Given below are two statements:
Statement I: When an electric discharge is passed through gaseous hydrogen, the hydrogen molecules dissociate and the energetically excited hydrogen atoms produce electromagnetic radiation of discrete frequencies.
Statement II: The frequency of second line of Balmer series obtained from He^+$He^{+}$ is equal to that of first line of Lyman series obtained from hydrogen atom. In the light of the above statements, choose the correct answer from the options given below:
A.textBoth Statement I and Statement II are true$\text{Both Statement I and Statement II are true}$
B.textBoth Statement I and Statement II are false$\text{Both Statement I and Statement II are false}$
C.textStatement I is false but Statement II is true$\text{Statement I is false but Statement II is true}$
D.textStatement I is true but Statement II is false$\text{Statement I is true but Statement II is false}$
Solution & Explanation
### Related Formula
frac1lambda = RZ^2left(frac1n_1^2 - frac1n_2^2right)$$\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$
### Core Logic
Statement I is a factual description of how the hydrogen emission spectrum is obtained. Dissociation yields excited atoms that emit light at discrete frequencies. So Statement I is true.
For Statement II:
First line of Lyman series for H atom (Z=1, n_1=1, n_2=2$Z=1, n_1=1, n_2=2$):
frac1lambda_1 = R(1)^2left(frac11^2 - frac12^2right) = Rleft(1 - frac14right) = frac3R4$$\frac{1}{\lambda_1} = R(1)^2\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}$$
Second line of Balmer series for He^+$He^+$ (Z=2, n_1=2, n_2=4$Z=2, n_1=2, n_2=4$):
frac1lambda_2 = R(2)^2left(frac12^2 - frac14^2right) = 4Rleft(frac14 - frac116right) = 4Rleft(frac316right) = frac3R4$$\frac{1}{\lambda_2} = R(2)^2\left(\frac{1}{2^2} - \frac{1}{4^2}\right) = 4R\left(\frac{1}{4} - \frac{1}{16}\right) = 4R\left(\frac{3}{16}\right) = \frac{3R}{4}$$
Since frac1lambda_1 = frac1lambda_2$\frac{1}{\lambda_1} = \frac{1}{\lambda_2}$, their wavelengths (and therefore frequencies) are exactly the same. Statement II is true.
### Step 1: Final Conclusion
Both statements are correct.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Structure of Atom
Keywords:#Balmer series He+#Lyman series H atom#JEE Main 2026 Morning Q70#Structure of Atom JEE Main 2026
More Structure of Atom Previous-Year Questions — Page 2
Q40jee_main_2025_08_april_eveningQuantum Numbers and Electronic Configuration
Identify the correct statements for an element possessing atomic number 9:
A. There can be 5 electrons for which m_s = +frac12$m_s = +\frac{1}{2}$ and 4 electrons for which m_s = -frac12$m_s = -\frac{1}{2}$.
B. There is only one electron in the p_z$p_z$ orbital.
C. The last electron goes into an orbital described by quantum parameters n = 2$n = 2$ and l = 1$l = 1$.
D. The sum of angular nodes of all populated atomic orbitals is 1.
Choose the correct answer from the options given below:
A.textC and D Only$\text{C and D Only}$
B.textA and C Only$\text{A and C Only}$
C.textA, C and D Only$\text{A, C and D Only}$
D.textA and B Only$\text{A and B Only}$
Solution
### Core Logic
The element with atomic number 9 is **Fluorine (F)**. Let's write out its ground state electronic configuration:
F (Z=9) = 1s^2 \, 2s^2 \, 2p^5$$F (Z=9) = 1s^2 \, 2s^2 \, 2p^5$$
Let's systematically audit each statement:
* **Statement A**: Within the total population of 9 electrons, the pairing distribution across shells shows: 1s^2$1s^2$ (one up, one down), 2s^2$2s^2$ (one up, one down), 2p^5$2p^5$ (three up, two down). Summing up-spins (m_s = +frac12$m_s = +\frac{1}{2}$) yields 1 + 1 + 3 = 5$1 + 1 + 3 = 5$ electrons. Down-spins (m_s = -frac12$m_s = -\frac{1}{2}$) yield 1 + 1 + 2 = 4$1 + 1 + 2 = 4$ electrons. **Statement A is fully correct.** Orbital spin configuration box diagram for Fluorine atom
* **Statement B**: By Hund's Rule, the 5 electrons in the 2p subshell occupy the degenerate p_x, p_y, p_z$p_x, p_y, p_z$ states. This produces two fully-filled sub-orbitals and one half-filled sub-orbital. The unpaired slot can reside arbitrarily in *any* of the three orbitals (p_x, p_y$p_x, p_y$, or p_z$p_z$) due to spatial symmetry. It is not constrained to p_z$p_z$. **Statement B is incorrect.**
* **Statement C**: The highest energy valence electron enters the 2p subshell, which is defined by principal number n = 2$n = 2$ and azimuthal index l = 1$l = 1$. **Statement C is fully correct.**
* **Statement D**: Angular nodes are given directly by the quantum number l$l$. For s-orbitals (1s, 2s$1s, 2s$), angular nodes = 0$= 0$. For each of the three populated p-orbitals (2p$2p$), angular nodes = 1$= 1$. The sum total of angular nodes across all orbitals is 0 + 0 + 3 = 3$0 + 0 + 3 = 3$. **Statement D is incorrect.**
### Pattern Recognition
Total angular nodes equals the total number of p-electrons' spatial orientation count, not simply the subshell boundary value. Recognizing that degenerate p-orbitals share uniform probability status exposes the restriction in Statement B instantly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Structure of Atom
Q47jee_main_2025_08_april_eveningBohr's Model
The energy of an electron in the first Bohr orbit of the Hydrogen atom is -13.6 text eV$-13.6 \text{ eV}$. The magnitude of the energy value of an electron in the first excited state of the textBe^3+$\text{Be}^{3+}$ ion is _________ eV (as the nearest integer value).
Numerical Answer.Answer: 54 to 54
Solution
### Related Formula
Bohr energy level formula for hydrogenic species:
E_n = -13.6 times fracZ^2n^2 quad texteV$$E_n = -13.6 \times \frac{Z^2}{n^2} \quad \text{eV}$$
where:
Z$Z$ = atomic number of the species
n$n$ = principal quantum number of the orbit
### Execution
Step 1: Identify the parameters for the first excited state of textBe^3+$\text{Be}^{3+}$:
* For Beryllium (textBe$\text{Be}$), the atomic number is Z = 4$Z = 4$.
* The term 'first excited state' refers to the second energy level, so n = 2$n = 2$.
Step 2: Substitute these values into the Bohr energy equation:
E_textBe^3+ = -13.6 times frac4^22^2 = -13.6 times frac164$$E_{\text{Be}^{3+}} = -13.6 \times \frac{4^2}{2^2} = -13.6 \times \frac{16}{4}$$E_textBe^3+ = -13.6 times 4 = -54.4 text eV$$E_{\text{Be}^{3+}} = -13.6 \times 4 = -54.4 \text{ eV}$$
Step 3: Extract the magnitude and round to the nearest integer value:
|E_textBe^3+| = 54.4 approx 54$$|E_{\text{Be}^{3+}}| = 54.4 \approx 54$$
### Pattern Recognition
For the first excited state of Beryllium (Z=4, n=2$Z=4, n=2$), the term fracZ^2n^2 = frac4^22^2 = frac164 = 4$\frac{Z^2}{n^2} = \frac{4^2}{2^2} = \frac{16}{4} = 4$. Thus, the energy value is exactly 4 times that of the ground-state hydrogen atom (13.6 times 4 = 54.4$13.6 \times 4 = 54.4$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Structure of Atom
Q27jee_main_2025_29_jan_eveningBohr Model for Hydrogen-like Species
For hydrogen like species, which of the following graphs provides the most appropriate representation of E vs Z plot for a constant n?
[E: Energy of the stationary state, Z: atomic number, n: principal quantum number]
A. Graph (1)
B. Graph (2)
C. Graph (3)
D. Graph (4)
Solution
### Related Formula
E_n = -13.6 fracZ^2n^2text eV$$E_n = -13.6 \frac{Z^2}{n^2}\text{ eV}$$
### Core Logic
For a constant principal quantum number n$n$, the energy E$E$ is directly proportional to -Z^2$-Z^2$.
This represents a quadratic relation where the curve is a downward-opening parabola starting from the origin in the negative energy region.
### Pattern Recognition
Since energy values are inherently negative for bound states, as Z$Z$ increases, E$E$ becomes rapidly more negative following a parabolic curve.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Structure of Atom
Given below are two statements:
Statement (I): It is impossible to specify simultaneously with arbitrary precision, both the linear momentum and the position of a particle.
Statement (II) If the uncertainty in the measurement of position and uncertainty in measurement of momentum are equal for an electron, then the uncertainty in the measurement of velocity is gesqrtfrachpitimesfrac12m$gesqrt{\frac{h}{pi}}\times\frac{1}{2m}$
In the light of the above statements, choose the correct answer from the options given below:
A. Statement I is true but Statement II is false.
B. Both Statement I and Statement II are true.
C. Statement I is false but Statement II is true.
D. Both Statement I and Statement II are false.
Solution
### Related Formula
Delta x cdot Delta p ge frach4pi$$Delta x cdot Delta p ge \frac{h}{4pi}$$
### Core Logic
Statement I is a verbatim definition of Heisenberg's Uncertainty Principle, hence it is completely true.
For Statement II, we are given that Delta x = Delta p$Delta x = Delta p$:
Delta p cdot Delta p ge frach4pi implies (Delta p)^2 ge frach4pi $$Delta p cdot Delta p ge \frac{h}{4pi} implies (Delta p)^2 ge \frac{h}{4pi} $$Delta p ge sqrtfrach4pi = frac12sqrtfrachpi$$Delta p ge sqrt{\frac{h}{4pi}} = \frac{1}{2}sqrt{\frac{h}{pi}}$$
Since Delta p = m cdot Delta v$Delta p = m cdot Delta v$:
m cdot Delta v ge frac12sqrtfrachpi implies Delta v ge frac12msqrtfrachpi$$m cdot Delta v ge \frac{1}{2}sqrt{\frac{h}{pi}} implies Delta v ge \frac{1}{2m}sqrt{\frac{h}{pi}}$$
This perfectly matches Statement II, so it is also true.
### Pattern Recognition
When solving inequality bounds for identical uncertainties, always substitute directly to obtain a clean quadratic form before taking the square root.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Structure of Atom
Q38jee_main_2025_28_jan_morningQuantum Numbers and Orbital Energies
In a multielectron atom, which of the following orbitals described by three quantum numbers with have same energy in absence of electric and magnetic fields?
A. n = 1, l = 0, m_l = 0$n = 1, l = 0, m_{l} = 0$
B. n = 2, l = 0, m_l = 0$n = 2, l = 0, m_{l} = 0$
C. n = 2, l = 1, m_l = 1$n = 2, l = 1, m_{l} = 1$
D. n = 3, l = 2, m_l = 1$n = 3, l = 2, m_{l} = 1$
E. n = 3, l = 2, m_l = 0$n = 3, l = 2, m_{l} = 0$
Choose the correct answer from the options given below:
A.textA and B only$\text{A and B only}$
B.textB and C only$\text{B and C only}$
C.textC and D only$\text{C and D only}$
D.textD and E only$\text{D and E only}$
Solution
### Core Logic
In multi-electron systems, the energy profile depends on both the primary shell (n$n$) and azimuthal subshell (l$l$) quantum values via the (n+l)$(n+l)$ rule. Orbitals sharing the exact same n$n$ and l$l$ values are energy-degenerate as long as external field metrics are zero.
Let us map each set:
- **A:** 1s$1s$
- **B:** 2s$2s$
- **C:** 2p$2p$
- **D:** 3d$3d$ (m_l = 1$m_l = 1$)
- **E:** 3d$3d$ (m_l = 0$m_l = 0$)
Since both D and E describe subcomponents of the same 3d$3d$ subshell (n=3, l=2$n=3, l=2$), they share identical energies, establishing perfect degeneracy.
### Pattern Recognition
Sees: Orbitals matching energy level in multi-electron space.
Trap: Confusing single-electron hydrogen atoms (where energy depends solely on n$n$) with multielectron atoms.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Structure of Atom
More Structure of Atom Questions — jee_main_2026_21_jan_morning
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