CBSE · NCERT · Class 11 Chemistry · Chapter 2

NCERT Solutions: Class 11 Chemistry Chapter 2 - Structure of Atom

61 textbook Q&A61 verifiedFree Content

Chapter-wise NCERT intext questions and exercise answers for Structure of Atom, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
Sections in this chapter
Exercises 61
Your Progress - Chapter 20% complete
1Exercises61 questions
Q.2.1(i) Calculate the number of electrons which will together weigh one gram. (ii) Calculate the mass and charge of one mole of electrons.v
Solution

Mass of one electron = 9.109 x 10^-31 kg. Number weighing 1 g = 10^-3/(9.109 x 10^-31) = 1.10 x 10^27. One mole of electrons has mass = (6.022 x 10^23)(9.109 x 10^-31 kg) = 5.49 x 10^-7 kg = 5.49 x 10^-4 g. Charge = (6.022 x 10^23)(-1.602 x 10^-19 C) = -9.65 x 10^4 C.

Answer:

(i) 1.10 x 10^27 electrons. (ii) Mass = 5.49 x 10^-4 g; charge = -9.65 x 10^4 C.

✓ Verified and validated
Q.2.2(i) Calculate the total number of electrons present in one mole of methane. (ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of 14C. (Assume that mass of a neutron = 1.675 x 10^-27 kg). (iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3 at STP. Will the answer change if the temperature and pressure are changed?v
Solution

CH4 has 10 electrons per molecule, so 1 mol CH4 has 10NA = 6.022 x 10^24 electrons. In 7 mg of 14C, moles of carbon atoms = 0.007/14 = 5.0 x 10^-4 mol. Each 14C atom has 8 neutrons, so moles of neutrons = 0.004 mol and number = 0.004NA = 2.41 x 10^21. Their mass is (2.41 x 10^21)(1.675 x 10^-27) = 4.03 x 10^-6 kg. In 34 mg NH3, moles NH3 = 0.034/17 = 0.002 mol. Each molecule has 10 protons, so moles of protons = 0.020 mol and number = 1.20 x 10^22. Their mass is about (1.20 x 10^22)(1.673 x 10^-27) = 2.01 x 10^-5 kg. Changing temperature or pressure does not change the number of molecules in a given amount.

Answer:

(i) 6.022 x 10^24 electrons. (ii) 2.41 x 10^21 neutrons; 4.03 x 10^-6 kg. (iii) 1.20 x 10^22 protons; 2.01 x 10^-5 kg. No, it depends on amount of NH3, not on temperature or pressure.

✓ Verified and validated
Q.2.4Write 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 = 9v
Solution

The complete nuclear symbol is written with mass number A and atomic number Z. Z = 17 is chlorine, Z = 92 is uranium and Z = 4 is beryllium. Therefore the symbols are ^35_17Cl, ^233_92U and ^9_4Be.

Answer:

(i) 35Cl. (ii) 233U. (iii) 9Be.

✓ Verified and validated
Q.2.5Yellow light emitted from a sodium lamp has a wavelength (lambda) of 580 nm. Calculate the frequency (nu) and wave number of the yellow light.v
Solution

lambda = 580 nm = 5.80 x 10^-7 m. Frequency nu = c/lambda = (3.00 x 10^8)/(5.80 x 10^-7) = 5.17 x 10^14 s^-1. Wave number = 1/lambda = 1.72 x 10^6 m^-1 = 1.72 x 10^4 cm^-1.

Answer:

Frequency = 5.17 x 10^14 s^-1; wave number = 1.72 x 10^6 m^-1, or 1.72 x 10^4 cm^-1.

✓ Verified and validated
Q.2.6Find energy of each of the photons which (i) correspond to light of frequency 3 x 10^15 Hz. (ii) have wavelength of 0.50 A.v
Solution

Use E = hnu = hc/lambda. (i) E = (6.626 x 10^-34)(3.0 x 10^15) = 1.99 x 10^-18 J. (ii) 0.50 A = 0.50 x 10^-10 m, so E = (6.626 x 10^-34)(3.0 x 10^8)/(0.50 x 10^-10) = 3.98 x 10^-15 J.

Answer:

(i) 1.99 x 10^-18 J. (ii) 3.98 x 10^-15 J.

✓ Verified and validated
Q.2.7Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0 x 10^-10 s.v
Solution

Frequency nu = 1/T = 1/(2.0 x 10^-10) = 5.0 x 10^9 s^-1. Wavelength lambda = c/nu = (3.0 x 10^8)/(5.0 x 10^9) = 6.0 x 10^-2 m. Wave number = 1/lambda = 16.7 m^-1.

Answer:

Frequency = 5.0 x 10^9 s^-1; wavelength = 6.0 x 10^-2 m; wave number = 16.7 m^-1.

✓ Verified and validated
Q.2.8What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?v
Solution

lambda = 4000 pm = 4.000 x 10^-9 m. Energy of one photon = hc/lambda = (6.626 x 10^-34)(3.0 x 10^8)/(4.000 x 10^-9) = 4.97 x 10^-17 J. Number of photons = 1/(4.97 x 10^-17) = 2.01 x 10^16.

Answer:

2.01 x 10^16 photons.

✓ Verified and validated
Q.2.9A 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.v
Solution

Photon energy in eV = 1240/lambda(nm). Here lambda = 400 nm, so E = 1240/400 = 3.10 eV. Kinetic energy = 3.10 - 2.13 = 0.97 eV = 1.55 x 10^-19 J. From KE = (1/2)mv^2, v = sqrt(2KE/m_e) = sqrt[2(1.55 x 10^-19)/(9.109 x 10^-31)] = 5.84 x 10^5 m s^-1.

Answer:

(i) 3.10 eV. (ii) 0.97 eV. (iii) 5.84 x 10^5 m s^-1.

✓ Verified and validated
Q.2.10Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol^-1.v
Solution

Energy per photon = hc/lambda = (6.626 x 10^-34)(3.0 x 10^8)/(242 x 10^-9) = 8.21 x 10^-19 J. Per mole, energy = 8.21 x 10^-19 x 6.022 x 10^23 = 4.95 x 10^5 J mol^-1 = 495 kJ mol^-1.

Answer:

4.95 x 10^2 kJ mol^-1.

✓ Verified and validated
Q.2.11A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 um. Calculate the rate of emission of quanta per second.v
Solution

lambda = 0.57 um = 5.7 x 10^-7 m. Energy per photon = hc/lambda = 3.49 x 10^-19 J. A 25 W source emits 25 J s^-1, so number of photons per second = 25/(3.49 x 10^-19) = 7.17 x 10^19.

Answer:

7.17 x 10^19 photons s^-1.

✓ Verified and validated
Q.2.12Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 A. Calculate threshold frequency and work function (W0) of the metal.v
Solution

At threshold, kinetic energy is zero, so nu0 = c/lambda. lambda = 6800 A = 6.800 x 10^-7 m. Hence nu0 = (3.0 x 10^8)/(6.800 x 10^-7) = 4.41 x 10^14 s^-1. Work function W0 = hnu0 = (6.626 x 10^-34)(4.41 x 10^14) = 2.92 x 10^-19 J = 1.82 eV.

Answer:

Threshold frequency = 4.41 x 10^14 s^-1; work function = 2.92 x 10^-19 J, or 1.82 eV.

✓ Verified and validated
Q.2.13What 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?v
Solution

For hydrogen, 1/lambda = R(1/n1^2 - 1/n2^2) = R(1/2^2 - 1/4^2) = R(3/16). With R = 1.097 x 10^7 m^-1, 1/lambda = 2.06 x 10^6 m^-1, so lambda = 4.86 x 10^-7 m = 486 nm.

Answer:

486 nm.

✓ Verified and validated
Q.2.14How 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).v
Solution

For hydrogen, ionisation energy from level n is 13.6/n^2 eV. For n = 5, E = 13.6/25 = 0.544 eV = 8.72 x 10^-20 J. From n = 1 the ionisation energy is 13.6 eV, so the n = 5 value is 1/25 as large.

Answer:

0.544 eV per atom, or 8.72 x 10^-20 J per atom. This is 1/25 of the ground-state ionisation energy.

✓ Verified and validated
Q.2.15What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?v
Solution

The maximum number of lines produced from level n is n(n - 1)/2. For n = 6, number of lines = 6 x 5/2 = 15.

Answer:

15 emission lines.

✓ Verified and validated
Q.2.16(i) The energy associated with the first orbit in the hydrogen atom is -2.18 x 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.v
Solution

For hydrogen, En = E1/n^2. Thus E5 = (-2.18 x 10^-18)/25 = -8.72 x 10^-20 J atom^-1. Radius rn = n^2a0, where a0 = 0.529 A. Therefore r5 = 25(0.529 A) = 13.2 A = 1.32 x 10^-9 m.

Answer:

(i) -8.72 x 10^-20 J atom^-1. (ii) 1.32 x 10^-9 m, or 13.2 A.

✓ Verified and validated
Q.2.17Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.v
Solution

The longest wavelength in the Balmer series corresponds to the smallest energy transition ending at n = 2, i.e. n = 3 to n = 2. Thus wave number = R(1/2^2 - 1/3^2) = R(5/36) = (1.097 x 10^7)(5/36) = 1.52 x 10^6 m^-1.

Answer:

1.52 x 10^6 m^-1, or 1.52 x 10^4 cm^-1.

✓ Verified and validated
Q.2.18What 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.v
Solution

The given ground-state energy is -2.18 x 10^-11 erg = -2.18 x 10^-18 J. E5 = E1/25 = -8.72 x 10^-20 J. Energy required = E5 - E1 = 2.09 x 10^-18 J. If the electron returns from n = 5 to n = 1, the same energy is emitted as a photon. lambda = hc/E = (6.626 x 10^-34)(3.0 x 10^8)/(2.09 x 10^-18) = 9.50 x 10^-8 m = 95.0 nm.

Answer:

Energy absorbed = 2.09 x 10^-18 J; wavelength emitted = 95.0 nm.

✓ Verified and validated
Q.2.19The electron energy in hydrogen atom is given by En = (-2.18 x 10^-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?v
Solution

For n = 2, E2 = -2.18 x 10^-18/4 = -5.45 x 10^-19 J. To remove the electron to n = infinity, the required energy is 5.45 x 10^-19 J. The longest wavelength has exactly this energy: lambda = hc/E = (6.626 x 10^-34)(3.0 x 10^8)/(5.45 x 10^-19) = 3.65 x 10^-7 m = 3.65 x 10^-5 cm.

Answer:

Energy required = 5.45 x 10^-19 J; longest wavelength = 3.65 x 10^-5 cm.

✓ Verified and validated
Q.2.20Calculate the wavelength of an electron moving with a velocity of 2.05 x 10^7 m s^-1.v
Solution

Use de Broglie's equation lambda = h/mv. With h = 6.626 x 10^-34 J s, m_e = 9.109 x 10^-31 kg and v = 2.05 x 10^7 m s^-1, lambda = 6.626 x 10^-34/[(9.109 x 10^-31)(2.05 x 10^7)] = 3.55 x 10^-11 m.

Answer:

3.55 x 10^-11 m.

✓ Verified and validated
Q.2.21The mass of an electron is 9.1 x 10^-31 kg. If its K.E. is 3.0 x 10^-25 J, calculate its wavelength.v
Solution

Momentum p = sqrt(2mK) = sqrt[2(9.1 x 10^-31)(3.0 x 10^-25)] = 7.39 x 10^-28 kg m s^-1. Hence lambda = h/p = (6.626 x 10^-34)/(7.39 x 10^-28) = 8.96 x 10^-7 m.

Answer:

8.96 x 10^-7 m.

✓ Verified and validated
Q.2.22Which of the following are isoelectronic species i.e., those having the same number of electrons? Na+, K+, Mg2+, Ca2+, S2-, Arv
Solution

Count electrons by adding electrons for negative charge and subtracting for positive charge. Sodium ion has 10 electrons and magnesium ion has 10 electrons. Potassium ion, calcium ion, sulphide ion and argon each have 18 electrons.

Answer:

Na+ and Mg2+ are isoelectronic with 10 electrons. K+, Ca2+, S2- and Ar are isoelectronic with 18 electrons.

✓ Verified and validated
Q.2.23(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 (c) 3p5? (iii) Which atoms are indicated by the following configurations? (a) [He] 2s1 (b) [Ne] 3s2 3p3 (c) [Ar] 4s2 3d1.v
Solution

H- has 2 electrons. Na+, O2- and F- each have 10 electrons, so their configuration is the neon configuration. For outermost electrons, 3s1 corresponds to Na (Z = 11), 2p3 to N (Z = 7), and 3p5 to Cl (Z = 17). The given noble-gas configurations identify Li, P and Sc respectively.

Answer:

(i) H-: 1s2; Na+: 1s2 2s2 2p6; O2-: 1s2 2s2 2p6; F-: 1s2 2s2 2p6. (ii) 11, 7, 17. (iii) Li, P, Sc.

✓ Verified and validated
Q.2.24What is the lowest value of n that allows g orbitals to exist?v
Solution

For a shell with principal quantum number n, allowed l values are 0 to n - 1. A g orbital has l = 4. Therefore the lowest n for which l = 4 is possible is n = 5.

Answer:

n = 5.

✓ Verified and validated
Q.2.25An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron.v
Solution

For a 3d orbital, the principal quantum number is n = 3 and d means l = 2. The magnetic quantum number can have all integral values from -l to +l, so ml = -2, -1, 0, +1, +2.

Answer:

n = 3, l = 2, and ml = -2, -1, 0, +1 or +2.

✓ Verified and validated
Q.2.27Give the number of electrons in the species H2-, H2 and O2+v
Solution

A neutral H2 molecule has 2 electrons. Adding one negative charge gives H2- with 3 electrons. A neutral O2 molecule has 16 electrons, and O2+ has lost one electron, so it has 15 electrons.

Answer:

H2- has 3 electrons, H2 has 2 electrons and O2+ has 15 electrons.

✓ Verified and validated
Q.2.28(i) An atomic orbital has n = 3. What are the possible values of l and ml ? (ii) List the quantum numbers (ml and l) of electrons for 3d orbital. (iii) Which of the following orbitals are possible? 1p, 2s, 2p and 3fv
Solution

For a given n, l can have values from 0 to n - 1. For each l, ml ranges from -l to +l. Thus for n = 3 the allowed l values are 0, 1 and 2. A d orbital has l = 2. The orbital 1p is impossible because n = 1 allows only l = 0; 3f is impossible because f means l = 3, but n = 3 allows only l = 0, 1, 2.

Answer:

(i) l = 0, 1, 2; for l = 0, ml = 0; for l = 1, ml = -1, 0, +1; for l = 2, ml = -2, -1, 0, +1, +2. (ii) For 3d, l = 2 and ml = -2, -1, 0, +1, +2. (iii) 2s and 2p are possible; 1p and 3f are not possible.

✓ Verified and validated
Q.2.29Using s, p, d notations, describe the orbital with the following quantum numbers. (a) n=1, l=0; (b) n = 3; l=1 (c) n = 4; l =2; (d) n=4; l=3.v
Solution

The letter notation is l = 0 -> s, l = 1 -> p, l = 2 -> d, l = 3 -> f. Combine this with the given n value to name the orbital.

Answer:

(a) 1s. (b) 3p. (c) 4d. (d) 4f.

✓ Verified and validated
Q.2.30Explain, giving reasons, which of the following sets of quantum numbers are not possible. (a) n = 0, l = 0, ml = 0, ms = + 1/2 (b) n = 1, l = 0, ml = 0, ms = - 1/2 (c) n = 1, l = 1, ml = 0, ms = + 1/2 (d) n = 2, l = 1, ml = 0, ms = - 1/2 (e) n = 3, l = 3, ml = -3, ms = + 1/2 (f) n = 3, l = 1, ml = 0, ms = + 1/2v
Solution

n must be a positive integer, so n = 0 in (a) is impossible. For a given n, l can range only from 0 to n - 1; hence for n = 1, l = 1 in (c) is impossible, and for n = 3, l = 3 in (e) is impossible. In (b), (d) and (f), l is allowed, ml lies between -l and +l, and ms is +/-1/2.

Answer:

Not possible: (a), (c) and (e). Possible: (b), (d) and (f).

✓ Verified and validated
Q.2.31How many electrons in an atom may have the following quantum numbers? (a) n = 4, ms = - 1/2 (b) n = 3, l = 0v
Solution

For n = 4, the shell can hold 2n^2 = 32 electrons. Half can have ms = -1/2, so 16 electrons may have n = 4 and ms = -1/2. For n = 3, l = 0 is the 3s subshell, which has one orbital and can hold 2 electrons.

Answer:

(a) 16 electrons. (b) 2 electrons.

✓ Verified and validated
Q.2.32Show 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.v
Solution

Bohr's quantisation condition is mvr = nh/2π. Multiplying by 2π gives 2πr = nh/(mv). From de Broglie's relation, λ = h/(mv). Therefore 2πr = nλ. Thus the circumference of an allowed Bohr orbit contains an integral number n of electron wavelengths.

Answer:

The circumference is 2πr = nλ, so it is an integral multiple of the de Broglie wavelength.

✓ Verified and validated
Q.2.33What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum ?v
Solution

For a hydrogen-like ion, wave number is proportional to Z^2(1/n1^2 - 1/n2^2). For He+, Z = 2 and the 4 -> 2 transition gives 4R(1/2^2 - 1/4^2) = 4R(3/16) = 3R/4. In hydrogen, R(1/1^2 - 1/2^2) = 3R/4, so the same wavelength corresponds to transition n = 2 to n = 1.

Answer:

The n = 2 to n = 1 transition in hydrogen.

✓ Verified and validated
Q.2.34Calculate 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 x 10^-18 J atom^-1v
Solution

For hydrogen-like species, ionisation energy from the ground state is Z^2 times the hydrogen value. For He+, Z = 2. Therefore energy required = 2^2(2.18 x 10^-18) = 8.72 x 10^-18 J atom^-1.

Answer:

8.72 x 10^-18 J atom^-1.

✓ Verified and validated
Q.2.35If 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.v
Solution

Diameter of one carbon atom = 0.15 nm = 1.5 x 10^-10 m. Length = 20 cm = 0.20 m. Number of atoms = 0.20/(1.5 x 10^-10) = 1.33 x 10^9.

Answer:

1.33 x 10^9 carbon atoms.

✓ Verified and validated
Q.2.362 x10^8 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.v
Solution

Diameter of one atom = total length/number of atoms = 2.4 cm/(2 x 10^8) = 1.2 x 10^-8 cm = 1.2 x 10^-10 m. Radius = half of diameter = 6.0 x 10^-11 m = 60 pm.

Answer:

6.0 x 10^-11 m, or 60 pm.

✓ Verified and validated
Q.2.37The diameter of zinc atom is 2.6 A. Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.v
Solution

Diameter = 2.6 A, so radius = 1.3 A = 130 pm. Also 2.6 A = 2.6 x 10^-8 cm. Number of atoms in 1.6 cm = 1.6/(2.6 x 10^-8) = 6.15 x 10^7.

Answer:

(a) 130 pm. (b) 6.15 x 10^7 atoms.

✓ Verified and validated
Q.2.38A certain particle carries 2.5 x 10^-16C of static electric charge. Calculate the number of electrons present in it.v
Solution

Number of electrons = total charge/magnitude of electronic charge = (2.5 x 10^-16)/(1.602 x 10^-19) = 1.56 x 10^3.

Answer:

1.56 x 10^3 electrons, approximately 1560 electrons.

✓ Verified and validated
Q.2.39In 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^-18C, calculate the number of electrons present on it.v
Solution

Number of electrons = |q|/e = (1.282 x 10^-18)/(1.602 x 10^-19) = 8.00.

Answer:

8 electrons.

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

Heavy nuclei such as gold have large positive charge and large mass, so alpha-particles experience strong repulsion and the target nucleus recoils little. A light aluminium nucleus has smaller nuclear charge and is less massive, so the repulsion is weaker and recoil is more appreciable. Therefore fewer alpha-particles would be scattered through large angles or bounced back.

Answer:

Large-angle deflections and back scattering would be much less frequent.

✓ Verified and validated
Q.2.41Symbols 79/35Br and 79Br can be written, whereas symbols 35/79Br and 35Br are not acceptable. Answer briefly.v
Solution

In nuclide notation A/Z X, A is mass number and Z is atomic number. For bromine, Z is fixed at 35. The isotope with mass number 79 is therefore written as ^79_35Br. Because the symbol Br already identifies Z = 35, ^79Br is also meaningful. The alternatives with 35 as mass number or 79 as atomic number are not acceptable.

Answer:

Mass number is written as the upper left superscript and atomic number as the lower left subscript. Since bromine has Z = 35 and the isotope has A = 79, 79/35Br and 79Br are acceptable, but 35/79Br and 35Br place or imply the wrong mass number.

✓ Verified and validated
Q.2.42An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.v
Solution

Let the number of protons be Z. Then neutrons = 1.317Z. Mass number A = Z + 1.317Z = 2.317Z = 81, so Z = 34.96 ≈ 35. The element with atomic number 35 is bromine. Hence the atomic symbol is ^81_35Br.

Answer:

^81_35Br.

✓ Verified and validated
Q.2.43An ion with mass number 37 possesses one unit of negative charge. If the ion conatins 11.1% more neutrons than the electrons, find the symbol of the ion.v
Solution

Let protons be Z. A singly negative ion has electrons = Z + 1. Neutrons = 1.111(Z + 1). Since A = Z + neutrons = 37, Z + 1.111(Z + 1) = 37. This gives Z ≈ 17, so the ion is chlorine with mass number 37 and charge -1: ^37_17Cl-.

Answer:

^37_17Cl-.

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

Let protons be Z. A 3+ ion has electrons = Z - 3. Neutrons = 1.304(Z - 3). Since A = Z + neutrons = 56, Z + 1.304(Z - 3) = 56. Thus 2.304Z = 59.912 and Z ≈ 26. The element is iron, so the ion is ^56_26Fe3+.

Answer:

^56_26Fe3+.

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

In the electromagnetic spectrum, radio waves have the lowest frequencies among these, followed by microwaves, visible light, X-rays and cosmic rays. Amber light is visible light, so it lies above microwaves and below X-rays.

Answer:

FM radio < microwave oven < amber light < X-rays < cosmic rays.

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

lambda = 616 nm = 6.16 x 10^-7 m. Frequency nu = c/lambda = 3.0 x 10^8/(6.16 x 10^-7) = 4.87 x 10^14 s^-1. Distance in 30 s = ct = (3.0 x 10^8)(30) = 9.0 x 10^9 m. Energy per quantum = hnu = (6.626 x 10^-34)(4.87 x 10^14) = 3.23 x 10^-19 J. Number of quanta for 2 J = 2/(3.23 x 10^-19) = 6.19 x 10^18.

Answer:

(a) 4.87 x 10^14 s^-1. (b) 9.0 x 10^9 m. (c) 3.23 x 10^-19 J. (d) 6.19 x 10^18 quanta.

✓ Verified and validated
Q.2.48In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 x 10^-18 J from the radiations of 600 nm, calculate the number of photons received by the detector.v
Solution

Energy of one photon is E = hc/lambda = (6.626 x 10^-34)(3.0 x 10^8)/(600 x 10^-9) = 3.31 x 10^-19 J. Number of photons = total energy/energy per photon = (3.15 x 10^-18)/(3.31 x 10^-19) = 9.51.

Answer:

About 9.51 photons, so approximately 10 photons.

✓ Verified and validated
Q.2.50The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calcualte the frequency of each transition and energy difference between two excited states.v
Solution

Frequency is nu = c/lambda. For 589 nm, nu1 = 3.0 x 10^8/(589 x 10^-9) = 5.093 x 10^14 Hz. For 589.6 nm, nu2 = 3.0 x 10^8/(589.6 x 10^-9) = 5.088 x 10^14 Hz. Energy difference = h(nu1 - nu2) = (6.626 x 10^-34)(5.18 x 10^11) = 3.43 x 10^-22 J.

Answer:

Frequencies are 5.09 x 10^14 Hz and 5.09 x 10^14 Hz, more precisely 5.093 x 10^14 Hz and 5.088 x 10^14 Hz. Energy difference = 3.43 x 10^-22 J.

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

At threshold, W0 = hc/lambda0 = hnu0. With W0 = 1.9 eV, lambda0 = 1240/1.9 = 653 nm and nu0 = W0/h = 4.59 x 10^14 s^-1. For 500 nm radiation, photon energy = 1240/500 = 2.48 eV. Kinetic energy = 2.48 - 1.90 = 0.58 eV = 9.3 x 10^-20 J. From KE = (1/2)mv^2, v = sqrt(2KE/m_e) = 4.5 x 10^5 m s^-1.

Answer:

(a) 653 nm. (b) 4.59 x 10^14 s^-1. For 500 nm radiation, kinetic energy = 0.58 eV = 9.3 x 10^-20 J and velocity = 4.5 x 10^5 m s^-1.

✓ Verified and validated
Q.2.53The 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.v
Solution

Photon energy = hc/lambda = 1240/256.7 = 4.83 eV. The stopping potential gives maximum kinetic energy = 0.35 eV. Therefore work function = 4.83 - 0.35 = 4.48 eV = 7.18 x 10^-19 J.

Answer:

4.48 eV, or 7.18 x 10^-19 J.

✓ Verified and validated
Q.2.54If the photon of the wavelength 150 pm strikes an atom and one of tis inner bound electrons is ejected out with a velocity of 1.5 x 10^7 m s^-1, calculate the energy with which it is bound to the nucleus.v
Solution

Photon energy = hc/lambda = (6.626 x 10^-34)(3.0 x 10^8)/(150 x 10^-12) = 1.325 x 10^-15 J. Kinetic energy of the ejected electron = (1/2)mv^2 = 0.5(9.109 x 10^-31)(1.5 x 10^7)^2 = 1.025 x 10^-16 J. Binding energy = photon energy - kinetic energy = 1.22 x 10^-15 J = 7.63 x 10^3 eV.

Answer:

1.22 x 10^-15 J, or 7.63 keV.

✓ Verified and validated
Q.2.55Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represeted as v = 3.29 x 10^15 (Hz) [1/3^2 - 1/n^2] Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.v
Solution

The frequency for 1285 nm is nu = c/lambda = 3.0 x 10^8/(1285 x 10^-9) = 2.33 x 10^14 Hz. Therefore nu/(3.29 x 10^15) = 0.07096 = 1/9 - 1/n^2. Hence 1/n^2 = 0.11111 - 0.07096 = 0.04015, so n^2 ≈ 24.9 and n ≈ 5. A wavelength of 1285 nm is in the infrared region.

Answer:

n = 5; the line lies in the infrared region.

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

For hydrogen, rn = n^2a0 with a0 = 52.9 pm. The initial radius 1.3225 nm = 1322.5 pm = 25a0, so ni = 5. The final radius 211.6 pm = 4a0, so nf = 2. Thus the transition is 5 -> 2. Wave number = R(1/2^2 - 1/5^2), giving lambda = 4.34 x 10^-7 m = 434 nm. Since the final level is n = 2, it belongs to the Balmer series and lies in the visible region.

Answer:

434 nm; Balmer series; visible region.

✓ Verified and validated
Q.2.57Dual 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 x 10^6 ms^-1, calculate de Broglie wavelength associated with this electron.v
Solution

Use lambda = h/mv. lambda = 6.626 x 10^-34/[(9.109 x 10^-31)(1.6 x 10^6)] = 4.55 x 10^-10 m.

Answer:

4.55 x 10^-10 m.

✓ Verified and validated
Q.2.58Similar 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.v
Solution

From lambda = h/mv, v = h/(m lambda). For a neutron, m = 1.675 x 10^-27 kg and lambda = 800 pm = 8.00 x 10^-10 m. Hence v = 6.626 x 10^-34/[(1.675 x 10^-27)(8.00 x 10^-10)] = 4.94 x 10^2 m s^-1.

Answer:

4.94 x 10^2 m s^-1.

✓ Verified and validated
Q.2.59If the velocity of the electron in Bohr’s first orbit is 2.19 x 10^6 ms^-1, calculate the de Broglie wavelength associated with it.v
Solution

lambda = h/mv = 6.626 x 10^-34/[(9.109 x 10^-31)(2.19 x 10^6)] = 3.32 x 10^-10 m.

Answer:

3.32 x 10^-10 m.

✓ Verified and validated
Q.2.60The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 x 10^5 ms^-1. If the hockey ball of mass 0.1 kg is moving with this velocity, calcualte the wavelength associated with this velocity.v
Solution

For the hockey ball, lambda = h/mv = 6.626 x 10^-34/[(0.1)(4.37 x 10^5)] = 1.52 x 10^-38 m.

Answer:

1.52 x 10^-38 m.

✓ Verified and validated
Q.2.62The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists: 1. n = 4, l = 2, ml = -2, ms = -1/2 2. n = 3, l = 2, ml = 1, ms = +1/2 3. n = 4, l = 1, ml = 0, ms = +1/2 4. n = 3, l = 2, ml = -2, ms = -1/2 5. n = 3, l = 1, ml = -1, ms = +1/2 6. n = 4, l = 1, ml = 0, ms = +1/2v
Solution

Use the (n + l) rule. Electron 5 is 3p, with n + l = 4. Electrons 2 and 4 are 3d, with n + l = 5 and the same subshell energy. Electrons 3 and 6 are 4p, with n + l = 5 but higher n than 3d, so they have higher energy than 3d and equal energy to each other. Electron 1 is 4d, with n + l = 6, so it is highest among these.

Answer:

Increasing energy: 5 < 2 = 4 < 3 = 6 < 1.

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

Outer-shell electrons are shielded by inner-shell electrons and are farther from the nucleus. Therefore a 4p electron in bromine is shielded more strongly than 2p and 3p electrons and experiences the lowest effective nuclear charge.

Answer:

An electron in the 4p orbital experiences the lowest effective nuclear charge.

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

An orbital closer to the nucleus or with greater penetration experiences larger effective nuclear charge. 2s is closer than 3s. For the same n, penetration decreases in the order s > p > d > f, so 4d experiences larger effective nuclear charge than 4f, and 3p experiences larger effective nuclear charge than 3d.

Answer:

(i) 2s. (ii) 4d. (iii) 3p.

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

Al and Si have their unpaired electrons in the same 3p subshell, but silicon has one more proton in the nucleus than aluminium. The additional nuclear charge is not completely cancelled by shielding, so the 3p electrons in Si experience greater effective nuclear charge.

Answer:

The 3p electrons in silicon experience more effective nuclear charge.

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

P is [Ne]3s2 3p3, so three 3p electrons are unpaired. Si is [Ne]3s2 3p2, so two are unpaired. Cr is [Ar]3d5 4s1, giving six unpaired electrons. Fe is [Ar]3d6 4s2, giving four unpaired 3d electrons. Kr has a closed-shell configuration, so it has no unpaired electron.

Answer:

(a) P: 3. (b) Si: 2. (c) Cr: 6. (d) Fe: 4. (e) Kr: 0.

✓ Verified and validated
Q.2.67(a) How many subshells are associated with n = 4 ? (b) How many electrons will be present in the subshells having ms value of -1/2 for n = 4 ?v
Solution

For n = 4, possible l values are 0, 1, 2 and 3, corresponding to 4s, 4p, 4d and 4f: four subshells. The n = 4 shell contains n^2 = 16 orbitals and 2n^2 = 32 electrons. One electron in each orbital can have ms = -1/2, so 16 electrons can have ms = -1/2.

Answer:

(a) 4 subshells. (b) 16 electrons.

✓ Verified and validated
Back: Ch.1
Next: Ch.3