Why does the $4s$ orbital fill before the $3d$ orbital according to the $(n+l)$ rule, yet during ionization, electrons are lost from $4s$ first?

During electron filling in neutral atoms, the $(n+l)$ rule governs energy: for $4s$, $n+l = 4+0 = 4$; for $3d$, $n+l = 3+2 = 5$. Hence, $4s$ is filled first. However, once $3d$ receives electrons, the increased nuclear screening and spatial distribution cause the $3d$ subshell to drop lower in energy than $4s$. Consequently, $4s$ electrons occupy the outermost spatial boundary and are removed first during ionization.

How do radial nodes differ from angular nodes, and how do they appear on wave density graphs?

Radial nodes are spherical surfaces centered at the nucleus where probability density $4\pi r^2 \psi^2 = 0$, calculated as $n - l - 1$. Angular nodes are planes or cones passing through the nucleus where wave amplitude is zero, given by $l$. On a radial probability plot ($4\pi r^2 \psi^2$ vs $r$), radial nodes appear as points where the curve touches the horizontal axis (excluding $r=0$).

What is the strategic approach to solve quantum number sets speed-questions in JEE/NEET without silly errors?

Apply three strict sequential inequalities: $n \ge 1$ (positive integer), $0 \le l \le n-1$ (integer), and $-l \le m_l \le +l$ (integer). Finally, verify $m_s = +\frac{1}{2}$ or $-\frac{1}{2}$. If any constraint breaks (e.g., $l = n$ or $|m_l| > l$), reject the set instantly without spending time on further analysis.

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Question #1NEET 2024

Calculate the energy required to excite an electron in a $\text{Li}^{2+}$ ion from its ground state ( $n=1$ ) to the first excited state ( $n=2$ ).

Solution Locked

Step-by-Step Solution & Concept

Step 1

State the Bohr energy formula for hydrogenic species: $E_n = -13.6 \times \frac{Z^2}{n^2} \text{ eV}$ . For $\text{Li}^{2+}$ , atomic number $Z = 3$ .

Step 2

Calculate energies of the initial state $n=1$ and first excited state $n=2$ : $E_1 = -13.6 \times \frac{3^2}{1^2} = -122.4 \text{ eV}$ and $E_2 = -13.6 \times \frac{3^2}{2^2} = -30.6 \text{ eV}$ .

Step 3

Determine excitation energy $\Delta E = E_2 - E_1 = -30.6 \text{ eV} - (-122.4 \text{ eV}) = 91.8 \text{ eV}$ .

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Question #2JEE Main 2023

An electron accelerated through a potential difference of $V \text{ volts}$ has a de Broglie wavelength of $\lambda_1$ . If the potential difference is increased to $4V \text{ volts}$ , the new wavelength becomes $\lambda_2$ . Determine the ratio $\frac{\lambda_1}{\lambda_2}$ .

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Step-by-Step Solution & Concept

Step 1

Relate de Broglie wavelength to accelerating potential $V$ using $\lambda = \frac{h}{\sqrt{2mqV}}$ , showing that $\lambda \propto \frac{1}{\sqrt{V}}$ .

Step 2

Set up ratio $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}} = \sqrt{\frac{4V}{V}}$ .

Step 3

Simplify to obtain $\frac{\lambda_1}{\lambda_2} = \sqrt{4} = 2$ , yielding the ratio $2 : 1$ .

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Mastering Structure of Atom: Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Bohr Orbit Energy

Rydberg Formula for Spectral Transitions

de Broglie Wavelength

Heisenberg Uncertainty Principle

Orbital Angular Momentum & Node Formulas

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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