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Given below are two statements: Statement I: Most of the mass of the atom and all its positive charge are concentrated in a tiny nucleus and the electrons revolve around it, is Rutherford's model. Statement II: An atom is a spherical cloud of positive charges with electrons embedded in it, is a special case of Rutherford's model. In the light of the above statements, choose the most appropriate from the options given below:

Solution & Explanation

### Core Logic * **Statement I**: According to Rutherford's planetary model of the atom, almost the entire mass of an atom and all of its positive charge are concentrated in a centrally located small volume called the nucleus, with electrons revolving around it. Thus, Statement I is **true**. * **Statement II**: Thomson's model of the atom (plum-pudding model) describes the atom as a spherical cloud of positive charge with electrons embedded in it. This model was proposed before Rutherford's nuclear model and is not a special case of Rutherford's model. Thus, Statement II is **false**. ### Step 1: Conclusion Therefore, Statement I is true but Statement II is false. ### Pattern Recognition Understand the evolutionary timeline of atomic models: 1. Thomson's Plum Pudding Model (spherical cloud with embedded electrons) 2. Rutherford's Nuclear Model (planetary orbital model with tiny centralized nucleus) They are conceptually distinct, making Statement II false. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

Reference Study Guides

Atoms

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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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