 × 10-19 J he
(a) W0 = \(\frac{h c}{\lambda_{0}}\)
λ0 = \(\frac{h c}{\mathrm{~W}_{0}}\) = \(\frac{6.626 \times 10^{-34} \mathrm{Js} \times 3 \times 10^{8} \mathrm{~ms}^{-1}}{3.04 \times 10^{-19} \mathrm{~J}}\)
λ0 = 6.54 × 10-7 m
= 654 nm
Thus, threshold wavelength of caesium is 654 nm

(b) Threshold frequency,
V0 = \(\frac{c}{\lambda_{0}}\) = \(\frac{3 \times 10^{8} \mathrm{~ms}^{-1}}{6.54 \times 10^{-7} \mathrm{~m}}\) = 4.59 × 1014 s-1
(c) Energy of photon of radiation of wavelength 500 nm
= \(\frac{h c}{\lambda}\) = \(\frac{6.626 \times 10^{-34} \mathrm{Js} \times 3 \times 10^{8} \mathrm{~ms}^{-1}}{500 \times 10^{-9} \mathrm{~m}}\) = 3.98 × 10-19 J
K.E. = hv – hv0 = 3.68 × 10-19 J – 3.04 × 10-19 J = 9.4 × 10-20 J
MP Board Class 11th Chemistry Solutions Chapter 2 Structure of Atom 3

Question 52.
Following results were observed when sodium metal is irradiated with different wavelengths. Calculate
(a) threshold wavelength and
(b) Planck’s constant.

λ(nm) 500 450 400
u x 10<sup>-6</sup> (ms<sup>-1</sup>) 2.55 4.35 5.20

Solution.
Suppose threshold wavelength = λ0 nm
0 × 10-9 m
Then h(v-v0) = \(\frac{1}{2}\) mv2 or hc \(\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)\) = \(\frac{1}{2}\) mv2
Substituting the given results of the three experiments, we get
\(\frac{h c}{10^{-9}}\left(\frac{1}{500}-\frac{1}{\lambda_{0}}\right)\) = \(\frac{1}{2} m\left(2.55 \times 10^{6}\right)^{2}\) ………………………. (i)
\(\frac{h c}{10^{-9}}\left(\frac{1}{450}-\frac{1}{\lambda_{0}}\right)\) = \(\frac{1}{2} m\left(4.35 \times 10^{6}\right)^{2}\) …………………………. (ii)
\(\frac{h c}{10^{-9}}\left(\frac{1}{400}-\frac{1}{\lambda_{0}}\right)\) = \(\frac{1}{2} m\left(5.20 \times 10^{6}\right)^{2}\) ………………………………….. (iii)
Dividing eqn. (ii) by eqn. (i), we get
MP Board Class 11th Chemistry Solutions Chapter 2 Structure of Atom 4
or 1.619 λ0 = 859.5
λ0 = 531 nm
Substituting this value in eqn. (iii), we get
\(\frac{h \times\left(3 \times 10^{8}\right)}{10^{-9}}\left(\frac{1}{400}-\frac{1}{531}\right)\) = \(\frac{1}{2}\left(9.11 \times 10^{-31}\right)\left(5.20 \times 10^{6}\right)^{2}\)
or h = 6.66 × 10-34 Js.

Question 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.
Since the photoelectron can be stopped by applying a voltage of 0.35 eV
therefore, K.E. of the photoelectron is equal to 0.35 eV.
K.E. = 0.35 eV = 0.35 × 1.6 × 1019 J = 0.56 × 1019 J
hv = \(\frac{h c}{\lambda}\) = \(\frac{6.626 \times 10^{-34} \mathrm{Js} \times 3 \times 10^{8} \mathrm{~ms}^{-1}}{256.7 \times 10^{-9} \mathrm{~m}}\)
= 7.74 × 10-19 J
hv = W0 + K.E.
W0 = hv – K.E.
= 7.74 × 10-19 J – 0.56 × 1019 J
= 7.18 × 10-19 J
Thus, the work function for silver is 7.18 × 10-19 J.

Question 54.
If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 × 107ms-1, calculate the energy with which it is bound to the nucleus.
Solution.
λ = 150 pm = 150 × 10-12 m
Energy of photon =\(\frac{h c}{\lambda}\) = \(\frac{6.626 \times 10^{-34} \mathrm{Js} \times 3 \times 10^{8} \mathrm{~ms}^{1}}{150 \times 10^{-12} \mathrm{~m}}\)
= 1.33 × 1015 J
Kinetic energy of the ejected electron
= \(\frac{1}{2}\) mv2 = \(\frac{1}{2}\) × 9.11 × 10-31 × (1.5 × 107)2 = 1.025 × 10-16 J
Binding energy = Energy of incident photon – Kinetic energy
= 1.33 × 10-15 J – 1.025 × 10-16 J
= 1.23 × 10-15 J.

Question 55.
Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 × 1015 (Hz) [1/32 – 1/n2].
Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.
Solution:
V = \(\frac{c}{\lambda}\) = \(\frac{3.0 \times 10^{8} \mathrm{~ms}^{-1}}{1285 \times 10^{-9} \mathrm{~m}} \) = 3.29 × 1015 \(\left(\frac{1}{3^{2}}-\frac{1}{n^{2}}\right)\)
or \(\frac{1}{n^{2}}\) = \(\frac{1}{9}-\frac{3.0 \times 10^{8}}{1285 \times 10^{-9}} \times \frac{1}{3.29 \times 10^{15}}\)
= 0.111 -0.071 = 0.04 = \(\frac{1}{25}\)
or n2 = 25 or n = 5
The radiation corresponding to 1285 nm lies in the infrared region.

Question 56.
Calculate the wavelength for the emission transition if it starts from the orbit having radius of 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.
Solution.
Radius of nth orbit of H-like particles
= \(\frac{0.529 n^{2}}{Z}\) Å = \(\frac{52.9 n^{2}}{\mathrm{Z}}\) pm
r1 = 1.3225 nm = 1322.5 pm = 52.9 n12 pm
n12 = \(\frac{1322.5}{52.9}\) = 25
n1 = 5
r2 = 211.6 pm = 52.9 n22 pm
n22 = \(\frac{2116}{52.9}\) = 4
n2 = 2
Thus, n2 = 2,n1 = 5. Therefore the transition is from 5th orbit to 2nd orbit. It belongs to
Balmer series.
MP Board Class 11th Chemistry Solutions Chapter 2 Structure of Atom 5
= 434 nm
It lies in the visible region.

Question 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 1.6 × 106 ms-1, calculate de-Broglie wavelength associated with this electron.
Solution.
Velocity, υ = 1.6 × 106ms-1
Mass of electron, m = 9.11 × 10-31 kg
λ = \(\frac{h}{m v}\)
= \(\frac{6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-1}}{9.11 \times 10^{-31} \mathrm{~kg} \times 1.6 \times 10^{6} \mathrm{~ms}^{-1}}\)
= 4.55 × 10-10 m
= 455 pm.

Question 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.
λ = 800 pm = 8.00 × 10-10 m
Mass of neutron, m = 1.675 × 10-27 kg
λ = \(\frac{h}{m v}\)
υ = \(\frac{h}{m \lambda}\) = \(\frac{6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-1}}{1.675 \times 10^{-27} \mathrm{~kg} \times 8.00 \times 10^{-10} \mathrm{~m}}\)
= 4.95 × 102 ms-1.

Question 59.
If the velocity of the electron in Bohr’s first orbit is 2.19 × 106 ms-1, calculate the de-Broglie wavelength associated with it.
Solution.
υ = 2.19 × 106ms-1; m = 9.11 × 10-31 kg
λ = \(\frac{h}{n v}\) = \(\frac{6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-1}}{9.11 \times 10^{-31} \mathrm{~kg} \times 2.19 \times 10^{6} \mathrm{~ms}^{-1}}\)
= 3.32 × 10-10 m
= 332 pm.

Question 60.
The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × 105 ms-1. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.
Solution.
Velocity of the hockey ball = 4.37 × 105 ms-1
Mass of the ball = 0.1 kg
λ = \(\frac{h}{m v}\) = \(\frac{6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-1}}{0.1 \mathrm{~kg} \times 4.37 \times 10^{5} \mathrm{~ms}^{-1}}\)
= 1.52 × 10-38 m
This wavelength is too small to be detected by any means.

MP Board Class 11th Chemistry Solutions Chapter 2 Structure of Atom

Question 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 h/4π × 0.05 nm, is there any problem in defining this value.
Solution.
Δx = 0.002 nm = 2 × 10-3 nm
= 2 × 10-3 × 10-9 m = 2 × 10-12 m h
Δx × Δp = \(\frac{h}{4 \pi}\)
∴ Δp = \(\frac{h}{4 \pi \Delta x}\) = \(\frac{6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-1}}{4 \times 3.14 \times 2 \times 10^{-12} \mathrm{~m}}\)
= 2.638 × 10-23 kg ms-1
MP Board Class 11th Chemistry Solutions Chapter 2 Structure of Atom 6
= 1.055 × 10-24 kg ms-1
It cannot be defined as the actual magnitude of the momentum is smaller than the uncertainty in momentum.

Question 62.
The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. List if any of these combination (s) has/have the same energy:
MP Board Class 11th Chemistry Solutions Chapter 2 Structure of Atom 7
solution:
Apply Bohr Bury rules Their energies will be in the order : (v) <(ii) = (iv) <(vi) = (iii) < (i).

Question 63.
The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbitals, 6 electrons in 3p orbitals and 5 electrons in 4p orbitals. Which of these electrons experiences the lowest effective nuclear charge?
Solution.
4p electrons having largest value of n are farthest from the nucleus and hence experience the lowest effective nuclear charge. For orbitals of the same type of subshell greater the value of n smaller is the effective nuclear charge.

Question 64.
Among the following pairs of orbitals which orbital will experience the large effective nuclear charge?
(a) 2s and 3s, (6) 4d and 4f, (c) 3d and 3p.
Solution.
(a) 2s (b) 4d (c) 3p.

Question 65.
The unpaired electrons in A1 and Si are present in 3p orbital. Which electron will experience more effective charge from the nucleus?
Solution.
Silicon has greater nuclears charge (+ 14) than aluminium (+ 13). Hence, the unpaired 3 p electron in case of silicon will experience more effective nuclear charge.

Question 66.
Indicate the number of unpaired electrons in:
(i) P (ii) Si (iii) Cr (iv) Fe (v) Kr.
Solution.
(i) 15P: 1s2, 2s2, 2p6, 3s2, 3px1, 3py1, 3pz1
There are 3 unpaired electrons in P atom.
(ii) 14Si: 1s2, 2s2, 2p6, 3s2, 3px1, 3py1
There are 2 unpaired electrons in Si atom.
(iii) 24Cr: 1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d5
There are 6 unpaired electrons in Cr atom.
(iv) 26Fe: [Ar] 4s2 3d6
There are 4 unpaired electrons in Fe atom.
(v) 36Kr: [Ar] 4s2 3c10 4p6
There are no unpaired electrons i l Kr atom.

Question 67.
(a) How many sub-shells are associated with n = 4?
(b) How many electrons will be present in the sub-shells having ms value of – 1/2 for n = 4?
Solution.
(a) For n = 4, l can have values 0, 1, 2, 3. Thus, there are four sub-shells in n = 4 energy level. These four sub-shells are 4s, 4p, 4d and 4f.
(b) For n = 4, the number of orbitals is = (4)2 = 16. Each orbital can have one electron with ms = –\(\frac{1}{2} \) . Thus, there are 16 electrons in sub-shells having n = 4 and ms = –\(\frac{1}{2}\)

MP Board Class 11 Chemistry Structure of Atom Important Questions and Answers

Question 1.
By whom was the atomic model first proposed?
Answer:
John Dalton.

Question 2.
In which part of an atom are neutrons present?
Answer:
Neutrons are present in the nucleus of the atom.

Question 3.
What is the important feature of Thomson atomic model?
Answer:
The important feature of Thomson atomic model is that the mass of an atom is considered to be evenly distributed over the atom.

Question 4.
What are a-particles?
Answer:
a-particles are high energy, positively charged (+ 2 charge) helium ions.

Question 5.
What are electromagnetic radiations?
Answer:
Ordinary light, X-rays, y-rays, infra-red rays etc., are called electromagnetic radiations.

MP Board Class 11th Chemistry Solutions Chapter 2 Structure of Atom

Question 6.
How are wavelength frequency and velocity of a wave related?
Answer:
Wavelength × frequency – velocity.
λ × v = v

Question 7.
Define wave number.
Answer:
Wavenumber is the number of wavelengths per unit of length. It is represented by \(\bar{v}\) .
\(\bar{v}\) = \(\frac{1}{\lambda}\).

Question 8.
Name any two phenomenon that can be explained by wave nature of electromagnetic radiation.
Answer:
Diffraction and interference.

Question 9.
Name any two observations that cannot be explained by wave nature of electromagnetic radiation.
Answer:
(i) Radiations from a hot body
(ii) Photoelectric effect.

Question 10.
Define a quantum.
Answer:
Quantum is the smallest quantity of energy that can be emitted or absorbed in the form of electromagnetic radiations.

Question 11.
What is photoelectric effect?
Answer:
The process of ejection of electrons from the surface of c< lain metals (like Cs, K, Rb etc.). When they are exposed to light is known as photoelectric effect.

Question 12.
What is meant by diffraction?
Answer:
The bending of waves around an obstacle is called diffraction.

Question 13.
What are isotones?
Answer:
Isotones are the atoms of different elements (atomic no. different) having the same number of neutrons e.g. 613C and 714N.

Question 14.
What is meant by quantization of energy?
Answer:
Quantization of energy means that energy cannot be continuously emitted or absorbed to have any arbitrary value but can change only discontinuously to have some specific value.

Question 15.
Write down the Balmer formula.
Answer:
\(\bar{v}\) =\(\frac{1}{\lambda}\) = 1.097 × 107\(\left[\frac{1}{2^{2}}-\frac{1}{n^{2}}\right]\) m-1
Where n is some integral whole number greater than 2.

Question 16.
What is threshold frequency?
Answer:
Threshold frequency is the minimum frequency of electric magnetic radiation that is capable of causing an emission of electron from the surface of a metal.

Question 17.
What is Zeeman effect?
Answer:
It refers to the splitting up of a spectral line into close paced lines under the influence of a magnetic field.

Question 18.
What is Stark effect?
Answer:
It refers to the splitting up of a spectral line into closely lines under the influence of an electric field.

MP Board Class 11th Chemistry Solutions Chapter 2 Structure of Atom

Question 19.
How is work functions related to threshold frequency?
Answer:
Work function = Planck’s constant × Threshold frequency
i. e., W0 = h × v0.

Question 20.
What is an emission spectrum?
Answer:
The spectrum of radiation emitted by a substance that has absorbed energy is called an emission spectrum.

Question 21.
What is the name given to the study of emission and absorption spectrum?
Answer:
Spectroscopy.

Question 22.
What is the value of Rydberg’s constant?
Answer:
It is 109, 677 cm-1.

Question 23.
Write down de Broglie’s equation.
Answer:
λ = \(\frac{h}{mv} \) or λ = \(\frac{h}{p}\)
where p = momentum of particle.

Question 24.
What type of wave is a de Broglie’s wave?
Answer:
It is a matter-wave.

Question 25.
What are isoelectronic ions?
Answer:
The isoelectronic ions are those which have the same number of electrons, e.g. O2-, Na+, Al3+ and Ne are isoelectronic because all these have 10 electrons.

Question 26.
State (n + 1) rule.
Answer:
The sum of quantum numbers (n + l) determine the energy of the orbital. The lower the (n + l) sum the lower is energy. In case two (n + l) sums are same the orbital having lower n value has lower energy.

Question 27.
What is probability density?
Answer:
It is the probability of finding an electron at a point within an atom. It is given by φ2 and is always positive.

Question 28.
How can we distinguish atomic orbitals?
Answer:
Atomic orbitals are precisely distinguished by quantum number.

Question 29.
What is the lowest value of n for a shell to have g-orbitals?
Answer:
n =5.

Question 30.
What are quantum numbers?
Answer:
These are tin; four numbers is that we need to completely identify an electron in a multi-electron atom.

Question 31.
Which of the following orbitals are not possible?
1s, 1p, 2s, 2p, 3d, 3f
Answer:
1p and 3f are not possible.

MP Board Class 11th Chemistry Solutions Chapter 2 Structure of Atom

Question 32.
What are degenerate orbitals?
Answer:
Orbitals having equal energy are called degenerate orbitals.

Question 33.
How many electrons are present in H2+ and H2?
Answer:
One and two electrons respectively.

Question 34.
Name the five series of lines observed in a spectrum of hydrogen.
Answer:
These are called Lyman, Balmer, Paschen, Bracket and Pfund series.

Question 35.
What were the failures of Bohr’s model?
Answer:
The failures of Bohr’s theory are:
1. Bohr’s model was quite successful in accounting for the main features of the hydrogen spectrum and also for the spectrum of He+ ion (having only one electron). However, it failed to predict the energy states of more complicated atoms.
2. Further if there are only just a few stationary states in an atom, the spectra of elements should consists of very few lines. But a detailed examination of spectra of elements containing many electrons, reveals the presence of a series of lines.

Question 36.
Write the electronic configurations of the following:
(i) Cu2+ = (Z for Cu = 29)
(ii) Cr3+ = (Z for Cr = 24)
(iii) Na+ = (Z for Na =11)
Answer:
(i) 1s2 2s2 2p6 3s2 3p6 3d9
(ii) 1s2 2s2 2p6 3s2 3p6 3d3.
(iii) 1s2 2s2 2p6.

MP Board Class 11th Chemistry Solutions Chapter 2 Structure of Atom

Question 37.
What are the important features of quantum mechanical model of an atom?
Answer:
The following are the important features of quantum mechanical model of an atom.

  • The energy of electron in an atom is quantised.
  • The existence of quantised electronic energy levels is a direct result of wave like properties of electrons.
  • Both exact position and exact velocity of an electron in an atom cannot be determined simultaneously.
  • An atomic orbital is 3-dimentional space around nucleus where probability of finding electron is maximum.

Question 38.
What is difference between electromagnetic wave and matter wave?
Answer:

Electromagnetic wave Matter-wave
1. They are associated with electrical and magnetic fields which are at right angles to each. 1. They are not associated with electrical and magnetic field.
2. They can pass through vacuum. 2. They cannot pass through vacuum.
3. They travel with same velocity i.e. equal to the velocity of light. 3. Travel with different velocities, less than light.