Question

1.Calculate the energy required to remove the electron from a hydrogen atom in its ground state.
2.Calculate the energy required to excite the hydrogen electron from level n = 1 to level n= 2.
Also calculate the wavelength of light that must be absorbed by a hydrogen atom in its ground
state to reach this excited state.
3.Ahydrogen atom absorbs a photon of visible light, and its electron enters the n =4 energy
level. Calculate a). the change in energy of the atom and b). the wavelength (in nm) of the
photon.
4.A hydrogen atom with its electronin n=6 energy level emits a photon of IR light. Calculate
a). the change in energy of the atom and b). the wavelength (in angstroms)
5.An excited hydrogen atom emits light with a frequency of 1.141 x 1014s
-1as its electron moves
from one energy level to the energy level for which n = 4. (a) Calculate the energy change of
the electron. (b) In what energy level did it begin?
6.What amount of energy is emitted when the electron of an excited hydrogen atom falls from
the 3rdenergy level to the energy level closest to the nucleus?

Answer 1

To answer your questions regarding electronic transitions in a hydrogen atom, we can use the following equations and constants:

1. **Energy required to remove an electron (Ionization Energy)**: The energy of an electron in a hydrogen atom in the \(n\)th energy level is given by the formula:

\[
E_n = -\frac{13.6 \, \text{eV}}{n^2}
\]

For the ground state (\(n = 1\)):

\[
E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV}
\]

To remove the electron (ionization), we need to provide energy equal to the absolute value of this energy:

\[
\text{Energy required} = 13.6 \, \text{eV}
\]

2. **Energy to excite from \(n=1\) to \(n=2\)**: The energy difference between these levels is given by:

\[
\Delta E = E_2 - E_1 = -\frac{13.6 \, \text{eV}}{2^2} - \left(-\frac{13.6 \, \text{eV}}{1^2}\right) = -\frac{13.6}{4} + 13.6 = 10.2 \, \text{eV}
\]

**Wavelength of light absorbed**:

\[
\lambda = \frac{hc}{\Delta E}
\]

Where \(h = 6.626 \times 10^{-34}\) J·s (Planck's constant), and \(c = 3.00 \times 10^8\) m/s (speed of light). First, convert \(10.2 \, \text{eV}\) to Joules (1 eV = \(1.602 \times 10^{-19}\) J):

\[
\Delta E = 10.2 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} = 1.632 \times 10^{-18} \, \text{J}
\]

Now substitute into the wavelength formula:

\[
\lambda = \frac{(6.626 \times 10^{-34} \, \text{J·s})(3.00 \times 10^8 \, \text{m/s})}{1.632 \times 10^{-18} \, \text{J}} \approx 1.215 \times 10^{-6} \, \text{m} = 1215 \, \text{nm}
\]

3. **Hydrogen atom absorbs a photon to reach \(n=4\)**:
- **Change in energy**:

\[
\Delta E = E_4 - E_1 = -\frac{13.6 \, \text{eV}}{4^2} - \left(-\frac{13.6 \, \text{eV}}{1^2}\right) = -\frac{13.6}{16} + 13.6 = 12.1 \, \text{eV}
\]

- **Wavelength of the photon absorbed**:

\[
\lambda = \frac{hc}{\Delta E} \approx \frac{(6.626 \times 10^{-34} \, \text{J·s})(3.00 \times 10^8 \, \text{m/s})}{12.1 \times 1.602 \times 10^{-19} \, \text{J}} \approx 1.024 \times 10^{-7} \, \text{m} = 102.4 \, \text{nm}
\]

4. **Hydrogen atom in \(n=6\) emits IR light**:
- **Change in energy (to \(n=5\))**:

\[
\Delta E = E_5 - E_6 = -\frac{13.6}{5^2} - \left(-\frac{13.6}{6^2}\right) = -\frac{13.6}{25} + \frac{13.6}{36} \approx 0.434 \, \text{eV}
\]

- **Wavelength of the photon emitted**:

\[
\lambda = \frac{hc}{\Delta E} \approx \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{0.434 \times 1.602 \times 10^{-19}} \approx 2.865 \times 10^{-6} \, \text{m} = 2865 \, \text{nm} = 286.5 \, \text{µm}
\]

5. **Electron moves to \(n=4\) with a frequency of \(1.141 \times 10^{14} \, s^{-1}\)**:
- **Energy change**:

\[
E = h \nu = (6.626 \times 10^{-34})(1.141 \times 10^{14}) \approx 7.57 \times 10^{-20} \, \text{J} \approx 4.72 \, \text{eV}
\]

- **Initial energy level**: To find the initial energy level \(n\):

\[
\Delta E = E_n - E_4 \Rightarrow E_n = E_4 + 4.72 \approx -3.88 \, \text{eV} \Rightarrow n \approx 2
\]

(Solving \(E_n = -\frac{13.6 \, \text{eV}}{n^2}\) gives \(n \approx 2\))

6. **Energy emitted from \(n=3\) to \(n=1\)**:

\[
\Delta E = E_1 - E_3 = -\frac{13.6 \, \text{eV}}{1^2} - (-\frac{13.6 \, \text{eV}}{3^2}) = -13.6 + \frac{13.6}{9} = -13.6 + 1.51 = 12.09 \, \text{eV}
\]

In summary, the calculations are structured following the principles of quantum mechanics related to the hydrogen atom, utilizing the energy levels of the hydrogen atom defined by Bohr's model. The results include changes in energy, wavelength conversions, and frequency calculations.