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Chapter 6: Problem 39

(a) Using Equation 6.5 , calculate the energy of an electron in the hydrogen atom when \(n=3\) and when \(n=6\). Calculate the wavelength of the radiation released when an electron moves from \(n=6\) to \(n=3 .(\mathbf{b})\) Is this line in the visible region of the electromagnetic spectrum?

Short Answer

Expert verified
The electron energy is \(-1.51\) eV for \(n=3\) and \(-0.378\) eV for \(n=6\). The energy difference between these two energy levels is \(1.132\) eV. The wavelength of the radiation released when the electron moves from \(n=6\) to \(n=3\) is approximately \(1.095 \times 10^{-6} \text{ m}\) or \(1095 \text{ nm}\), which is not within the visible region of the electromagnetic spectrum (380 nm to 750 nm).

Step by step solution

01

Calculate the electron energy for \(n=3\) and \(n=6\) using Equation 6.5

To find the energy of the electron for \(n=3\) and \(n=6\), we will use Equation 6.5: \[E_n = -\frac{13.6 \text{ eV}}{n^2}\] For \(n=3\): \[E_3 = -\frac{13.6 \text{ eV}}{(3)^2} = -\frac{13.6 \text{ eV}}{9} = -1.51 \text{ eV}\] For \(n=6\): \[E_6 = -\frac{13.6 \text{ eV}}{(6)^2} = -\frac{13.6 \text{ eV}}{36} = -0.378 \text{ eV}\] The electron energy is \(-1.51\) eV for \(n=3\) and \(-0.378\) eV for \(n=6\).
02

Calculate the energy difference between the two energy levels

Now, we need to find the energy difference between \(n=6\) and \(n=3\). For this, simply subtract the energy at \(n=3\) from the energy at \(n=6\): \[\Delta E = E_6 - E_3 = (-0.378 - (-1.51)) \text{ eV} = 1.132 \text{ eV}\] The energy difference between these two energy levels is \(1.132\) eV.
03

Calculate the wavelength of the radiation released

To calculate the wavelength of the radiation released, we will use the energy difference and the Planck's constant (\(h = 4.14 \times 10^{-15} \text{ eV s}\)): \[\lambda = \frac{hc}{\Delta E}\] First, convert the energy difference to Joules: \[\Delta E = 1.132 \text{ eV} \times \frac{1.6 \times 10^{-19} \text{ J}}{1 \text{ eV}} = 1.8112 \times 10^{-19} \text{ J}\] Now, calculate the wavelength: \[\lambda = \frac{(6.63 \times 10^{-34} \text{ J s})(3 \times 10^8 \text{ m/s})}{1.8112 \times 10^{-19} \text{ J}} = 1.095 \times 10^{-6} \text{ m}\] The wavelength of the radiation released is approximately \(1.095 \times 10^{-6} \text{ m}\) or \(1095 \text{ nm}\).
04

Determine if the wavelength is within the visible spectrum

The visible spectrum ranges from approximately 380 nm to 750 nm. Since the calculated wavelength is approximately \(1095 \text{ nm}\), it is not within the visible range. Therefore, the radiation released when the electron moves from \(n=6\) to \(n=3\) is not within the visible region of the electromagnetic spectrum.

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Most popular questions from this chapter

Molybdenum metal must absorb radiation with an energy higher than $7.22 \times 10^{-19} \mathrm{~J}$ ( "energy threshold") before it can eject an electron from its surface via the photoelectric effect. (a) What is the frequency threshold for emission of electrons? (b) What wavelength of radiation will provide a photon of this energy? (c) If molybdenum is irradiated with light of wavelength of \(240 \mathrm{nm}\), what is the maximum possible velocity of the emitted electrons?

One type of sunburn occurs on exposure to UV light of wavelength in the vicinity of \(325 \mathrm{nm}\) (a) What is the energy of a photon of this wavelength? (b) What is the energy of a mole of these photons? (c) How many photons are in a \(1.00 \mathrm{~mJ}\) burst of this radiation? \((\mathbf{d})\) These \(\mathrm{UV}\) photons can break chemical bonds in your skin to cause sunburn-a form of radiation damage. If the \(325-\mathrm{nm}\) radiation provides exactly the energy to break an average chemical bond in the skin, estimate the average energy of these bonds in \(\mathrm{kJ} / \mathrm{mol}\).

Write the condensed electron configurations for the following atoms and indicate how many unpaired electrons each has: $(\mathbf{a}) \mathrm{Mg},(\mathbf{b}) \mathrm{Ge},(\mathbf{c}) \mathrm{Br},(\mathbf{d}) \mathrm{V},(\mathbf{e}) \mathrm{Y},(\mathbf{f}) \mathrm{Lu} .$

Sketch the shape and orientation of the following types of orbitals: \((\mathbf{a}) s,(\mathbf{b}) p_{z},(\mathbf{c}) d_{x y}\).

Consider a transition in which the electron of a hydrogen atom is excited from \(n=1\) to \(n=\infty\). (a) What is the end result of this transition? (b) What is the wavelength of light that must be absorbed to accomplish this process? (c) What will occur if light with a shorter wavelength than that in part (b) is used to excite the hydrogen atom? (d) How are the results of parts \((\mathrm{b})\) and \((\mathrm{c})\) related to the plot shown in Exercise \(6.88 ?\)
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