as n gets larger what happens to the energy of the atom and orbitals

In this section we will discuss the energy level of the electron of a hydrogen atom, and how it changes every bit the electron undergoes transition. Co-ordinate to Bohr's theory, electrons of an atom revolve effectually the nucleus on certain orbits, or electron shells. Each orbit has its specific energy level, which is expressed equally a negative value. This is because the electrons on the orbit are "captured" by the nucleus via electrostatic forces, and impedes the freedom of the electron. The orbits closer to the nucleus have lower energy levels because they interact more than with the nucleus, and vice versa. Bohr named the orbits as $\text{K }(n=i), \text{L }(n=2), \text{M }(n=iii), \text{Due north }(n=4), \text{O }(n=5), \cdots$ in order of increasing altitude from the nucleus. Note that $northward$ refers to the principal breakthrough number.

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The energy of the electron of a monoelectronic atom depends simply on which vanquish the electron orbits in. The energy level of the electron of a hydrogen atom is given by the post-obit formula, where $north$ denotes the principal quantum number: $E_n=-\frac{1312}{northward^2}\text{ kJ/mol}.$ For a single electron instead of per mole, the formula in eV (electron volts) is also widely used: $E_n=-\frac{13.6}{n^2}\text{ eV}.$

Discover that the energy level is always negative, and increases equally $northward.$ Since $north$ tin can only take on positive integers, the free energy level of the electron can but accept on specific values such as $E_1=-xiii.6\text{ eV},$ $E_2=-3.39\text{ eV},$ $E_3=-1.51\text{ eV}, \cdots$ and so on. Thus, we can say that the free energy level of an electron is quantized, rather than continuous. The figure below shows the electron energy level diagram of a hydrogen atom. Notice how the lines become closer as $n$ increases.

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For atoms other than hydrogen, we merely multiply $-\frac{1312}{n^2}\text{ kJ/mol}$ or $-\frac{xiii.6}{northward^2}\text{ eV}$ past $Z_{\text{eff}}^2,$ where $Z_{\text{eff}}$ refers to the effective nuclear charge. Keep in mind that this rule can only be applied to monatomic atoms (or ions) such equally $\ce{H}, \ce{He+}, \ce{Li}^{two+}.$

Notice the ionization energy of hydrogen.

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Ionization energy is the energy needed to accept away an electron from an atom. It is equivalent to the energy needed to excite an electron from $n=ane$ (ground state) to $northward=\infty,$ which is $E_{\infty}-E_1=1312\text{ kJ/mol},$ or $E_{\infty}-E_1=13.6\text{ eV}.\ _\square$

In chemistry, energy is a measure out of how stable a substance is. The lower the free energy level of an electron, the more stable the electron is. Thus an electron would be in its most stable land when it is in the K shell $(n=1).$ For this reason, we refer to $n=one$ as the basis state of the electron. If the electron is in whatever other crush, nosotros say that the electron is in excited land.

Information technology is quite obvious that an electron at ground state must gain energy in society to become excited. Too, an electron at a college energy level releases free energy every bit it falls down to a lower energy level. Using the formula above, we can summate how much free energy is absorbed/released during the transition of an electron. The energy alter during the transition of an electron from $n=n_1$ to $north=n_2$ is $\Delta E=E_{2}-E_{1}=thirteen.6\times\left(\frac{one}{n_1^ii}-\frac{1}{n_2^2}\right)\text{ eV}.$ Obviously, a positive energy change means that the electron absorbs energy, while a negative energy change implies a release of energy from the electron. Annotation that the formula is the energy per mole, rather than that of a single photon.

During transition, an electron absorbs/releases energy is in the form of calorie-free energy. The energy of the photon $E$ captivated/released during the transition is equal to the energy change $\Delta E$ of the electron. Using the properties of DeBroglie waves, we can calculate the wavelength and frequency of the following formula: $E=h\nu=h\frac{c}{\lambda},$ where $h=6.63\times10^{-34}\text{ J}\cdot\text{s}$ denotes Planck's abiding, $\nu$ denotes frequency, $\lambda$ denotes wavelength, and $c=iii.00\times10^8\text{ m/southward}$ denotes the speed of light. Combining this formula with the $\Delta Due east$ formula higher up gives the famous Rydberg formula: $\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{one}{n_2^two}\correct)\text{ k}^{-1},$ where $R=ane.097\times10^7\text{ m}^{-one}$ is the Rydberg constant. Using the Rydberg formula, we can compute the wavelength of the light the electron absorbs/releases, which ranges from ultraviolet to infrared.

Because the value of $\frac{i}{n^2}$ substantially decreases as $due north$ increases, the value of the free energy change or wavelength depends on the smaller between $n_1$ and $n_2.$ For this reason, the low-cal emission by the autumn of the free energy level of an electron can be categorized into several groups. If an electron falls from any $n\ge2$ to $northward=1,$ then the wavelength calculated using the Rydberg formula gives values ranging from 91 nm to 121 nm, which all autumn under the domain of ultraviolet. As this was discovered past a scientist named Theodore Lyman, this kind of electron transition is referred to as the Lyman series. Similarly, any electron transition from $north\ge3$ to $n=2$ emits visible lite, and is known equally the Balmer series. Electron transition from $n\ge4$ to $n=3$ gives infrared, and this is referred to as the Paschen series.

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Since the energy level of the electron of a hydrogen cantlet is quantized instead of continuous, the spectrum of the lights emitted past the electron via transition is also quantized. In other words, the wavelength $\lambda$ can but take on specific values since $n_1$ and $n_2$ are integers. As a result, the electron transition gives spectral lines as shown in the correct figure below (showing only visible low-cal, or Balmer series). Note how this differs to the continuous spectrum shown in the left figure below. Running sunlight through a prism would give a continuous spectrum.

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When analyzing spectral lines, nosotros must approach them from the right side. This is because the lines become closer and closer every bit the wavelength decreases within a series, and information technology is harder to tell them autonomously. The line with the longest wavelength inside a series corresponds to the electron transition with the everyman energy within that serial. Hence in the effigy above, the ruby line indicates the transition from $n=3$ to $n=2,$ which is the transition with the lowest energy inside the Balmer serial.

Recall that the free energy level of the electron of an atom other than hydrogen was given by $E_n=-\frac{1312}{due north^2}\cdot Z_{\text{eff}}^2\text{ kJ/mol}.$ Since each chemical element has a unique $Z_{\text{eff}}$ value, the spectral lines of each element would be different. Therefore spectral lines tin be idea of the "fingerprints" of an element, and be used to identify an element.

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The figure above shows the spectrum of Balmer series. Which of the post-obit electron transitions corresponds to the turquoise line $(\lambda\approx485\text{ nm})$ in the figure to a higher place?

(A) $northward=2\rightarrow n=i$
(B) $n=3\rightarrow n=1$
(C) $n=3\rightarrow n=two$
(D) $n=4\rightarrow n=2$

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Detect that the red line has the longest wavelength inside the Balmer series. Since a longer wavelength means smaller free energy, the red line correspond to the transition which emits the lowest energy within the Balmer series, which is $due north=3\rightarrow n=2.$ The turquoise line indicates the transition with the 2d lowest energy within the Balmer series, which is $north=4\rightarrow north=2.$ Therefore our answer is (D). $_\square$

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Source: https://brilliant.org/wiki/energy-level-and-transition-of-electrons/

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