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NCERT Solutions for class 12 Physics | Chapter 12 - Atoms

Free download in PDF Class 12 Physics Chapter 12 Atoms Multiple Choice Questions and Answers for Board, JEE, NEET, AIIMS, JIPMER, IIT-JEE, AIEE and other competitive exams. MCQ Questions for Class 12 Physics with Answers were prepared based on the latest exam pattern. These short solved questions or quizzes are provided by Gkseries. These will help the students for preparation of their examination./p>

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(1) In the ground state in …A… electrons are in stable equilibrium while in …B… electrons always experiences a net force. Here, A and B refer to
[A] Dalton’s theory, Rutherford model
[B] Rutherford’s model, Bohr’s model
[C] Thomson’s model, Rutherford’s model
[D] Rutherford’s model, Thomson’s model
Answer: Thomson’s model, Rutherford’s model
(2) The angular momentum of the electron in hydrogen atom in the ground state is
[A] 2h
[B] h/2
[C] h/2π
[D] h/4π
Answer: h/2π
(3) As the quantum number increases, the difference of energy between consecutive energy levels
[A] remain the same
[B] increases
[C] decreases
[D] sometimes increases and sometimes decreases.
Answer: decreases
(4) In a Rutherford scattering experiment when a projectile of charge Z1 and mass M1 approaches a target nucleus of charge Z2 and mass M2, the distance of closest approach is r0. The energy of the projectile is
[A] directly proportional to Z1 Z2
[B] inversely proportional to Z1
[C] directly proportional to mass M1
[D] directly proportional to M1 × M2
Answer: directly proportional to Z1 Z2
(5) Rutherford’s α-particle experiment showed that the atoms have
[A] Proton
[B] Nucleus
[C] Neutron
[D] Electrons
Answer: Nucleus
(6) The Rutherford α-particle experiment shows that most of the α-particles pass through almost unscattered while some are scattered through large angles. What information does
[A] Atom is hollow.
[B] The whole mass of the atom is concentrated in a small centre called nucleus
[C] Nucleus is positively charged
[D] All of the above
Answer: All of the above
(7) According to Bohr’s model of hydrogen atom
[A] the linear velocity of the electron is quantised.
[B] the angular velocity of the electron is quantised.
[C] the linear momentum of the electron is quantised.
[D] the angular momentum of the electron is quantised.
Answer: the angular momentum of the electron is quantised.
(8) According to the Bohr theory of H-atom, the speed of the electron, its energy and the radius of its orbit varies with the principal quantum number n, respectively, as
[A] 1/n, n2, 1/n2
[B] n, 1/n2, n2
[C] n, 1/n2, 1/n2
[D] 1/n, 1/n2, 1/n2
Answer: 1/n, 1/n2, 1/n2
(9) The first spectral series was disscovered by
[A] Balmer
[B] Lyman
[C] Paschen
[D] Pfund
Answer: Balmer
(10) When an electron jumps from the fourth orbit to the second orbit, one gets the
[A] second line of Paschen series
[B] second line of Balmer series
[C] first line of Pfund series
[D] second line of Lyman series
Answer: second line of Balmer series
(11) The Balmer series for the H-atom can be observed
[A] if we measure the frequencies of light emitted when an excited atom falls to the ground state
[B] if we measure the frequencies of light emitted due to transitions between excited states and the first excited state
[C] in any transition in a H-atom
[D] None of these
Answer: if we measure the frequencies of light emitted due to transitions between excited states and the first excited state
(12) In Balmer series of emission spectrum of hydrogen, first four lines with different wavelength Hα Hβ Hγ and Hδ are obtained. Which line has maximum frequency out of these?
[A] Hα 
[B] Hβ 
[C] Hγ
[D] Hδ
Answer: Hδ
(13) In which of the following series, does the 121.5 nm line of the spectrum of the hydrogen atom lie ?
[A] Lyman series
[B] Balmer series
[C] Paschen series
[D] Brackett series.
Answer: Lyman series
(14) Which of the following series in the spectrum of hydrogen atom lies in the visible region of the electromagnetic spectrum?
[A] Paschen series
[B] Balmer series
[C] Lyman series
[D] Brackett series
Answer: Balmer series
(15) The shortest wavelength in Balmer’s series for Hydrogen atom is …A… and this is obtained by substituting …B … in Balmer’s formula. Here, A and B refer to
[A] 656.3 nm, n = 3
[B] 486.1 nm, n = 4
[C] 410.2 nm, n = 5
[D] 364.6 nm, n = ∞
Answer: 364.6 nm, n = ∞
(16) As an electron makes a transition from an excited state to the ground state of a hydrogen – like atom/ion
[A] kinetic energy decreases, potential energy increases but total energy remains same
[B] kinetic energy and total energy decrease but potential energy increases
[C] its kinetic energy increases but potential energy and total energy decrease
[D] kinetic energy, potential energy and total energy decrease
Answer: its kinetic energy increases but potential energy and total energy decrease
(17) In a hydrogen atom, which of the following electronic transitions would involve the maximum energy change
[A] n = 2 to n = 1
[B] n = 3 to n = 1
[C] n = 4 to n = 2
[D] n = 3 to n = 2
Answer: n = 3 to n = 1
(18) Hydrogen atom excites energy level from fundamental state to n = 3. Number of spectral lines according to Bohr, is
[A] 4
[B] 3
[C] 1
[D] 2
Answer: 3
(19) The transition from the state n = 4 to n = 3 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from
[A] 2 → 1
[B] 3 → 2
[C] 4 → 2
[D] 5 → 4
Answer: 5 → 4
(20) For a given value of n, the number of electrons in an orbit is
[A] n
[B] n2
[C] 2n2
[D] 2n
Answer: 2n2

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