BITSAT Physics Atoms MCQs

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 2^nd 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 E_A < E_B < E_C. 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 R_H = 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