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BITSAT/BITSAT Physics: Atoms Questions
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Question: Hydrogen (H), deuterium (D), singly ionized helium (He+) and doubly ionized lithium (Li++) all have one electron around the nucleus. Consider n = 2 to n = 1 transition. The wavelengths of emitted radiations are λ1, λ2, λ3 and λ4 respectively. Then approximately:
- a) λ1 = λ2 = 4 λ3 = 9 λ4
- b) 4 λ1 = 2λ2 = 2 λ3 = λ4
- c) λ1 = 2 λ2 = 2√2 λ3 = 3√2 λ4
- d) λ1 = λ2 = 2 λ3 = 3√2 λ4
Answer: λ1 = λ2 = 4 λ3 = 9 λ4
Question: If the series limit wavelength of Lyman series for the hydrogen atom is 912 Å, then the series limit wavelength for Balmer series of hydrogen atoms is
- a) 912 Å
- b) 912 × 2 Å
- c) 912 × 4 Å
- d)
Answer: 912 × 4 Å
Question: One of the lines in the emission spectrum of Li2+ has the same wavelength as that of the 2nd line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is n = 12→ n = x. Find the value of x.
- a) 8
- b) 6
- c) 7
- d) 5
Answer: 6
Question: Energy required for the electron excitation in Li++ from the first to the third Bohr orbit is
- a) 36.3 eV
- b) 108.8 eV
- c) 122.4 eV
- d) 12.1 eV
Answer: 108.8 eV
Question: The angular momentum of electron in nth orbit is given by
- a) nh
- b)
- c)
- d)
Answer:
Question: The energy of electron in the nth orbit of hydrogen atom is expressed as 
The shortest and longest wavelength of Lyman series will be
- a) 910 Å, 1213 Å
- b) 5463 Å, 7858 Å
- c) 1315 Å, 1530 Å
- d) None of these
Answer: 910 Å, 1213 Å
Question: If the Kα radiation of Mo (Z = 42) has a wavelength of 0.71 Å, calculate wavelength of the corresponding radiation of Cu, i.e., Kα for Cu (Z = 29) assuming σ= 1.
- a) 1.52 Å
- b) 5.14 Å
- c) 3.02 Å
- d) 0.52 Å
Answer: 1.52 Å
Question: The energies of energy levels A, B and C for a given atom are in the sequence EA < EB < EC. If the radiations of wavelengths λ1, λ2 and λ3 are emitted due to the atomic transitions C to B, B to A and C to A respectively then which of the following relations is correct ?
- a) λ1 + λ2 +λ3 = 0
- b) λ3 = λ1 2 + λ2
- c) λ3 = λ1 + λ2
- d)
Answer:
Question: The third line of Balmer series of an ion equivalnet to hydrogen atom has wavelength of 108.5 nm. The ground state energy of an electron of this ion will be
- a) 3.4 eV
- b) 13.6 eV
- c) 54.4 eV
- d) 122.4 eV
Answer: 54.4 eV
Question: Hydrogen atom in ground state is excited by a monochromatic radiation of λ = 975 Å. Number of spectral lines in the resulting spectrum emitted will be
- a) 3
- b) 2
- c) 6
- d) 10
Answer: 6
Question: Taking Rydberg’s constant RH = 1.097 × 107m, first and second wavelength of Balmer series in hydrogen spectrum is
- a) 2000 Å, 3000 Å
- b) 1575 Å, 2960 Å
- c) 6529 Å, 4280 Å
- d) 6552 Å, 4863 Å
Answer: 6552 Å, 4863 Å
Question: In which of the following series, does the 121.5nm line of the spectrum of the hydrogen atom lie?
- a) Lyman series
- b) Balmer series
- c) Paschen series
- d) Brackett series.
Answer: Lyman series
Question: The ionisation potential of H-atom is 13.6 V. When it is excited from ground state by monochromatic radiations of 970.6 Å, the number of emission lines will be (according to Bohr’s theory)
- a) 10
- b) 8
- c) 6
- d) 4
Answer: 6
Question: The angular momentum of an electron in first orbit of Li++ ion is –
- a)
- b)
- c)
- d)
Answer:
Question: The de Broglie wavelength of the electron in the first Bohr orbit of the hydrogen atom is
- a) Equal to the diameter of the first orbit
- b) Equal to the circumference of the first orbit
- c) Equal to half the circumference of the first orbit
- d) Independent of the size of the first orbit
Answer: Equal to the circumference of the first orbit
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Practice MCQs: Atoms (BITSAT)
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Core Objective Practice Sets for Atoms
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You can get most exhaustive BITSAT Physics Atoms MCQs for free on StudiesToday.com. These MCQs for BITSAT Physics are updated for the 2026-27 academic session as per BITSAT examination standards.
Yes, our BITSAT Physics Atoms MCQs include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the BITSAT paper is now competency-based.
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