Does the hydrogen atom "expand" or "contract" when an electron is excited from the \(n=1\) state to the \(n=3\) state?

Short Answer

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When an electron in a hydrogen atom is excited from the \(n=1\) state to the \(n=3\) state, the average distance of the electron from the nucleus increases, as calculated using the formula \(r_n = a_0n^2\). Since the average distance is greater in the \(n=3\) state compared to the \(n=1\) state, the hydrogen atom "expands" during this transition.

Step by step solution

01

Recall the Bohr Model for Hydrogen Atom

The Bohr model is a simplified model of a hydrogen atom where the electron orbits around the nucleus in specific energy levels. Each energy level is represented by an integer number called the principal quantum number, denoted by n.
02

Understand the meaning of "expanding" and "contracting"

As an electron moves from a lower energy level to a higher energy level, its average distance from the nucleus, also known as the radius of the orbital, tends to increase. If the average distance (radius) increases, then the hydrogen atom is said to "expand." Conversely, if the average distance (radius) decreases, the hydrogen atom is said to "contract."
03

Calculate the average distance of the electron in energy levels n=1 and n=3

The average distance of the electron in the nth orbital can be calculated using the formula: \(r_n = a_0n^2\), where \(a_0\) is the Bohr radius and roughly equals \(5.29 \times 10^{-11}\) meters. For n=1: \(r_1 = a_0(1)^2 = a_0\) For n=3: \(r_3 = a_0(3)^2 = 9a_0\) Now we need to compare these distances.
04

Compare the average distances to determine if the hydrogen atom expands or contracts

Since \(r_3 = 9a_0\) is larger than \(r_1 = a_0\), it means that the electron is on average farther from the nucleus when it is in the n=3 state than when it is in the n=1 state. Therefore, the hydrogen atom "expands" when an electron is excited from the n=1 state to the n=3 state.

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