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Chapter 6: Problem 105

A 4.653-\mum emission line of atomic hydrogen corresponds to transition between the states \(n_{f}=5\) and \(n_{i} .\) Find \(n_{i}\).

Short Answer

Expert verified
The initial energy level (\(n_i\)) for a hydrogen atom transition with an emission line wavelength of \(4.653 \ \mu m\) and a final energy level (\(n_f\)) of 5 is approximately 6.

Step by step solution

01

Understand the Rydberg Formula

The Rydberg formula can be expressed as follows: \(\frac{1}{\lambda} = R_H ( \frac{1}{n_f^2} - \frac{1}{n_i^2})\) where \(\lambda\) is the wavelength of the hydrogen emission line, \(R_H\) is the Rydberg constant for hydrogen (approximately \(1.097373 \times 10^7 m^{-1}\)), and \(n_i\) and \(n_f\) represent the initial and final energy levels, respectively. In this problem, we are given the emission line's wavelength (\(\lambda = 4.653 \ \mu m\)) and the final energy level (\(n_f = 5\)). We need to find the value of \(\textit{n}_{i}\), the initial energy level.
02

Convert the Wavelength to Meters

The wavelength given in the problem is in micrometers (\(\mu m\)), so we need to convert it to meters for consistency with the SI units used in the Rydberg constant: \(\lambda = 4.653 \ \mu m = 4.653 \times 10^{-6} \ m\)
03

Rearrange the Rydberg Formula to Solve for \(n_i^2\)

Rearrange the Rydberg formula to isolate \(\textit{n}_{i}^2\): \(n_i^2 = \frac{1}{ \frac{1}{n_f^2} - \frac{1}{\lambda R_H} }\)
04

Insert Known Values into the Formula and Solve for \(n_i^2\)

Now insert the known values \(\lambda = 4.653 \times 10^{-6} \ m\), \(R_H = 1.097373 \times 10^7 m^{-1}\), and \(n_f = 5\) into the formula and solve for \(n_i^2\): \(n_i^2 = \frac{1}{ \frac{1}{5^2} - \frac{1}{(4.653 \times 10^{-6} \ m)(1.097373 \times 10^7 m^{-1})} }\) Calculate the denominator of the fraction: \(\frac{1}{25} - \frac{1}{(4.653 \times 10^{-6} \ m)(1.097373 \times 10^7 m^{-1})} \approx 0.039816\) Now, calculate the value of \(n_i^2\): \(n_i^2 = \frac{1}{0.039816} \approx 25.126\)
05

Determine \(n_i\) from \(n_i^2\)

Since we have found the value of \(n_i^2\), we need to determine \(n_i\): \(n_i = \sqrt{25.126} \approx 5.01\) Since energy levels are whole numbers, round the value of \(n_i\) up: \(n_i \approx 6\) Thus, the initial energy level (\(n_i\)) is approximately 6.

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