Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The Bohr model is applicable to hydrogen and hydrogen-like atoms only. Reason R: The formulation of Bohr model does not include repulsive force between electrons. In the light of the above statements, choose the correct answer from the options given below :

Solution & Explanation

### Related Formula Bohr's electrostatic centripetal balance equation: fracm v^2r = frac14pivarepsilon_0 fracZ e^2r^2 This basic equation matches ONLY a single electron orbiting a nucleus of charge +Ze. ### Core Logic Assertion Analysis: - Bohr's model matches single-electron species (such as mathrmH, mathrmHe^+, mathrmLi^2+, etc.). Hence, Assertion A is true. Reason Analysis: - The model is restricted because it models ONLY the attractive force between the positive nucleus and one orbiting electron. It cannot handle multi-electron systems due to the presence of inter-electron repulsive forces, which are not integrated into Bohr's simple formulation. Hence, Reason R is true. Connection Check: - The lack of repulsive forces is exactly why the model fails for multi-electron species and remains applicable only to single-electron (hydrogen-like) species. Thus, R is the correct explanation of A. ### Pattern Recognition For single vs. multi-electron systems in atomic physics: Bohr model = single-electron ONLY. Quantum mechanics (Schrodinger) is required for multi-electron systems due to electron-electron interactions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

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More Atoms Previous-Year Questions — Page 4

Q36 jee_main_2024_30_jan_morning Bohr Model and Electron Energy
The ratio of the magnitude of the kinetic energy to the potential energy of an electron in the 5^textth excited state of a hydrogen atom is :
  • A. 4
  • B. frac14
  • C. frac12
  • D. 1

Solution

### Related Formula textKE = -E textPE = 2E textKE = frac12 |textPE| ### Core Logic In any allowed Bohr orbit (for any value of principal quantum number n), the relationship between kinetic energy (KE), potential energy (PE), and total energy (E) strictly obeys the virial theorem for a Coulombic force field. |textPE| = 2 times textKE ### Step 1: Formulate the Ratio The question asks for the ratio of the magnitude of KE to the magnitude of PE. fractextKE|textPE| = frac12 This ratio is independent of the orbit state. Even though it is the 5^textth excited state, the ratio remains frac12. ### Pattern Recognition Energy relationships in Bohr orbits (and planetary motion): K = -E = -U/2. The state number (n) is given solely as a distractor. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms
Q53 jee_main_2024_30_jan_morning Hydrogen Energy Levels and Transitions
A electron of hydrogen atom on an excited state is having energy E_n = -0.85 mathrm~eV. The maximum number of allowed transitions to lower energy level is ....
Numerical Answer. Answer: 6 to 6

Solution

### Related Formula E_n = -frac13.6n^2 mathrm~eV textNumber of transitions = fracn(n - 1)2 ### Core Logic First, identify the principal quantum number n corresponding to the energy -0.85 mathrm~eV. Then, use the combinatorics formula to find the total possible downward emission transitions. ### Step 1: Find Quantum State E_n = -frac13.6n^2 -0.85 = -frac13.6n^2 n^2 = frac13.60.85 = 16 n = 4 ### Step 2: Calculate Transitions Maximum number of transitions from n=4 to lower levels (n=3, 2, 1): = fracn(n - 1)2 = frac4(4 - 1)2 = frac122 = 6 ### Pattern Recognition Energy states in Hydrogen are heavily standardized: n=1 rightarrow -13.6, n=2 rightarrow -3.4, n=3 rightarrow -1.51, n=4 rightarrow -0.85. Recognize -0.85 mathrm~eV as state 4 immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms
Q36 jee_main_2024_31_jan_morning Hydrogen Spectrum
If the wavelength of the first member of Lyman series of hydrogen is lambda. The wavelength of the second member will be
  • A. frac2732 lambda
  • B. frac3227lambda
  • C. frac275 lambda
  • D. frac527lambda

Solution

### Related Formula frac1lambda = R Z^2 left[ frac1n_1^2 - frac1n_2^2 right] ### Core Logic For the first member of the Lyman series of hydrogen (n_1 = 1, n_2 = 2): frac1lambda = frac13.6 Z^2hc left[ frac11^2 - frac12^2 right] frac1lambda = frac13.6 Z^2hc left[ frac34 right] dots dots (texti) ### Step 2: Second Member Calculation For the second member of the Lyman series (n_1 = 1, n_2 = 3): frac1lambda' = frac13.6 Z^2hc left[ frac11^2 - frac13^2 right] frac1lambda' = frac13.6 Z^2hc left[ frac89 right] dots dots (textii) ### Step 3: Ratio On dividing equation (i) by (ii): fraclambda'lambda = frac3/48/9 = frac34 times frac98 fraclambda'lambda = frac2732 lambda' = frac2732 lambda ### Pattern Recognition Rydberg ratios between members of the same series are purely derived from the bracket terms [1/n_1^2 - 1/n_2^2]. For Lyman 1st and 2nd, the ratio is (3/4) / (8/9) = 27/32. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

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