Q33.An electron projected perpendicular to a uniform magnetic field B moves in a circle. If Bohr's quantization is applicable, then the radius of the electronic orbit in the first excited state is : (1) √ πeeBh (2) √2hπeB (3) √ 2πeBh (4) √4πeBh
What This Question Tests
This multi-concept question combines the classical motion of a charged particle in a magnetic field with Bohr's quantization condition for angular momentum to determine the radius of the orbit.
Concepts Tested
Formulas Used
F_magnetic = qvB = mv^2/r
L = mvr = nh/2π
📚 NCERT Sections This Tests
12.6 — (A) Using The Bohr’S Model Calculate The Speed Of The Electron In A
Physics Class 12 · Chapter 12
12.6 (a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels.
12.7 — The Radius Of The Innermost Electron Orbit Of A Hydrogen Atom Is
Physics Class 12 · Chapter 12
12.7 The radius of the innermost electron orbit of a hydrogen atom is 5.3×10–11 m. What are the radii of the n = 2 and n =3 orbits?
4.11 — In A Chamber, A Uniform Magnetic Field Of 6.5 G (1 G = 10–4 T) Is
Physics Class 11 · Chapter 4
4.11 In a chamber, a uniform magnetic field of 6.5 G (1 G = 10–4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 106 m s–1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e = 1.5 × 10–19 C, me = 9.1×10–31 kg)
📋 Question Details
- Chapter
- Atoms
- Topic
- Bohr's Quantization and Circular Motion in Magnetic Field
- Year
- 2025
- Shift
- 22 Jan Shift 2
- Q Number
- Q33
- Type
- MCQ
- NCERT Ref
- Class 12 Physics Ch 4: Moving Charges and Magnetism; Class 12 Physics Ch 12: Atoms
More from this Chapter
Q70.If 13.6eV energy is required to ionize the hydrogen atom, then the energy required to remove an electron from n = 2 is (1) 10.2eV (2) 0eV (3) 3.4eV (4) 6.8eV
Q79.In a hydrogen atom, if energy of an electron in ground state is 13.6 eV, then that in the 2nd excited state is JEE Main 2002 JEE Main Previous Year Paper (1) 1.51 eV (2) 3.4 eV (3) 6.04 eV (4) 13.6 eV
Q69.The wavelengths involved in the spectrum of deuterium (21D) are slightly different from that of hydrogen spectrum, because (1) the size of the two nuclei are different (2) the nuclear forces are different in the two cases (3) the masses of the two nuclei are different (4) the attraction between the electron and the nucleus is different in the two cases
Q70.If the binding energy of the electron in a hydrogen atom is 13.6eV, the energy required to remove the electron from the first excited state of Li++ is (1) 30.6eV (2) 13.6eV (3) 3.4eV (4) 122.4eV