Structure Of Atom MCQs with Answers for CBSE JEE Chemistry

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

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Practice Structure Of Atom MCQs for JEE Chemistry

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Multiple Choice Questions

Question 1: A transition in the spectrum of He\(^+\) that has the same wavelength as the first line of the Lyman series of hydrogen is
(a) n = 4 to n = 2
(b) n = 3 to n = 1
(c) n = 2 to n = 1
(d) n = 3 to n = 2

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Answer: (a) n = 4 to n = 2

Explanation:
1. H first Lyman line: \(1-1/4 = 3/4\)
2. For He\(^+\), \(4(1/n_1^2-1/n_2^2) = 3/4\), so the bracket is 3/16, which is \(1/4-1/16\): 4 to 2.

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 2: The ionisation potential of the hypothetical one-electron atom with ground-state energy \(-15.6\) eV is
(a) 5.3 V
(b) 10.3 V
(c) 20.9 V
(d) 15.6 V

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Answer: (d) 15.6 V

Explanation:
1. The energy needed to remove the electron from n = 1 to infinity is 15.6 eV, which corresponds to 15.6 V.
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 3: The ratio of the de Broglie wavelengths of two electrons accelerated from rest through 50 V and 200 V is
(a) 1 : 2
(b) 3 : 10
(c) 2 : 1
(d) 10 : 3

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Answer: (c) 2 : 1

Explanation:
1. \(\lambda\propto1/\sqrt V\), so \(\lambda_{50}/\lambda_{200} = \sqrt{200/50} = 2\).
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 4: If the radius of the first Bohr orbit of hydrogen is x, the de Broglie wavelength of the electron in the 3rd orbit is
(a) \(3\pi x\)
(b) \(9\pi x/2\)
(c) \(\pi x/2\)
(d) \(6\pi x\)

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Answer: (d) \(6\pi x\)

Explanation:
1. \(r_3 = 9x\) and \(2\pi r_3 = 3\lambda\), so \(\lambda = 2\pi r_3/3 = 2\pi(9x)/3 = 6\pi x\).
2. Only option (d) agrees with this working.

Question 5: The de Broglie wavelength of the electron in the second orbit of Li\(^{2+}\) equals that of the electron in
(a) n = 3 of H
(b) n = 6 of Be\(^{3+}\)
(c) n = 4 of C\(^{5+}\)
(d) n = 3 of He\(^+\)

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Answer: (c) n = 4 of C\(^{5+}\)

Explanation:
1. \(\lambda = 2\pi r/n\propto n/Z\)
2. Li\(^{2+}\) (n = 2): 2/3
3. C\(^{5+}\) (n = 4): 4/6 = 2/3
4. H (3): 3; Be\(^{3+}\) (6): 1.5; He\(^+\) (3): 1.5.

Question 6: The uncertainty in the momentum of an electron is \(1.0\times10^{-5}\) kg m s\(^{-1}\) (\(h = 6.626\times10^{-34}\) J s). The minimum uncertainty in its position is
(a) \(1.05\times10^{-28}\) m
(b) \(5.27\times10^{-30}\) m
(c) \(1.05\times10^{-26}\) m
(d) \(5.25\times10^{-28}\) m

Show Answer & Explanation

Answer: (b) \(5.27\times10^{-30}\) m

Explanation:
1. \(\Delta x = h/(4\pi\Delta p) = 6.626\times10^{-34}/(4\pi\times10^{-5}) = 5.27\times10^{-30}\) m.
2. Only option (b) agrees with this working.

Question 7: The uncertainties in position and velocity of a particle are 0.1 nm and \(5.27\times10^{-24}\) m s\(^{-1}\). Taking the minimum product, its mass (\(h = 6.625\times10^{-34}\) J s) is nearest to
(a) 0.01 kg
(b) 1 kg
(c) \(10^{-3}\) kg
(d) 0.1 kg

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Answer: (d) 0.1 kg

Explanation:
1. \(m = h/(4\pi\,\Delta x\,\Delta v) = 6.625\times10^{-34}/(4\pi\times10^{-10}\times5.27\times10^{-24}) = 0.100\) kg.
2. Only option (d) agrees with this working.

Question 8: For an electron to have the same de Broglie wavelength as a proton (1836 times heavier), the electron must have
(a) the same speed as the proton
(b) a speed 1/1836 that of the proton
(c) a speed 1836 times that of the proton
(d) a kinetic energy equal to the proton's

Show Answer & Explanation

Answer: (c) a speed 1836 times that of the proton

Explanation:
1. \(\lambda = h/mv\) is the same when \(m_ev_e = m_pv_p\)
2. Since \(m_p = 1836m_e\), \(v_e = 1836v_p\).

Question 9: A particle Y has 25% of the mass and 75% of the speed of particle X, which has de Broglie wavelength 1 A. The de Broglie wavelength of Y is nearest to
(a) 3 A
(b) 5.33 A
(c) 6.88 A
(d) 48 A

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Answer: (b) 5.33 A

Explanation:
1. \(\lambda\propto1/mv\), so \(\lambda_Y = 1/(0.25\times0.75) = 5.33\) A.
2. Only option (b) agrees with this working.

Question 10: The uncertainty in position of a particle is twice the uncertainty in its momentum. The minimum uncertainty in its velocity is
(a) \(\dfrac{1}{m}\sqrt{\dfrac{h}{2\pi}}\)
(b) \(\dfrac{1}{2m}\sqrt{\dfrac{h}{\pi}}\)
(c) \(\dfrac{1}{2m}\sqrt{\dfrac{h}{2\pi}}\)
(d) \(\dfrac{h}{4\pi m}\)

Show Answer & Explanation

Answer: (c) \(\dfrac{1}{2m}\sqrt{\dfrac{h}{2\pi}}\)

Explanation:
1. \(\Delta x = 2\Delta p\) and \(\Delta x\Delta p = h/4\pi\) give \(2(\Delta p)^2 = h/4\pi\), \(\Delta p = \sqrt{h/8\pi} = \tfrac12\sqrt{h/2\pi}\)
2. Divide by m.

Question 11: If the uncertainty in the position of an electron moving at 300 m/s is to be accurate to 0.001% of the speed, the uncertainty in position is nearest to (\(h/4\pi m = 5.8\times10^{-5}\) m\(^2\) s\(^{-1}\) for the electron)
(a) \(5.76\times10^{-2}\) m
(b) \(1.92\times10^{-2}\) m
(c) \(3.84\times10^{-2}\) m
(d) \(19.2\times10^{-2}\) m

Show Answer & Explanation

Answer: (b) \(1.92\times10^{-2}\) m

Explanation:
1. \(\Delta v = 10^{-5}\times300 = 3\times10^{-3}\) m/s
2. \(\Delta x = 5.8\times10^{-5}/3\times10^{-3} = 1.93\times10^{-2}\) m.

Question 12: The de Broglie wavelength of a tennis ball of mass 60 g moving at 10 m/s is of the order of
(a) \(10^{-33}\) m
(b) \(10^{-31}\) m
(c) \(10^{-16}\) m
(d) \(10^{-25}\) m

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Answer: (a) \(10^{-33}\) m

Explanation:
1. \(\lambda = 6.63\times10^{-34}/(0.06\times10) = 1.1\times10^{-33}\) m.
2. Only option (a) agrees with this working.

Question 13: The kinetic energy of an electron whose de Broglie wavelength is 10 nm is nearest to
(a) \(4.8\times10^{-21}\) J
(b) \(2.4\times10^{-29}\) J
(c) \(4.8\times10^{-29}\) J
(d) \(2.4\times10^{-21}\) J

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Answer: (d) \(2.4\times10^{-21}\) J

Explanation:
1. \(KE = h^2/2m\lambda^2 = (6.626\times10^{-34})^2/(2\times9.1\times10^{-31}\times10^{-16}) = 2.4\times10^{-21}\) J.
2. Only option (d) agrees with this working.

Question 14: A moving body of mass 0.1 mg has de Broglie wavelength \(3.31\times10^{-29}\) m. Its kinetic energy is nearest to
(a) \(2.0\times10^{-6}\) J
(b) \(2.0\times10^{-3}\) J
(c) \(1.0\times10^{-3}\) J
(d) \(4.0\times10^{-3}\) J

Show Answer & Explanation

Answer: (b) \(2.0\times10^{-3}\) J

Explanation:
1. \(v = h/m\lambda = 6.626\times10^{-34}/(10^{-7}\times3.31\times10^{-29}) = 200\) m/s
2. \(KE = \tfrac12\times10^{-7}\times200^2 = 2\times10^{-3}\) J.

Question 15: The speed of an electron in the 2nd Bohr orbit of an atom is \(1.1\times10^{6}\) m/s. Its speed in the 3rd orbit is
(a) \(7.33\times10^{5}\) m/s
(b) \(3.3\times10^{6}\) m/s
(c) \(2.2\times10^{6}\) m/s
(d) \(3.67\times10^{5}\) m/s

Show Answer & Explanation

Answer: (a) \(7.33\times10^{5}\) m/s

Explanation:
1. \(v\propto1/n\), so \(v_3 = 1.1\times10^{6}\times2/3 = 7.33\times10^{5}\) m/s.
2. Only option (a) agrees with this working.

Question 16: An electron falls from a level where its de Broglie wavelength is \(\lambda_1\) to a level where it is \(\lambda_2\), in a hydrogen-like atom. The wavelength of the photon emitted is
(a) \(\dfrac{4mc}{h}\dfrac{\lambda_1^2-\lambda_2^2}{\lambda_1\lambda_2}\)
(b) \(\dfrac{h}{2mc}\dfrac{\lambda_1^2\lambda_2^2}{\lambda_1^2-\lambda_2^2}\)
(c) \(\dfrac{2mc}{h}(\lambda_1^2-\lambda_2^2)\)
(d) \(\dfrac{2mc}{h}\dfrac{\lambda_1^2\lambda_2^2}{\lambda_1^2-\lambda_2^2}\)

Show Answer & Explanation

Answer: (d) \(\dfrac{2mc}{h}\dfrac{\lambda_1^2\lambda_2^2}{\lambda_1^2-\lambda_2^2}\)

Explanation:
1. KE \(= h^2/2m\lambda^2\) and \(\Delta E\) equals the KE change (since \(E = -KE\)): \(\Delta E = (h^2/2m)(1/\lambda_2^2-1/\lambda_1^2)\) for the lower level having the smaller \(\lambda\)
2. Then \(\lambda_{ph} = hc/\Delta E = \dfrac{2mc}{h}\dfrac{\lambda_1^2\lambda_2^2}{\lambda_1^2-\lambda_2^2}\).

Question 17: The correct time-independent Schrodinger equation for an electron (total energy E, potential energy V) is
(a) \(\dfrac{\partial^2\psi}{\partial x^2}+\dfrac{\partial^2\psi}{\partial y^2}+\dfrac{\partial^2\psi}{\partial z^2}+\dfrac{8\pi^2m}{h^2}(E-V)\psi = 0\)
(b) \(\dfrac{\partial^2\psi}{\partial x^2}+\dfrac{\partial^2\psi}{\partial y^2}+\dfrac{\partial^2\psi}{\partial z^2}+\dfrac{8\pi^2m}{h}(E-V)\psi = 0\)
(c) \(\dfrac{\partial^2\psi}{\partial x^2}+\dfrac{\partial^2\psi}{\partial y^2}+\dfrac{\partial^2\psi}{\partial z^2}+\dfrac{8\pi^2h^2}{m}(E-V)\psi = 0\)
(d) \(\dfrac{\partial^2\psi}{\partial x^2}+\dfrac{\partial^2\psi}{\partial y^2}+\dfrac{\partial^2\psi}{\partial z^2}-\dfrac{8\pi^2m}{h^2}(E-V)\psi = 0\)

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Answer: (a) \(\dfrac{\partial^2\psi}{\partial x^2}+\dfrac{\partial^2\psi}{\partial y^2}+\dfrac{\partial^2\psi}{\partial z^2}+\dfrac{8\pi^2m}{h^2}(E-V)\psi = 0\)

Explanation:
1. The standard form is \(\nabla^2\psi+\dfrac{8\pi^2m}{h^2}(E-V)\psi = 0\); the factor is \(8\pi^2m/h^2\) with a plus sign.
2. Only option (a) agrees with this working.