LEVELJEE Main
Visualized Solution
The Sigma Insight: Bohr's Atomic Model and Energy Levels
The Second Excited State
Let's begin by visualizing the Hydrogen atom and the Lithium ion () side by side. The problem states that both are in the second excited state.
It is crucial to remember the terminology of energy states. The ground state corresponds to . The first state above it is the first excited state (), which means the second excited state corresponds to a principal quantum number of for both atoms.
Bohr's Quantization of Angular Momentum
First, let's compare their electronic angular momenta. According to Bohr's quantization postulate, the angular momentum of an electron in the orbit is an integral multiple of . The formula is given by:
Notice how this formula depends only on the principal quantum number . It does not depend on the atomic number or the mass of the nucleus. Since both the Hydrogen atom and the Lithium ion have their electron in the orbit, we can directly conclude that their angular momenta must be exactly the same.
The heavier, more highly charged nucleus of Lithium doesn't change the angular momentum of the electron in a given orbit.
Energy Dependence on Atomic Number
Now, let's evaluate their energies. The total energy of an electron in the orbit of a hydrogen-like species is given by the expression:
Here, the atomic number plays a crucial role. For Hydrogen, the atomic number is . But for the Lithium ion, is . Because the magnitude of energy is directly proportional to (for a constant ), we can compare them easily.
This means the energy magnitude for Lithium will be three squared, which is nine times the energy magnitude of Hydrogen. Therefore, , which implies .
Conclusion
Putting it all together, the angular momenta are strictly equal because they depend only on , but the magnitude of energy for the Lithium ion is strictly greater than that of the Hydrogen atom because it scales with . This perfectly matches option (b).
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