To calculate the De Broglie wavelength of an electron in the 3rd orbit of a hydrogen atom, we need to go through several steps. Let's break down the calculations step by step.
### Step 1: Principal Quantum Number
The electron is in the 3rd orbit of a hydrogen atom, which means the principal quantum number [tex]\( n \)[/tex] is 3.
### Step 2: Energy of the Electron in the 3rd Orbit
The energy of an electron in the nth orbit of a hydrogen atom is given by:
[tex]\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \][/tex]
For [tex]\( n = 3 \)[/tex]:
[tex]\[ E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.511 \, \text{eV} \][/tex]
### Step 3: Convert Energy to Joules
To convert electron volts (eV) to Joules (J), use the conversion factor:
[tex]\[ 1 \, \text{eV} = 1.60218 \times 10^{-19} \, \text{J} \][/tex]
Therefore,
[tex]\[ E_3 \approx -1.511 \, \text{eV} \times 1.60218 \times 10^{-19} \, \text{J/eV} \approx -2.421072 \times 10^{-19} \, \text{J} \][/tex]
### Step 4: Calculate the Velocity of the Electron
Using the relationship between kinetic energy and velocity, where the kinetic energy [tex]\( E_k \)[/tex] is the magnitude of [tex]\( E_3 \)[/tex], we have:
[tex]\[ E_k = \frac{1}{2} mv^2 \][/tex]
Solving for velocity [tex]\( v \)[/tex]:
[tex]\[ v = \left( \frac{2E_k}{m} \right)^{0.5} \][/tex]
Here, the mass of the electron [tex]\( m \)[/tex] is:
[tex]\[ m = 9.10938356 \times 10^{-31} \, \text{kg} \][/tex]
Substitute [tex]\( E_k = 2.421072 \times 10^{-19} \, \text{J} \)[/tex]:
[tex]\[ v = \left( \frac{2 \times 2.421072 \times 10^{-19} \, \text{J}}{9.10938356 \times 10^{-31} \, \text{kg}} \right)^{0.5} \approx 729078.608 \, \text{m/s} \][/tex]
### Step 5: Calculate the De Broglie Wavelength
The De Broglie wavelength [tex]\( \lambda \)[/tex] is given by:
[tex]\[ \lambda = \frac{h}{mv} \][/tex]
Where [tex]\( h \)[/tex] is Planck's constant:
[tex]\[ h = 6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s} \][/tex]
Substitute the values:
[tex]\[ \lambda = \frac{6.62607015 \times 10^{-34}}{9.10938356 \times 10^{-31} \times 729078.608} \approx 9.976833 \times 10^{-10} \, \text{m} \][/tex]
### Step 6: Convert Wavelength to Angstroms
To convert from meters to angstroms, use the conversion factor:
[tex]\[ 1 \, \text{Å} = 10^{-10} \, \text{m} \][/tex]
Thus,
[tex]\[ \lambda \approx 9.976833 \times 10^{-10} \, \text{m} \times \frac{1 \, \text{Å}}{10^{-10} \, \text{m}} \approx 9.976833 \, \text{Å} \][/tex]
### Final Answer
The De Broglie wavelength of an electron in the 3rd orbit of a hydrogen atom is approximately [tex]\( 9.9768 \, \text{Å} \)[/tex].
