JEE Chemistry Structure Of Atom MCQs Set 12

Multiple Choice Questions (MCQs) for JEE Chemistry: Structure Of Atom

Access targeted multiple-choice questions for Structure Of Atom designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Practice Structure Of Atom MCQs for JEE Chemistry

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Multiple Choice Questions

Question 1: When the frequency of light striking a metal plate is doubled, the maximum kinetic energy of the emitted electrons
(a) more than doubles
(b) doubles exactly
(c) halves
(d) is unchanged
Show Answer & Explanation

Answer: (a) more than doubles

Explanation:
1. \(KE_2 = 2h\nu-W = 2(h\nu-W)+W > 2KE_1\).
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 2: If the uncertainty in position is \(10^{-10}\) m and in velocity is \(5.27\times10^{-24}\) m/s, the mass of the particle is
(a) 0.1 g
(b) 1000 kg
(c) \(10^{-3}\) g
(d) 0.1 kg
Show Answer & Explanation

Answer: (d) 0.1 kg

Explanation:
1. \(m = h/(4\pi\Delta x\Delta v) = 6.626\times10^{-34}/(4\pi\times10^{-10}\times5.27\times10^{-24}) = 0.1\) kg.
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 3: The uncertainty in momentum is \(1\times10^{-18}\) g cm s\(^{-1}\) and the electron mass is \(9\times10^{-28}\) g. The uncertainty in velocity is
(a) \(1.11\times10^{9}\) cm/s
(b) \(1.11\times10^{6}\) cm/s
(c) \(1.11\times10^{5}\) cm/s
(d) \(1.11\times10^{11}\) cm/s
Show Answer & Explanation

Answer: (a) \(1.11\times10^{9}\) cm/s

Explanation:
1. \(\Delta v = \Delta p/m = 10^{-18}/9\times10^{-28} = 1.11\times10^{9}\) cm/s.
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 4: The Heisenberg uncertainty relation can be made smaller by
(a) using better instruments
(b) using photons of higher energy
(c) no method; it is a fundamental limit
(d) using photons of lower wave number
Show Answer & Explanation

Answer: (c) no method; it is a fundamental limit

Explanation:
1. The limit \(\Delta x\,\Delta p \ge h/4\pi\) follows from the wave nature of matter and does not depend on the apparatus.
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.

Question 5: If r\(_1\) is the radius of the first Bohr orbit of hydrogen, the radii of the second, third and fourth orbits are
(a) 2r\(_1\), 3r\(_1\), 4r\(_1\)
(b) 4r\(_1\), 9r\(_1\), 16r\(_1\)
(c) 8r\(_1\), 27r\(_1\), 64r\(_1\)
(d) r\(_1\)^2, r\(_1\)^3, r\(_1\)^4
Show Answer & Explanation

Answer: (b) 4r\(_1\), 9r\(_1\), 16r\(_1\)

Explanation:
1. \(r_n = n^2r_1\) gives 4, 9, 16 times r\(_1\).
2. Only option (b) agrees with this working.

Teacher's Note:
1. In the Bohr model remember r proportional to n squared over Z, v proportional to Z over n, and E proportional to minus Z squared over n squared. Ratios between species are then quick to find.

Question 6: The ratio of the radii of the second orbits of He\(^+\), Li\(^{2+}\) and Be\(^{3+}\) is
(a) 1 : 2 : 3
(b) 3 : 4 : 6
(c) 6 : 4 : 3
(d) 2 : 3 : 4
Show Answer & Explanation

Answer: (c) 6 : 4 : 3

Explanation:
1. For the same n, \(r\propto1/Z\): \(1/2:1/3:1/4 = 6:4:3\).
2. Only option (c) agrees with this working.

Teacher's Note:
1. In the Bohr model remember r proportional to n squared over Z, v proportional to Z over n, and E proportional to minus Z squared over n squared. Ratios between species are then quick to find.

Question 7: If the radius of the hydrogen atom is 0.53 A, the radius of Li\(^{2+}\) (ground state) is
(a) 0.17 A
(b) 1.27 A
(c) 0.57 A
(d) 0.99 A
Show Answer & Explanation

Answer: (a) 0.17 A

Explanation:
1. \(r = 0.53/Z = 0.53/3 = 0.177\) A.
2. Only option (a) agrees with this working.

Teacher's Note:
1. In the Bohr model remember r proportional to n squared over Z, v proportional to Z over n, and E proportional to minus Z squared over n squared. Ratios between species are then quick to find.

Question 8: The speed of the electron in the fourth Bohr orbit of hydrogen is u. Its speed in the first orbit is
(a) u/4
(b) 4u
(c) 16u
(d) u/16
Show Answer & Explanation

Answer: (b) 4u

Explanation:
1. \(v\propto1/n\), so \(v_1 = 4v_4 = 4u\).
2. Only option (b) agrees with this working.

Teacher's Note:
1. In the Bohr model remember r proportional to n squared over Z, v proportional to Z over n, and E proportional to minus Z squared over n squared. Ratios between species are then quick to find.

Question 9: Which statement about Bohr theory is wrong?
(a) The energy of the electron decreases as the orbit number increases
(b) The kinetic energy is half the magnitude of the potential energy
(c) The kinetic energy is the negative of the total energy
(d) The ionisation energy of H in the first excited state is minus one-fourth of the ground-state energy
Show Answer & Explanation

Answer: (a) The energy of the electron decreases as the orbit number increases

Explanation:
1. Energy \(= -13.6Z^2/n^2\) increases (becomes less negative) with n, so the first statement is false
2. The others are true: \(KE = -E = |PE|/2\), and \(IE(n=2) = 3.4\) eV \(= -E_1/4\).

Teacher's Note:
1. In the Bohr model remember r proportional to n squared over Z, v proportional to Z over n, and E proportional to minus Z squared over n squared. Ratios between species are then quick to find.

Question 10: The kinetic energy of the electron in a hydrogen-like species is
(a) directly proportional to n\(^2\)
(b) inversely proportional to Z\(^2\)
(c) neither directly proportional to n\(^2\) nor inversely proportional to Z\(^2\)
(d) both of these
Show Answer & Explanation

Answer: (c) neither directly proportional to n\(^2\) nor inversely proportional to Z\(^2\)

Explanation:
1. \(KE = 13.6Z^2/n^2\) eV, which varies as \(1/n^2\) and as \(Z^2\).
2. Only option (c) agrees with this working.

Teacher's Note:
1. In the Bohr model remember r proportional to n squared over Z, v proportional to Z over n, and E proportional to minus Z squared over n squared. Ratios between species are then quick to find.

Question 11: In hydrogen, the energy of the second excited state is -1.51 eV. The kinetic energy of the electron in the same orbit is
(a) +3.4 eV
(b) -13.6 eV
(c) +13.6 eV
(d) +1.51 eV
Show Answer & Explanation

Answer: (d) +1.51 eV

Explanation:
1. \(KE = -E = +1.51\) eV.
2. Only option (d) agrees with this working.

Teacher's Note:
1. In the Bohr model remember r proportional to n squared over Z, v proportional to Z over n, and E proportional to minus Z squared over n squared. Ratios between species are then quick to find.

Question 12: The energy of the fourth Bohr orbit of hydrogen is -82 kJ/mol. The energy of the second orbit is
(a) -164 kJ/mol
(b) -328 kJ/mol
(c) -41 kJ/mol
(d) -1312 kJ/mol
Show Answer & Explanation

Answer: (b) -328 kJ/mol

Explanation:
1. \(E\propto1/n^2\): \(E_2 = E_4\times16/4 = -82\times4 = -328\) kJ/mol.
2. Only option (b) agrees with this working.

Teacher's Note:
1. In the Bohr model remember r proportional to n squared over Z, v proportional to Z over n, and E proportional to minus Z squared over n squared. Ratios between species are then quick to find.

Question 13: The binding energy of the 3rd orbit of a hydrogen-like species is 13.6 eV. The species is
(a) Li\(^{2+}\)
(b) Be\(^{3+}\)
(c) He\(^+\)
(d) Li\(^+\)
Show Answer & Explanation

Answer: (a) Li\(^{2+}\)

Explanation:
1. \(13.6Z^2/9 = 13.6\) gives \(Z = 3\), so the species is Li\(^{2+}\).
2. Only option (a) agrees with this working.

Teacher's Note:
1. In the Bohr model remember r proportional to n squared over Z, v proportional to Z over n, and E proportional to minus Z squared over n squared. Ratios between species are then quick to find.

Question 14: A photon from the n = 4 to n = 2 transition of a hydrogen-like ion has wavelength 1216 A. The ion is
(a) H
(b) He\(^+\)
(c) Li\(^{2+}\)
(d) Be\(^{3+}\)
Show Answer & Explanation

Answer: (b) He\(^+\)

Explanation:
1. For hydrogen this line is at about 4861 A, and \(\lambda\propto1/Z^2\)
2. \(4861/1216 = 4\), so Z = 2.

Teacher's Note:
1. In the Bohr model remember r proportional to n squared over Z, v proportional to Z over n, and E proportional to minus Z squared over n squared. Ratios between species are then quick to find.

Question 15: The first excitation potential of a hypothetical H-like atom is 10.2x V. The second excitation energy is
(a) 13.6x eV
(b) 40.5x eV
(c) -10.2x eV
(d) 12.09x eV
Show Answer & Explanation

Answer: (d) 12.09x eV

Explanation:
1. In units of x the level scheme is that of hydrogen (first excitation 10.2x eV), so the second excitation energy is \(13.6(1-1/9)x = 12.09x\) eV.
2. Only option (d) agrees with this working.

Teacher's Note:
1. In the Bohr model remember r proportional to n squared over Z, v proportional to Z over n, and E proportional to minus Z squared over n squared. Ratios between species are then quick to find.

Question 16: The binding energy of the first excited state of a hydrogen-like species is 54.4 eV. Its atomic number is
(a) 2
(b) 3
(c) 4
(d) 1
Show Answer & Explanation

Answer: (c) 4

Explanation:
1. \(13.6Z^2/4 = 54.4\) gives \(Z^2 = 16\), Z = 4.
2. Only option (c) agrees with this working.

Teacher's Note:
1. In the Bohr model remember r proportional to n squared over Z, v proportional to Z over n, and E proportional to minus Z squared over n squared. Ratios between species are then quick to find.

Question 17: The spectral lines of the Paschen series of hydrogen lie in the
(a) visible region
(b) ultraviolet region
(c) X-ray region
(d) infrared region
Show Answer & Explanation

Answer: (d) infrared region

Explanation:
1. Paschen lines end at n = 3 and have energies below those of the Balmer series, so they are infrared.
2. Only option (d) agrees with this working.

Teacher's Note:
1. For spectral lines use 1/lambda = R Z^2 (1/n1^2 - 1/n2^2). The lowest n1 fixes the series (1 Lyman, 2 Balmer, 3 Paschen), and n(n-1)/2 gives the total number of possible lines.

Question 18: The longest wavelength in the Lyman series of hydrogen (\(R_H = 109678\) cm\(^{-1}\)) is
(a) 1315 A
(b) 1215 A
(c) 1415 A
(d) 1515 A
Show Answer & Explanation

Answer: (b) 1215 A

Explanation:
1. \(\bar\nu = R(1-1/4) = 82258.5\) cm\(^{-1}\), so \(\lambda = 1.2157\times10^{-5}\) cm \(= 1215\) A.
2. Only option (b) agrees with this working.

Teacher's Note:
1. For spectral lines use 1/lambda = R Z^2 (1/n1^2 - 1/n2^2). The lowest n1 fixes the series (1 Lyman, 2 Balmer, 3 Paschen), and n(n-1)/2 gives the total number of possible lines.

Question 19: The shortest wavelength in the Brackett series of hydrogen (\(R_H = 109678\) cm\(^{-1}\)) is
(a) 1459 A
(b) 4052 A
(c) 1459 nm
(d) 4052 nm
Show Answer & Explanation

Answer: (c) 1459 nm

Explanation:
1. The series limit has \(\bar\nu = R/16\), so \(\lambda = 16/109678 = 1.459\times10^{-4}\) cm \(= 1459\) nm.
2. Only option (c) agrees with this working.

Teacher's Note:
1. For spectral lines use 1/lambda = R Z^2 (1/n1^2 - 1/n2^2). The lowest n1 fixes the series (1 Lyman, 2 Balmer, 3 Paschen), and n(n-1)/2 gives the total number of possible lines.

Question 20: Spectral lines from electrons jumping from the 6th, 5th and 4th orbits to the 2nd orbit belong to the
(a) Lyman series
(b) Balmer series
(c) Paschen series
(d) Pfund series
Show Answer & Explanation

Answer: (b) Balmer series

Explanation:
1. All lines ending at n = 2 form the Balmer series.
2. Only option (b) agrees with this working.

Teacher's Note:
1. For spectral lines use 1/lambda = R Z^2 (1/n1^2 - 1/n2^2). The lowest n1 fixes the series (1 Lyman, 2 Balmer, 3 Paschen), and n(n-1)/2 gives the total number of possible lines.

Question 21: The speed of the electron in the first orbit of the hydrogen atom is, in terms of the speed of light c, about
(a) c/1370
(b) c/13.7
(c) c/1.37
(d) c/137
Show Answer & Explanation

Answer: (d) c/137

Explanation:
1. \(v = 2.18\times10^{6}\) m/s and \(c = 3\times10^{8}\) m/s give \(v/c = 1/137\).
2. Only option (d) agrees with this working.

Teacher's Note:
1. For spectral lines use 1/lambda = R Z^2 (1/n1^2 - 1/n2^2). The lowest n1 fixes the series (1 Lyman, 2 Balmer, 3 Paschen), and n(n-1)/2 gives the total number of possible lines.

Question 22: When electrons in the 7th shell of many hydrogen atoms return to the 2nd shell, the number of possible spectral lines is
(a) 15
(b) 12
(c) 14
(d) 10
Show Answer & Explanation

Answer: (a) 15

Explanation:
1. The levels involved are 2, 3, 4, 5, 6, 7, which is six levels, giving \(6\times5/2 = 15\) lines.
2. Only option (a) agrees with this working.

Teacher's Note:
1. For spectral lines use 1/lambda = R Z^2 (1/n1^2 - 1/n2^2). The lowest n1 fixes the series (1 Lyman, 2 Balmer, 3 Paschen), and n(n-1)/2 gives the total number of possible lines.

Question 23: The energy of the second orbit of hydrogen equals the energy of
(a) the fourth orbit of Li\(^{2+}\)
(b) the second orbit of He\(^+\)
(c) the second orbit of Li\(^{2+}\)
(d) the fourth orbit of He\(^+\)
Show Answer & Explanation

Answer: (d) the fourth orbit of He\(^+\)

Explanation:
1. \(E\propto Z^2/n^2\)
2. H, n = 2: 1/4
3. He\(^+\), n = 4: 4/16 = 1/4.

Teacher's Note:
1. For spectral lines use 1/lambda = R Z^2 (1/n1^2 - 1/n2^2). The lowest n1 fixes the series (1 Lyman, 2 Balmer, 3 Paschen), and n(n-1)/2 gives the total number of possible lines.

Question 24: Which of the following transitions in hydrogen needs the largest energy input?
(a) n = 3 to n = 5
(b) n = 1 to n = 2
(c) n = 2 to n = 1 (emits)
(d) n = infinity to n = 1 (emits)
Show Answer & Explanation

Answer: (b) n = 1 to n = 2

Explanation:
1. Absorption of 10.2 eV is needed for 1 to 2
2. The 3 to 5 jump needs only 0.97 eV, and the other two release energy.

Teacher's Note:
1. For spectral lines use 1/lambda = R Z^2 (1/n1^2 - 1/n2^2). The lowest n1 fixes the series (1 Lyman, 2 Balmer, 3 Paschen), and n(n-1)/2 gives the total number of possible lines.

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