Question

A hydrogen atom exists in an excited state for typically 10−8 s. How many revolutions would an electron make in an n = 27 state before decaying? revolutions

Answer 1

An electron in a hydrogen atom in an n=27 state would make 162 revolutions before decaying. Earth would take approximately 162 * 365.25 days to orbit the sun this many times.

Step-by-step explanation:

In an excited state, a hydrogen atom can be represented by an electron orbiting the proton nucleus. The number of revolutions an electron makes in a given state can be determined by using the formula:

Number of Revolutions = (2π * n)

Given that n = 27, the electron would make 162 revolutions before decaying.

To determine the time it takes Earth to orbit the sun this many times:

Time = Number of Revolutions * Torbit

Where Torbit is the time it takes Earth to orbit the sun, which is approximately 365.25 days. Therefore, the time it takes Earth to orbit the sun this many times is approximately 162 * 365.25 days.