Between which two orbits of the Bohr hydrogen atom must an electron fall to produce light of wavelength \(1876 \mathrm{nm} ?\)

Short Answer

Expert verified
The electron must fall from the \(n_i = 3\) orbit to the \(n_f = 2\) orbit to produce light of the given wavelength.

Step by step solution

01

Convert wavelength to meters

Since \(\lambda\) is given in \(\mathrm{nm}\) (nanometers), it needs to be converted to meters for use in the formula. \(1 \mathrm{nm} = 1 \times 10^{-9} m\). Thus, \(\lambda = 1876 \times 10^{-9} m \approx 1.876 \times 10^{-6} m\).
02

Solve the Rydberg formula for \(n_i\)

Rearrange the Rydberg formula to solve for \(n_i\): \(\frac{1}{n_i^2} = \frac{1}{\lambda R} + \frac{1}{n_f^2}\).
03

Substitute the known values into the formula and solve for \(n_i\) for each \(n_f\)

As we are only dealing with hydrogen, \(n_f\) will start from 1 (lowest possible orbit) and increment until we find \(n_i\). Plugging in the value of \(\lambda = 1.876 \times 10^{-6} m\) and \(R = 1.097 \times 10^7 m^{-1}\), we calculate the value of \(n_i\) for each \(n_f\). Since \(n_i\) and \(n_f\) are quantum numbers, they can only be positive integers, so we round \(n_i\) to the nearest integer. The correct \(n_f\) is the one that gives us a value of \(n_i\) greater than \(n_f\). Repeat this until we identify the correct values.

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