JEE Chemistry Structure Of Atom MCQs Set 07

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Multiple Choice Questions

Question 1: The velocity of the electron in the first orbit of hydrogen is V. The velocity of the electron in the 3rd orbit of Li\(^{2+}\) is
(a) V/3
(b) V
(c) 3V
(d) 9V
Show Answer & Explanation

Answer: (b) V

Explanation:
1. \(v\propto Z/n\)
2. For H(n = 1) it is 1; for Li\(^{2+}\)(Z = 3, n = 3) it is 1
3. So the speed is V.

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 2: Which one of the following statements about hydrogen-like species is NOT correct?
(a) Potential energy of two opposite charges increases when they come closer
(b) Kinetic energy of the electron increases when it moves to a lower orbit
(c) Potential energy of the electron increases when it moves to a higher orbit
(d) An 11 eV photon cannot free the electron from the first excited state of He\(^+\)
Show Answer & Explanation

Answer: (a) Potential energy of two opposite charges increases when they come closer

Explanation:
1. Potential energy of opposite charges \(-kq_1q_2/r\) becomes more negative (decreases) as they approach, so the first statement is false
2. The other three are true: KE \(=-E\) rises as n falls; PE \(=2E\) rises with n; and the binding energy of n = 2 in He\(^+\) is 13.6 eV, more than 11 eV.

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 3: The ratio of the period of revolution of an electron in the n = 1 orbit to that in the n = 2 orbit of hydrogen is
(a) 1 : 2
(b) 1 : 4
(c) 2 : 1
(d) 1 : 8
Show Answer & Explanation

Answer: (d) 1 : 8

Explanation:
1. \(T\propto n^3\), so \(T_1:T_2 = 1:8\).
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 4: Two electrons in an atom move in circular Bohr orbits of radii R and 4R. The ratio of their periods of revolution is
(a) 1 : 4
(b) 1 : 8
(c) 4 : 1
(d) 8 : 1
Show Answer & Explanation

Answer: (b) 1 : 8

Explanation:
1. \(r\propto n^2\), so \(n = 1\) and \(n = 2\)
2. \(T\propto n^3\) gives 1 : 8.

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 5: The kinetic energy of the electron in the ground state of Li\(^{2+}\) is (r = radius of the orbit)
(a) \(+\dfrac{3e^2}{8\pi\varepsilon_0r}\)
(b) \(-\dfrac{3e^2}{8\pi\varepsilon_0r}\)
(c) \(+\dfrac{3e^2}{4\pi\varepsilon_0r}\)
(d) \(-\dfrac{3e^2}{4\pi\varepsilon_0r}\)
Show Answer & Explanation

Answer: (a) \(+\dfrac{3e^2}{8\pi\varepsilon_0r}\)

Explanation:
1. For nuclear charge Ze, \(KE = Ze^2/(8\pi\varepsilon_0r)\), positive
2. With Z = 3 this is \(3e^2/8\pi\varepsilon_0r\).

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: A hydrogen sample in excited state A absorbs photons of energy 2.55 eV and moves some electrons to state B. The orbit numbers of A and B are
(a) 1 and 3
(b) 3 and 5
(c) 2 and 5
(d) 2 and 4
Show Answer & Explanation

Answer: (d) 2 and 4

Explanation:
1. \(E_n = -13.6/n^2\): \(E_2 = -3.4\) and \(E_4 = -0.85\) eV, difference 2.55 eV. No other pair of the given levels has a gap of 2.55 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 7: The excitation energy of the first excited state of a hydrogen-like ion is 40.8 eV. The energy needed to remove the electron from the ground state is
(a) 54.4 eV
(b) 40.8 eV
(c) 13.6 eV
(d) 122.4 eV
Show Answer & Explanation

Answer: (a) 54.4 eV

Explanation:
1. \(\Delta E_{1\to2} = \tfrac34\times IE = 40.8\) eV, so \(IE = 54.4\) eV (this is He\(^+\), Z = 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 8: For the hydrogen atom, an electron jump 3 to 1 (x), 4 to 2 (y) and 5 to 3 (z) is considered. The photon of shortest wavelength is emitted in
(a) y
(b) z
(c) all are equal
(d) x
Show Answer & Explanation

Answer: (d) x

Explanation:
1. Energy gaps: x \(= 1-1/9 = 0.889\), y \(= 1/4-1/16 = 0.1875\), z \(= 1/9-1/25 = 0.071\) (in units of 13.6 eV)
2. The largest gap gives the shortest wavelength.

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: If the first ionisation potential of a hypothetical H-like atom is 16 V, the first excitation potential is
(a) 8 V
(b) 16 V
(c) 12 V
(d) 4 V
Show Answer & Explanation

Answer: (c) 12 V

Explanation:
1. First excitation energy \(= \tfrac34\times\) ionisation energy \(= 0.75\times16 = 12\) V.
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 10: An H-like ion has an orbit with a diameter of about 16.92 A in hydrogen. The maximum number of electrons that can be accommodated in that shell is
(a) 32
(b) 18
(c) 8
(d) 50
Show Answer & Explanation

Answer: (a) 32

Explanation:
1. \(r = 8.46\) A \(= 0.529n^2\), so \(n^2 = 16\) and n = 4
2. The shell holds \(2n^2 = 32\) electrons.

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: When an electron falls from level (n + 2) to level n in He\(^+\), the photon emitted has energy \(6.172\times10^{-19}\) J. The value of n is
(a) 2
(b) 4
(c) 5
(d) 3
Show Answer & Explanation

Answer: (d) 3

Explanation:
1. Energy \(= 3.85\) eV \(= 54.4(1/n^2-1/(n+2)^2)\), so the bracket is about 0.0708
2. For n = 3, \(1/9-1/25 = 0.0711\); n = 2 gives 0.188 and n = 4 gives 0.0347.

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 two longest wavelengths emitted when hydrogen atoms fall to n = 2 are nearest to
(a) 656 nm and 410 nm
(b) 486 nm and 434 nm
(c) 656 nm and 486 nm
(d) 122 nm and 103 nm
Show Answer & Explanation

Answer: (c) 656 nm and 486 nm

Explanation:
1. \(1/\lambda = R(1/4-1/9)\) gives 656 nm (3 to 2) and \(R(1/4-1/16)\) gives 486 nm (4 to 2)
2. These Balmer lines are the two longest.

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 13: A transition from n = 2 to n = 1 in a hydrogen-like ion emits a photon of wavelength \(3\times10^{-8}\) m. The atomic number of the ion is
(a) 3
(b) 2
(c) 4
(d) 1
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. \(1/\lambda = R_HZ^2(1-1/4)\), so \(Z^2 = 1/(3\times10^{-8}\times1.097\times10^{7}\times0.75) = 4.05 \approx 4\) and Z = 2 (He\(^+\)).
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 14: Three separate samples are taken: H atoms all in the first excited state, He\(^+\) ions all in the third excited state and Li\(^{2+}\) ions all in the fifth excited state. The total number of lines that the three samples produce (counting each sample separately) as the electrons return to the ground state is
(a) 21
(b) 28
(c) 22
(d) 36
Show Answer & Explanation

Answer: (c) 22

Explanation:
1. H, n = 2: \(2\times1/2 = 1\) line
2. He\(^+\), n = 4: \(4\times3/2 = 6\) lines
3. Li\(^{2+}\), n = 6: \(6\times5/2 = 15\) lines
4. Total \(1+6+15 = 22\).

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 15: The wavelength of a spectral line for an electronic transition is inversely proportional to
(a) the number of electrons making the transition
(b) the energy difference of the transition
(c) the nuclear charge only
(d) the velocity of the electron
Show Answer & Explanation

Answer: (b) the energy difference of the transition

Explanation:
1. \(\lambda = hc/\Delta E\), so \(\lambda\) varies inversely with the energy difference between the two levels.
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 16: If an electron in hydrogen can occupy n orbits, the total number of lines in the Lyman series is
(a) n - 1
(b) n
(c) n - 2
(d) n(n + 1)
Show Answer & Explanation

Answer: (a) n - 1

Explanation:
1. Each Lyman line ends at n = 1 and starts from one of the other n - 1 levels.
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 17: The number of visible lines (Balmer) when an electron returns from the 5th orbit to the ground state in hydrogen is
(a) 4
(b) 5
(c) 3
(d) 10
Show Answer & Explanation

Answer: (c) 3

Explanation:
1. Visible lines end at n = 2: 5-2, 4-2 and 3-2
2. That is 3 lines.

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: If the shortest wavelength in the Lyman series of H is x, the wavelength of the first line of the Balmer series of H is
(a) 9x/5
(b) 36x/5
(c) 5x/9
(d) 5x/36
Show Answer & Explanation

Answer: (b) 36x/5

Explanation:
1. Lyman limit: \(1/x = R\)
2. First Balmer line: \(1/\lambda = R(1/4-1/9) = 5R/36\), so \(\lambda = 36/5R = 36x/5\).

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: In a hydrogen sample, electrons de-excite from level n to level 1, and the total number of Balmer lines is two. The electrons are ionised from level n by photons of 13 eV. The kinetic energy of the ejected electrons is
(a) 12.15 eV
(b) 11.49 eV
(c) 12.46 eV
(d) 12.63 eV
Show Answer & Explanation

Answer: (a) 12.15 eV

Explanation:
1. Two Balmer lines (n to 2 and one more) means the highest level is n = 4 (4-2 and 3-2)
2. Binding energy \(= 13.6/16 = 0.85\) eV, so \(KE = 13-0.85 = 12.15\) eV.

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: The wave number of the first line of the Paschen series of Be\(^{3+}\) is
(a) \(7R/144\)
(b) \(7R/9\)
(c) \(7R/16\)
(d) \(R/144\)
Show Answer & Explanation

Answer: (b) \(7R/9\)

Explanation:
1. For Be\(^{3+}\) (Z = 4): \(\bar\nu = 16R(1/9-1/16) = 16R\times7/144 = 7R/9\).
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 Balmer lines of He\(^+\) (n to 2 transitions) are expected to lie in
(a) the ultraviolet region, because the energies are four times those in hydrogen
(b) the visible region, as for hydrogen
(c) the infrared region
(d) the X-ray region only
Show Answer & Explanation

Answer: (a) the ultraviolet region, because the energies are four times those in hydrogen

Explanation:
1. For He\(^+\), \(\bar\nu = 4R(1/4-1/n^2)\), four times as large as the hydrogen Balmer value
2. The first line \(5R/9\) has \(\lambda \approx 164\) nm, which is ultraviolet.

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: The wavelength of a spectral line in hydrogen corresponds to an electron falling from the tenth to the fifth level (\(R_H = 1.10\times10^{5}\) cm\(^{-1}\)). The line lies in the
(a) ultraviolet region
(b) visible region
(c) microwave region
(d) infrared region
Show Answer & Explanation

Answer: (d) infrared region

Explanation:
1. \(\bar\nu = 1.10\times10^{5}(1/25-1/100) = 3300\) cm\(^{-1}\), so \(\lambda = 3.03\times10^{-4}\) cm \(\approx 3030\) nm
2. This is infrared.

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: Hydrogen atoms in a sample jump from level x to a lower excited level and give a spectrum of only one series. One of the highest-energy photons has energy 0.6375 eV (which is 0.75 of 0.85 eV). The value of x is
(a) 6
(b) 8
(c) 5
(d) 7
Show Answer & Explanation

Answer: (b) 8

Explanation:
1. The series ends at n = 4 (\(E_4 = -0.85\) eV)
2. Highest energy photon from level x: \(0.85-13.6/x^2 = 0.6375\), so \(13.6/x^2 = 0.2125\) and \(x^2 = 64\), x = 8.

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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