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Chapter 6: Problem 37

(a) Using Equation \(6.5,\) calculate the energy of an electron in the hydrogen atom when \(n=2\) and when \(n=6 .\) Calculate the wavelength of the radiation released when an electron moves from \(n=6\) to \(n=2 .\) (b) Is this line in the visible region of the electromagnetic spectrum? If so, what color is it?

Short Answer

Expert verified
The energies of electrons in the hydrogen atom are -3.4 eV for n=2 and -0.378 eV for n=6. The wavelength of the released radiation when an electron moves from n=6 to n=2 is \(4.11\times10^{-7}\,\text{m}\), and its color is bluish-violet.

Step by step solution

01

Write down the equation for energy levels in a hydrogen atom

We are given Equation 6.5 for the energy levels in hydrogen atom, which is: \( E_n = -\frac{13.6\,\text{eV}}{n^2} \) where \(E_n\) is the energy of an electron at a particular energy level n.
02

Calculate the energy of an electron for n = 2 and n = 6

For n = 2, \(E_2 = -\frac{13.6\,\text{eV}}{(2)^2} = -\frac{13.6\,\text{eV}}{4} = -3.4\,\text{eV}\) For n = 6, \(E_6 = -\frac{13.6\,\text{eV}}{(6)^2} = -\frac{13.6\,\text{eV}}{36} \approx -0.378\,\text{eV}\)
03

Calculate the energy difference between n = 6 and n = 2

The energy difference ∆E between these two levels is given by: ∆E = E_6 - E_2 ≈ -0.378 eV - (-3.4 eV) ≈ 3.022 eV
04

Calculate the wavelength of the radiation released

The energy difference (∆E) is related to the wavelength of the released radiation (λ) by the formula: \( \Delta E = \frac{hc}{\lambda} \) where h is the Planck's constant (\(6.63\times10^{-34}\,\text{Js}\)), and c is the speed of light (\(3\times10^8\,\text{m/s}\)). Rearrange the formula to solve for λ: \( \lambda = \frac{hc}{\Delta E} \) To calculate the wavelength, we need to convert ∆E to joules: \( \Delta E = 3.022\,\text{eV} \times \frac{1.6\times10^{-19}\,\text{J}}{1\,\text{eV}} \approx 4.83\times10^{-19}\,\text{J} \) Now, we can calculate the wavelength λ: \( \lambda = \frac{(6.63\times10^{-34}\,\text{Js})(3\times10^8\,\text{m/s})}{4.83\times10^{-19}\,\text{J}} \approx 4.11\times10^{-7}\,\text{m} \)
05

Check if the calculated wavelength is in the visible region and determine its color

The visible region of the electromagnetic spectrum ranges approximately from \(4\times10^{-7}\,\text{m}\) (violet) to \(7\times10^{-7}\,\text{m}\) (red). Since the calculated wavelength \(4.11\times10^{-7}\,\text{m}\) lies within this range, it is in the visible region. This wavelength is close to the blue-violet boundary, so the color of the released radiation would appear as bluish-violet. In conclusion, the energies of electrons in the hydrogen atom are -3.4 eV for n=2 and -0.378 eV for n=6. The wavelength of the released radiation when an electron moves from n=6 to n=2 is \(4.11\times10^{-7}\,\text{m}\), and its color is bluish-violet.

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Most popular questions from this chapter

(a) The average distance from the nucleus of a 3 s electron in a chlorine atom is smaller than that for a \(3 p\) electron. In light of this fact, which orbital is higher in energy? (b) Would you expect it to require more or less energy to remove a 3 s electron from the chlorine atom, as compared with a \(2 p\) electron? Explain.

(a) Calculate the energies of an electron in the hydrogen atom for \(n=1\) and for \(n=\infty .\) How much energy does it require to move the electron out of the atom completely (from \(n=1\) to \(n=\infty),\) according to Bohr? Put your answer in \(\mathrm{kJ} / \mathrm{mol}\). (b) The energy for the process \(\mathrm{H}+\) energy \(\rightarrow \mathrm{H}^{+}+\mathrm{e}^{-}\) is called the ionization energy of hydrogen. The experimentally determined value for the ionization energy of hydrogen is \(1310 \mathrm{~kJ} / \mathrm{mol}\). How does this compare to your calculation?

Using Heisenberg's uncertainty principle, calculate the uncertainty in the position of (a) a 1.50 -mg mosquito moving at a speed of \(1.40 \mathrm{~m} / \mathrm{s}\) if the speed is known to within \(\pm 0.01 \mathrm{~m} / \mathrm{s} ;\) (b) a proton moving at a speed of \((5.00 \pm 0.01) \times 10^{4} \mathrm{~m} / \mathrm{s}\). (The mass of a proton is given in the table of fundamental constants in the inside cover of the text.)

A diode laser emits at a wavelength of \(987 \mathrm{nm}\). (a) In what portion of the electromagnetic spectrum is this radiation found? (b) All of its output energy is absorbed in a detector that measures a total energy of \(0.52 \mathrm{~J}\) over a period of \(32 \mathrm{~s}\). How many photons per second are being emitted by the laser?

As shown in the accompanying photograph, an electric stove burner on its highest setting exhibits an orange glow. (a) When the burner setting is changed to low, the burner continues to produce heat but the orange glow disappears. How can this observation be explained with reference to one of the fundamental observations that led to the notion of quanta? (b) Suppose that the energy provided to the burner could be increased beyond the highest setting of the stove. What would we expect to observe with regard to visible light emitted by the burner? [Section 6.2\(]\)
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