Question

Calculate the de Broglie wavelength associated with an electron revolving in the second excited state of hydrogen atom. The ground state energy of the hydrogen atom is – 13.6 eV. Take $m_e = 9 \times 10^{-31}$ kg, $h = 6.6 \times 10^{-34}$ J.s.

Hint:

An alternate, brilliantly faster method heavily utilizes Bohr's quantization condition $2\pi r = n\lambda$. If you remember the radius formula $r_n = 0.53 \text{ \AA} \times n^2$, you can cleanly bypass all heavy energy/joule conversions! $\lambda = 2\pi (0.53 \times 3^2) / 3 \approx 9.99 \text{ \AA} \approx 1 \text{ nm}$.

Answer 1

Concept:
• The quantized energy of an electron stably revolving in the $n$-th orbit of a standard hydrogen atom is robustly given by the formula $E_n = \frac{-13.6 \text{ eV}}{n^2}$.
• For any bound electron, its dynamic kinetic energy $K$ is exactly equal strictly to the positive magnitude of its total energy: $K = |E_n|$.
• The de Broglie wavelength uniquely associated with this revolving electron can be rapidly calculated using the standard matter-wave relation $\lambda = \frac{h}{\sqrt{2mK}}$.

Step 1: Identify the quantum state and calculate its energy
The problem specifically mentions the "second excited state".
In quantum mechanics, the ground state is $n=1$, the first excited state is $n=2$, and the second excited state strictly corresponds to $n=3$.
The total energy of the electron specifically in this $n=3$ state is:
\[ E_3 = \frac{-13.6}{3^2} \text{ eV} = \frac{-13.6}{9} \text{ eV} \]
\[ E_3 = -1.511 \text{ eV} \]
The kinetic energy $K$ is the absolute positive value of this total energy:
\[ K = 1.511 \text{ eV} \]

Step 2: Convert Kinetic Energy securely into Joules
To utilize standard SI physics formulas, we must absolutely convert this energy from electron-volts directly into standard Joules ($1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$):
\[ K = 1.511 \times 1.6 \times 10^{-19} \text{ J} \]
\[ K \approx 2.4176 \times 10^{-19} \text{ J} \]

Step 3: Calculate the de Broglie wavelength
We meticulously apply the de Broglie wavelength formula directly:
\[ \lambda = \frac{h}{\sqrt{2m_e K}} \]
Substitute the provided standard constants and our newly calculated kinetic energy:
\[ \lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times (9 \times 10^{-31}) \times (2.4176 \times 10^{-19})}} \]
First, carefully simplify the massive term securely trapped inside the square root:
\[ 2m_e K = 18 \times 10^{-31} \times 2.4176 \times 10^{-19} \]
\[ 2m_e K = 43.5168 \times 10^{-50} \]
Now, firmly take the mathematical square root of this value:
\[ \sqrt{43.5168 \times 10^{-50}} \approx 6.5967 \times 10^{-25} \text{ kg m/s} \]
Finally, cleanly divide Planck's constant by this calculated momentum:
\[ \lambda = \frac{6.6 \times 10^{-34}}{6.5967 \times 10^{-25}} \]
\[ \lambda \approx 1.0005 \times 10^{-9} \text{ m} \]

Step 4: Conclusion
The de Broglie wavelength strongly associated with the electron safely residing in the second excited state is mathematically evaluated to be approximately $1.0 \times 10^{-9} \text{ m}$ (or exactly $1 \text{ nm}$).