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The frequency of revolution of the electron in Bohr's orbit varies with n , the principal quantum number as

Solution & Explanation

### Related Formula The orbital frequency of revolution f of an electron is inversely proportional to its time period T: f = frac1T = fracv2pi r In Bohr's Atomic Model: * Velocity v propto fracZn * Radius r propto fracn^2Z ### Core Logic Substitute the proportional relationships of v and r into the frequency expression: f propto fracleft(frac1nright)n^2 implies f propto frac1n^3 ### Step 1: Verification Thus, the frequency varies inversely with the cube of the principal quantum number: f propto frac1n^3. ### Pattern Recognition Remember the sequence of powers of n in Bohr's model: radius expands as n^2, velocity drops as n^-1, angular momentum grows as n^1, and orbital time period or frequency changes as n^3 or n^-3 respectively. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

Reference Study Guides

More Atoms Previous-Year Questions — Page 2

Q20 2025 Hydrogen Spectrum
The number of spectral lines emitted by atomic hydrogen that is in the 4^textth energy level, is:
  • A. 6
  • B. 0
  • C. 3
  • D. 1

Solution

### Related Formula N = fracn(n - 1)2 where n is the principal quantum number of the starting energy level. ### Core Logic For a hydrogen sample initially in the n = 4 level, the possible downward transition pathways to reach the ground state (n=1) are:
Hydrogen Spectrum Transitions diagram for Q20 - JEE Main 2025 Evening
Hydrogen Spectrum Transitions diagram for Q20 - JEE Main 2025 Evening
Using the combination formula for all transitions: N = frac4(4 - 1)2 = frac4 times 32 = 6 Thus, 6 distinct spectral lines are generated. ### Pattern Recognition Think of it as counting combinations of transitions between levels: _nC_2. For n=4, _4C_2 = 6 lines. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms
Q5 2025 Bohr Model of the Hydrogen Atom
During the transition of electron from state A to state C of a Bohr atom, the wavelength of emitted radiation is 2000 Å and it becomes 6000 Å when the electron jumps from state B to state C. Then the wavelength of the radiation emitted during the transition of electrons from state A to state B is :-
  • A. 3000 Å
  • B. 6000 Å
  • C. 4000 Å
  • D. 2000 Å

Solution

### Related Formula The energy of the emitted photon during an atomic transition between energy states is given by: Delta E = frachclambda where h is Planck's constant, c is speed of light, and lambda is the photon wavelength. ### Core Logic Write equations for the energy transitions from the layout of levels
Bohr Model of the Hydrogen Atom diagram for Q5 - JEE Main 2025 Morning
Bohr Model of the Hydrogen Atom diagram for Q5 - JEE Main 2025 Morning
: E_A - E_C = frachclambda_AC = frachc2000text AA quad dots (i) E_B - E_C = frachclambda_BC = frachc6000text AA quad dots (ii) ### Step 1: Finding Transition A to B Subtracting equation (ii) from equation (i) gives the net transition energy from A to B : E_A - E_B = (E_A - E_C) - (E_B - E_C) frachclambda_AB = frachc2000 - frachc6000 frac1lambda_AB = frac3 - 16000 = frac26000 = frac13000 lambda_AB = 3000text AA ### Pattern Recognition Energy differences add linearly, which means their corresponding inverse wavelengths satisfy a parallel reciprocal subtraction rule: frac1lambda_AB = frac1lambda_AC - frac1lambda_BC. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

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