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Chapter 4: Problem 77

The electron in a hydrogen atom relaxes from the \(n=4\) shell to some lower- energy shell. The light emitted during the relaxation has a wavelength of \(1772.6 \mathrm{~nm}\). By calculating the energy of this light, determine the shell to which the electron relaxed. \(\left[1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}\right]\)

Short Answer

Expert verified
The energy of the emitted light is calculated as \(E = \frac{6.626 \times 10^{-34} J \cdot s \cdot 3 \times 10^8 m/s}{1772.6 \times 10^{-9} m}\). After converting the energy to eV, we use the Rydberg formula: \(\frac{1}{\lambda_{vac}} = R_H \left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right)\) to find the final shell \(n_f^2 = \left(\frac{1}{4^2} - \frac{1772.6 \times 10^{-9}}{1.097373 \times 10^{7}}\right)^{-1}\). Then, \(n_f = \sqrt{n_f^2}\), and round it to the nearest whole number. The electron relaxed to shell \(n_f = 2\).

Step by step solution

01

Calculate the energy of emitted light

First, we need to calculate the energy of the light emitted during the relaxation. The energy can be calculated using the formula E = h * c / λ, where h = 6.626 x 10^{-34} J s (Planck's constant), c = 3 x 10^8 m/s (speed of light), and λ = 1772.6 nm = 1772.6 x 10^{-9} m (wavelength). E = \(\frac{6.626 \times 10^{-34} J \cdot s \cdot 3 \times 10^8 m/s}{1772.6 \times 10^{-9} m}\)
02

Convert energy to eV

Now we need to convert the energy calculated in Step 1 from Joules to electron volts (eV). The conversion factor is given: [1 eV \( = 1.602 \times 10^{-19} J\)]. To convert the energy, divide it by the conversion factor: E_eV = \(\frac{E}{1.602 \times 10^{-19}}\)
03

Use the Rydberg formula to find the final shell

Now, we will use the Rydberg formula to find the final shell (n_f): \(\frac{1}{\lambda_{vac}} = R_H \left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right)\) where λ_vac is the vacuum wavelength (given), R_H is the Rydberg constant for hydrogen \(\left(1.097373 \times 10^{7} m^{-1}\right)\), n_i is the initial shell (\(n_i = 4\)), and n_f is the final shell (unknown). Rearrange the formula to solve for n_f: \(n_f^2 = \left(\frac{1}{n_i^2} - \frac{\lambda_{vac}}{R_H}\right)^{-1}\) Plug in the known values: \(n_f^2 = \left(\frac{1}{4^2} - \frac{1772.6 \times 10^{-9}}{1.097373 \times 10^{7}}\right)^{-1}\) Calculate the final shell (n_f) by taking the square root: \(n_f = sqrt(n_f^2)\)
04

Determine the shell to which the electron relaxed

Now that we have calculated n_f, round it to the nearest whole number to find the shell to which the electron relaxed.

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