JEE Chemistry Structure Of Atom MCQs Set 11

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Multiple Choice Questions

Question 1: A proton is best described as
(a) the nucleus of deuterium
(b) an ionised hydrogen molecule
(c) an alpha particle
(d) an ionised hydrogen atom (H\(^+\))
Show Answer & Explanation

Answer: (d) an ionised hydrogen atom (H\(^+\))

Explanation:
1. Ordinary hydrogen has one proton and one electron, so removing the electron leaves a bare proton.
2. Only option (d) agrees with this working.

Teacher's Note:
1. Know the standard values: nuclear radius about 1.3 fm times the cube root of A, and the charge and mass of the electron, proton and neutron. Quantities like e/m separate the particles quickly.

Question 2: Anode rays consist of particles that carry
(a) both mass and charge
(b) mass only
(c) charge only
(d) neither mass nor charge
Show Answer & Explanation

Answer: (a) both mass and charge

Explanation:
1. Anode rays are streams of positive ions, so they have mass and positive charge
2. They rotate a paddle wheel and are deflected by fields.

Teacher's Note:
1. Know the standard values: nuclear radius about 1.3 fm times the cube root of A, and the charge and mass of the electron, proton and neutron. Quantities like e/m separate the particles quickly.

Question 3: The ratio of the mass of an electron to the mass of a proton is nearest to
(a) 1 : 2
(b) 1 : 1
(c) 1 : 1837
(d) 1 : 3
Show Answer & Explanation

Answer: (c) 1 : 1837

Explanation:
1. \(m_p/m_e \approx 1837\).
2. Only option (c) agrees with this working.

Teacher's Note:
1. Know the standard values: nuclear radius about 1.3 fm times the cube root of A, and the charge and mass of the electron, proton and neutron. Quantities like e/m separate the particles quickly.

Question 4: The correct increasing order of e/m for He\(^+\), Ne\(^+\) and Ar\(^+\) is
(a) He\(^+\) < Ne\(^+\) < Ar\(^+\)
(b) Ar\(^+\) < Ne\(^+\) < He\(^+\)
(c) Ne\(^+\) < Ar\(^+\) < He\(^+\)
(d) all are equal
Show Answer & Explanation

Answer: (b) Ar\(^+\) < Ne\(^+\) < He\(^+\)

Explanation:
1. All carry the same charge, so e/m varies inversely as mass: Ar (40) is heaviest and He (4) is lightest, so Ar\(^+\) has the smallest e/m.
2. Only option (b) agrees with this working.

Teacher's Note:
1. Know the standard values: nuclear radius about 1.3 fm times the cube root of A, and the charge and mass of the electron, proton and neutron. Quantities like e/m separate the particles quickly.

Question 5: If two like charges are brought closer, the potential energy of the system
(a) decreases continuously
(b) first increases, then decreases
(c) increases continuously
(d) first decreases, then increases
Show Answer & Explanation

Answer: (c) increases continuously

Explanation:
1. \(PE = kq_1q_2/r\) is positive for like charges and rises as r falls.
2. Only option (c) agrees with this working.

Teacher's Note:
1. Know the standard values: nuclear radius about 1.3 fm times the cube root of A, and the charge and mass of the electron, proton and neutron. Quantities like e/m separate the particles quickly.

Question 6: When alpha particles are fired at a thin gold foil, only a few in a million are deflected because
(a) the nucleus occupies a tiny fraction of the volume of the atom
(b) the force of repulsion on the alpha particles is very weak
(c) there are many nuclei and only one electron
(d) alpha particles are heavier than the nucleus
Show Answer & Explanation

Answer: (a) the nucleus occupies a tiny fraction of the volume of the atom

Explanation:
1. Almost all of the atom is empty space; only an alpha particle that comes close to the tiny nucleus is scattered strongly.
2. Only option (a) agrees with this working.

Teacher's Note:
1. Know the standard values: nuclear radius about 1.3 fm times the cube root of A, and the charge and mass of the electron, proton and neutron. Quantities like e/m separate the particles quickly.

Question 7: Which of the following pairs consists of isoelectronic species: (I) Ar and S\(^{2-}\); (II) Fe and Ni\(^{2+}\)?
(a) Only pair I
(b) Both pairs
(c) Only pair II
(d) Neither pair
Show Answer & Explanation

Answer: (b) Both pairs

Explanation:
1. Ar and S\(^{2-}\) both have 18 electrons; Fe has 26 and Ni\(^{2+}\) has \(28-2 = 26\).
2. Only option (b) agrees with this working.

Teacher's Note:
1. Know the standard values: nuclear radius about 1.3 fm times the cube root of A, and the charge and mass of the electron, proton and neutron. Quantities like e/m separate the particles quickly.

Question 8: Which species has more electrons than neutrons?
(a) \(^{31}_{15}\)P\(^{3-}\)
(b) \(^{27}_{13}\)Al\(^{3+}\)
(c) \(^{2}_{1}\)H
(d) \(^{37}_{17}\)Cl\(^-\)
Show Answer & Explanation

Answer: (a) \(^{31}_{15}\)P\(^{3-}\)

Explanation:
1. P\(^{3-}\): 18 electrons, 16 neutrons
2. Al\(^{3+}\): 10 and 14
3. \(^2_1H\): 1 and 1
4. Cl\(^-\): 18 and 20.

Teacher's Note:
1. Know the standard values: nuclear radius about 1.3 fm times the cube root of A, and the charge and mass of the electron, proton and neutron. Quantities like e/m separate the particles quickly.

Question 9: The pairs CO\(_3^{2-}\)/NO\(_3^-\), PO\(_4^{3-}\)/SO\(_4^{2-}\) and CO/N\(_2\) are all
(a) isotonic pairs
(b) isobaric pairs
(c) isoelectronic pairs
(d) isotopic pairs
Show Answer & Explanation

Answer: (c) isoelectronic pairs

Explanation:
1. Electrons: 32 and 32; 50 and 50; 14 and 14
2. Each pair has equal electrons.

Teacher's Note:
1. Know the standard values: nuclear radius about 1.3 fm times the cube root of A, and the charge and mass of the electron, proton and neutron. Quantities like e/m separate the particles quickly.

Question 10: An ion A\(^{5+}\) is isoelectronic with N\(^{3-}\) and the atom A has number of neutrons equal to Z + 1. The molar mass of A is
(a) 30 g
(b) 29 g
(c) 32 g
(d) 31 g
Show Answer & Explanation

Answer: (d) 31 g

Explanation:
1. N\(^{3-}\) has 10 electrons, so A\(^{5+}\) has 10, A has 15 electrons and Z = 15
2. Neutrons \(= 16\)
3. Mass number \(= 31\).

Teacher's Note:
1. Know the standard values: nuclear radius about 1.3 fm times the cube root of A, and the charge and mass of the electron, proton and neutron. Quantities like e/m separate the particles quickly.

Question 11: Which pair are isotones?
(a) \(^{38}\)Ar and \(^{40}\)K
(b) \(^{40}\)K and \(^{40}\)Ca
(c) \(^{37}\)Cl and \(^{39}\)K
(d) \(^{39}\)K and \(^{40}\)K
Show Answer & Explanation

Answer: (c) \(^{37}\)Cl and \(^{39}\)K

Explanation:
1. Neutrons: Cl-37 \(= 20\); K-39 \(= 20\)
2. Ar-38 \(= 20\) but K-40 \(= 21\); K-40 \(= 21\), Ca-40 \(= 20\).

Teacher's Note:
1. Know the standard values: nuclear radius about 1.3 fm times the cube root of A, and the charge and mass of the electron, proton and neutron. Quantities like e/m separate the particles quickly.

Question 12: If the speed of an alpha particle approaching a nucleus is reduced to one third, the distance of closest approach changes by
(a) an increase of 800%
(b) an increase of 300%
(c) an increase of 600%
(d) a decrease of 80%
Show Answer & Explanation

Answer: (a) an increase of 800%

Explanation:
1. \(r_{min}\propto1/v^2\)
2. With \(v\to v/3\), \(r_{min}\to9r_{min}\), an increase of 800%.

Teacher's Note:
1. Know the standard values: nuclear radius about 1.3 fm times the cube root of A, and the charge and mass of the electron, proton and neutron. Quantities like e/m separate the particles quickly.

Question 13: A radio wave of frequency \(6\times10^{3}\) s\(^{-1}\) travels from Mars to Earth, a distance of \(8\times10^{7}\) km. The time taken is
(a) 2 min 46 s
(b) 4 min 26 s
(c) 10 min 46 s
(d) 5 min 30 s
Show Answer & Explanation

Answer: (b) 4 min 26 s

Explanation:
1. Time \(= 8\times10^{10}/3\times10^{8} = 266.7\) s \(= 4\) min 26.7 s
2. The frequency does not affect the time.

Teacher's Note:
1. For photon problems use E = hc/lambda = 12400/lambda(in A) eV and keep units consistent. In the photoelectric effect, KE(max) = h(nu - nu0), and intensity changes only the number of electrons.

Question 14: The wave number of radiation of frequency \(4\times10^{4}\) Hz is
(a) \(1.33\times10^{-7}\) cm\(^{-1}\)
(b) \(9\times10^{-11}\) cm\(^{-1}\)
(c) \(4\times10^{-5}\) cm\(^{-1}\)
(d) \(1.33\times10^{-6}\) cm\(^{-1}\)
Show Answer & Explanation

Answer: (d) \(1.33\times10^{-6}\) cm\(^{-1}\)

Explanation:
1. \(\bar\nu = \nu/c = 4\times10^{4}/3\times10^{10}\ \text{cm/s} = 1.33\times10^{-6}\) cm\(^{-1}\).
2. Only option (d) agrees with this working.

Teacher's Note:
1. For photon problems use E = hc/lambda = 12400/lambda(in A) eV and keep units consistent. In the photoelectric effect, KE(max) = h(nu - nu0), and intensity changes only the number of electrons.

Question 15: A 100 W bulb emits radiation of wavelength 400 nm. The number of photons emitted per second is nearest to
(a) \(2\times10^{15}\)
(b) \(2\times10^{10}\)
(c) \(2\times10^{20}\)
(d) \(2\times10^{5}\)
Show Answer & Explanation

Answer: (c) \(2\times10^{20}\)

Explanation:
1. Photon energy \(= 6.626\times10^{-34}\times3\times10^{8}/4\times10^{-7} = 4.97\times10^{-19}\) J. Rate \(= 100/4.97\times10^{-19} = 2\times10^{20}\) s\(^{-1}\).
2. Only option (c) agrees with this working.

Teacher's Note:
1. For photon problems use E = hc/lambda = 12400/lambda(in A) eV and keep units consistent. In the photoelectric effect, KE(max) = h(nu - nu0), and intensity changes only the number of electrons.

Question 16: The number of photons of wavelength 4000 pm needed to provide 1 J of energy is nearest to
(a) \(10^{16}\)
(b) \(5\times10^{16}\)
(c) \(9\times10^{16}\)
(d) \(2\times10^{16}\)
Show Answer & Explanation

Answer: (d) \(2\times10^{16}\)

Explanation:
1. \(\lambda = 4\times10^{-9}\) m, so \(E = 1.986\times10^{-25}/4\times10^{-9} = 4.97\times10^{-17}\) J. Number \(= 1/4.97\times10^{-17} = 2\times10^{16}\).
2. Only option (d) agrees with this working.

Teacher's Note:
1. For photon problems use E = hc/lambda = 12400/lambda(in A) eV and keep units consistent. In the photoelectric effect, KE(max) = h(nu - nu0), and intensity changes only the number of electrons.

Question 17: A dye absorbs \(n_1\) photons of wavelength 4000 A and emits \(n_2\) photons of wavelength 5000 A. If 40% of the absorbed energy is emitted, \(n_1/n_2\) is
(a) 1
(b) 2
(c) 3
(d) 4
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. Emitted energy \(= 0.4\times\) absorbed energy: \(n_2/5000 = 0.4\,n_1/4000\), so \(n_1/n_2 = 4000/(0.4\times5000) = 2\).
2. Only option (b) agrees with this working.

Teacher's Note:
1. For photon problems use E = hc/lambda = 12400/lambda(in A) eV and keep units consistent. In the photoelectric effect, KE(max) = h(nu - nu0), and intensity changes only the number of electrons.

Question 18: One quantum is absorbed per molecule of X\(_2\) to dissociate it into atoms. If the light has wavelength 1240 A, the bond energy of X\(_2\) is
(a) 20 J/mol
(b) 48 eV/molecule
(c) 10 eV/molecule
(d) 184 J/mol
Show Answer & Explanation

Answer: (c) 10 eV/molecule

Explanation:
1. \(E = 12400/1240 = 10\) eV per molecule.
2. Only option (c) agrees with this working.

Teacher's Note:
1. For photon problems use E = hc/lambda = 12400/lambda(in A) eV and keep units consistent. In the photoelectric effect, KE(max) = h(nu - nu0), and intensity changes only the number of electrons.

Question 19: As the momentum of an electron decreases, its de Broglie wavelength
(a) decreases
(b) increases
(c) remains the same
(d) cannot be predicted
Show Answer & Explanation

Answer: (b) increases

Explanation:
1. \(\lambda = h/p\), so a smaller momentum gives a larger wavelength.
2. Only option (b) agrees with this working.

Teacher's Note:
1. The de Broglie relation lambda = h/mv and the Heisenberg relation dx.dp = h/4pi are the two tools here; for an electron accelerated through V volts, lambda = sqrt(150/V) in angstrom.

Question 20: The de Broglie wavelength of a particle of mass 1 g moving at 100 m/s is
(a) \(6.63\times10^{-34}\) m
(b) \(6.63\times10^{-35}\) m
(c) \(6.63\times10^{-31}\) m
(d) \(6.63\times10^{-33}\) m
Show Answer & Explanation

Answer: (d) \(6.63\times10^{-33}\) m

Explanation:
1. \(\lambda = 6.63\times10^{-34}/(10^{-3}\times100) = 6.63\times10^{-33}\) m.
2. Only option (d) agrees with this working.

Teacher's Note:
1. The de Broglie relation lambda = h/mv and the Heisenberg relation dx.dp = h/4pi are the two tools here; for an electron accelerated through V volts, lambda = sqrt(150/V) in angstrom.

Question 21: A particle has momentum \(3.3\times10^{-24}\) kg m s\(^{-1}\) (\(h = 6.6\times10^{-34}\) J s). Its de Broglie wavelength is
(a) 2 A
(b) 0.02 A
(c) 90.5 A
(d) 500 A
Show Answer & Explanation

Answer: (a) 2 A

Explanation:
1. \(\lambda = 6.6\times10^{-34}/3.3\times10^{-24} = 2\times10^{-10}\) m \(= 2\) A.
2. Only option (a) agrees with this working.

Teacher's Note:
1. The de Broglie relation lambda = h/mv and the Heisenberg relation dx.dp = h/4pi are the two tools here; for an electron accelerated through V volts, lambda = sqrt(150/V) in angstrom.

Question 22: The ratio of the de Broglie wavelengths of a proton and an alpha particle of the same kinetic energy is
(a) 1 : 2
(b) 3 : 1
(c) 1 : 3
(d) 2 : 1
Show Answer & Explanation

Answer: (d) 2 : 1

Explanation:
1. \(\lambda = h/\sqrt{2mK}\) with equal K: \(\lambda_p/\lambda_\alpha = \sqrt{m_\alpha/m_p} = \sqrt4 = 2\).
2. Only option (d) agrees with this working.

Teacher's Note:
1. The de Broglie relation lambda = h/mv and the Heisenberg relation dx.dp = h/4pi are the two tools here; for an electron accelerated through V volts, lambda = sqrt(150/V) in angstrom.

Question 23: A 0.66 kg ball moves at 100 m/s (\(h = 6.6\times10^{-34}\) J s). Its de Broglie wavelength is
(a) \(6.6\times10^{-32}\) m
(b) \(1.0\times10^{-35}\) m
(c) \(6.6\times10^{-34}\) m
(d) \(1.0\times10^{-32}\) m
Show Answer & Explanation

Answer: (b) \(1.0\times10^{-35}\) m

Explanation:
1. \(\lambda = 6.6\times10^{-34}/(0.66\times100) = 10^{-35}\) m.
2. Only option (b) agrees with this working.

Teacher's Note:
1. The de Broglie relation lambda = h/mv and the Heisenberg relation dx.dp = h/4pi are the two tools here; for an electron accelerated through V volts, lambda = sqrt(150/V) in angstrom.

Question 24: A 1 kg ball has kinetic energy 0.5 J. Its de Broglie wavelength is
(a) \(13.2\times10^{-34}\) m
(b) \(10.4\times10^{-21}\) m
(c) \(6.626\times10^{-34}\) m
(d) \(6.626\times10^{-34}\) A
Show Answer & Explanation

Answer: (c) \(6.626\times10^{-34}\) m

Explanation:
1. \(v = \sqrt{2K/m} = 1\) m/s, so \(\lambda = h/mv = 6.626\times10^{-34}\) m.
2. Only option (c) agrees with this working.

Teacher's Note:
1. The de Broglie relation lambda = h/mv and the Heisenberg relation dx.dp = h/4pi are the two tools here; for an electron accelerated through V volts, lambda = sqrt(150/V) in angstrom.

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