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:
A.Both statement I and statement II are false.
B.Statement I is false but statement II is true
C.Statement I is true but statement II is false.
D.Both statement I and statement II are true.
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
### Related Formula
N = fracn(n - 1)2$N = \frac{n(n - 1)}{2}$
where n$n$ is the principal quantum number of the starting energy level.
### Core Logic
For a hydrogen sample initially in the n = 4$n = 4$ level, the possible downward transition pathways to reach the ground state (n=1$n=1$) are:
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$N = \frac{4(4 - 1)}{2} = \frac{4 \times 3}{2} = 6$
Thus, 6 distinct spectral lines are generated.
### Pattern Recognition
Think of it as counting combinations of transitions between levels: _nC_2$_nC_2$. For n=4$n=4$, _4C_2 = 6$_4C_2 = 6$ lines.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Atoms
Q5jee_main_2025_24_jan_morningBohr 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 $\Delta E = \frac{hc}{\lambda} $
where h$h$ is Planck's constant, c$c$ is speed of light, and lambda$\lambda$ 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 :
E_A - E_C = frachclambda_AC = frachc2000text AA quad dots (i)$E_{A} - E_{C} = \frac{hc}{\lambda_{AC}} = \frac{hc}{2000\text{ \AA}} \quad \dots (i)$E_B - E_C = frachclambda_BC = frachc6000text AA quad dots (ii)$E_{B} - E_{C} = \frac{hc}{\lambda_{BC}} = \frac{hc}{6000\text{ \AA}} \quad \dots (ii)$
### Step 1: Finding Transition A to B
Subtracting equation (ii)$(ii)$ from equation (i)$(i)$ gives the net transition energy from A to B :
E_A - E_B = (E_A - E_C) - (E_B - E_C)$E_{A} - E_{B} = (E_{A} - E_{C}) - (E_{B} - E_{C})$frachclambda_AB = frachc2000 - frachc6000$\frac{hc}{\lambda_{AB}} = \frac{hc}{2000} - \frac{hc}{6000}$frac1lambda_AB = frac3 - 16000 = frac26000 = frac13000$\frac{1}{\lambda_{AB}} = \frac{3 - 1}{6000} = \frac{2}{6000} = \frac{1}{3000}$lambda_AB = 3000text AA$\lambda_{AB} = 3000\text{ \AA}$
### Pattern Recognition
Energy differences add linearly, which means their corresponding inverse wavelengths satisfy a parallel reciprocal subtraction rule: frac1lambda_AB = frac1lambda_AC - frac1lambda_BC$\frac{1}{\lambda_{AB}} = \frac{1}{\lambda_{AC}} - \frac{1}{\lambda_{BC}}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Atoms
Q7jee_main_2025_28_jan_eveningBohr Model
The frequency of revolution of the electron in Bohr's orbit varies with n$n$ , the principal quantum number as
A.frac1n$\frac{1}{n}$
B.frac1mathrmn^3$\frac{1}{\mathrm{n}^3}$
C.frac1mathrmn^4$\frac{1}{\mathrm{n}^4}$
D.frac1n^2$\frac{1}{n^2}$
Solution
### Related Formula
The orbital frequency of revolution f$f$ of an electron is inversely proportional to its time period T$T$:
f = frac1T = fracv2pi r$f = \frac{1}{T} = \frac{v}{2\pi r}$
In Bohr's Atomic Model:
* Velocity v propto fracZn$v \propto \frac{Z}{n}$
* Radius r propto fracn^2Z$r \propto \frac{n^2}{Z}$
### Core Logic
Substitute the proportional relationships of v$v$ and r$r$ into the frequency expression:
f propto fracleft(frac1nright)n^2 implies f propto frac1n^3$f \propto \frac{\left(\frac{1}{n}\right)}{n^2} \implies f \propto \frac{1}{n^3}$
### Step 1: Verification
Thus, the frequency varies inversely with the cube of the principal quantum number: f propto frac1n^3$f \propto \frac{1}{n^3}$.
### Pattern Recognition
Remember the sequence of powers of n$n$ in Bohr's model: radius expands as n^2$n^2$, velocity drops as n^-1$n^{-1}$, angular momentum grows as n^1$n^1$, and orbital time period or frequency changes as n^3$n^3$ or n^-3$n^{-3}$ respectively.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Atoms
The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly:
A. 1.5 eV
B. 13.6 eV
C. 1.9 eV
D. 12.1 eV
Solution
### Related Formula
Bohr state energy level values:
E_n = -frac13.6n^2mathrm~eV$E_n = -\frac{13.6}{n^2}\mathrm{~eV}$
Transition excitation requirement:
Delta E = E_textfinal - E_textinitial$\Delta E = E_{\text{final}} - E_{\text{initial}}$
### Core Logic
To emit radiation in the Balmer series, the hydrogen electron must first be excited to at least the n=3$n=3$ shell. This allows it to jump down to n=2$n=2$ and produce the first spectral line of the Balmer series.
Energy required to transition from ground state (n=1$n=1$) to n=3$n=3$:
E_1 = -13.6mathrm~eV$E_1 = -13.6\mathrm{~eV}$E_3 = -frac13.63^2 = -1.51mathrm~eV$E_3 = -\frac{13.6}{3^2} = -1.51\mathrm{~eV}$
### Step 1: Calculate Energy Gap
Delta E = E_3 - E_1 = -1.51 - (-13.6) = 12.09mathrm~eV approx 12.1mathrm~eV$\Delta E = E_3 - E_1 = -1.51 - (-13.6) = 12.09\mathrm{~eV} \approx 12.1\mathrm{~eV}$
### Pattern Recognition
Balmer emissions always return down to n=2$n=2$. Thus, the initial excitation starting from ground state n=1$n=1$ must reach at least n=3$n=3$ to create a valid Balmer transition.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Atoms
More Atoms Questions — jee_main_2024_29_january_evening
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