To find the de Broglie wavelength of an electron in the second excited state of a hydrogen atom, we can follow these steps:
### Step 1: Identify the quantum number for the second excited state
The second excited state corresponds to the principal quantum number \( n = 3 \).
### Step 2: Calculate the energy of the electron in the second excited state
The energy of an electron in a hydrogen-like atom is given by the formula:
\[
E_n = -\frac{13.6 \, \text{eV}}{n^2}
\]
For \( n = 3 \):
\[
E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} = -1.51 \, \text{eV}
\]
### Step 3: Convert the energy from electron volts to joules
To convert electron volts to joules, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[
E_3 = -1.51 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = -2.416 \times 10^{-19} \, \text{J}
\]
### Step 4: Use the de Broglie wavelength formula
The de Broglie wavelength \( \lambda \) can be calculated using the formula:
\[
\lambda = \frac{h}{p}
\]
where \( p \) is the momentum. The momentum can also be expressed in terms of energy:
\[
p = \sqrt{2mE}
\]
Thus,
\[
\lambda = \frac{h}{\sqrt{2mE}}
\]
### Step 5: Substitute known values
Planck's constant \( h = 6.626 \times 10^{-34} \, \text{J s} \) and the mass of the electron \( m = 9.1 \times 10^{-31} \, \text{kg} \):
\[
\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 2.416 \times 10^{-19}}}
\]
### Step 6: Calculate the denominator
Calculating the denominator:
\[
2mE = 2 \times 9.1 \times 10^{-31} \times 2.416 \times 10^{-19} = 4.394 \times 10^{-49}
\]
Now take the square root:
\[
\sqrt{2mE} = \sqrt{4.394 \times 10^{-49}} \approx 6.64 \times 10^{-25}
\]
### Step 7: Calculate the de Broglie wavelength
Now substituting back into the wavelength formula:
\[
\lambda = \frac{6.626 \times 10^{-34}}{6.64 \times 10^{-25}} \approx 1.00 \times 10^{-9} \, \text{m}
\]
### Step 8: Convert to angstroms
Since \( 1 \, \text{angstrom} = 10^{-10} \, \text{m} \):
\[
\lambda \approx 10 \, \text{angstroms}
\]
### Final Answer
The de Broglie wavelength of the electron in the second excited state of the hydrogen atom is approximately \( 10 \, \text{angstroms} \).
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