Q25.If the maximum value of accelerating potential provided by a radio frequency oscillator is 12 kV . The number of revolution made by a proton in a cyclotron to achieve one sixth of the speed of light is: [mp = 1. 67 × 10−27 kg, e = 1. 6 × 10−19 C, Speed of light = 3 × 108 m s−1]
What This Question Tests
This question combines the kinetic energy calculation with the concept of energy gain per revolution in a cyclotron, requiring multiple steps and careful unit conversion.
Concepts Tested
Formulas Used
KE = ½ mv²
Energy gain per revolution = 2qV
N = KE / (2qV)
📚 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.
12.6 — (A) 2.18 × 106 M/S; 1.09 × 106 M/S; 7.27 × 105 M/S
Physics Class 12 · Chapter 12
12.6 (a) 2.18 × 106 m/s; 1.09 × 106 m/s; 7.27 × 105 m/s (b) 1.52 × 10–16 s; 1.22 × 10–15 s; 4.11 × 10–15 s.
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
- Magnetic Effects of Current
- Topic
- Cyclotron
- Year
- 2021
- Shift
- 26 Aug Shift 2
- Q Number
- Q25
- Type
- Numerical
- NCERT Ref
- Class 12 Physics Ch 4: Magnetic Effects of Current
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