Find de-Broglie wavelength of single electron in 2 nd orbit of hydrogen atom by two methods.

Explanation

To find the de-Broglie wavelength of a single electron in the second orbit of a hydrogen atom, we can use two different methods. Let's go through each method step by step.

### Method 1: Using Kinetic Energy

1. **Identify the Kinetic Energy of the Electron:**
The kinetic energy (KE) of an electron in the nth orbit of a hydrogen atom can be calculated using the formula:
\[ KE = \frac{13.6 \, \text{eV}}{n^2} \]
For the second orbit (n = 2):
\[ KE = \frac{13.6 \, \text{eV}}{2^2} = \frac{13.6 \, \text{eV}}{4} = 3.4 \, \text{eV} \]

2. **Use the de-Broglie Wavelength Formula:**
The de-Broglie wavelength (\(\lambda\)) can be calculated using the formula:
\[ \lambda = \sqrt{\frac{150}{KE}} \]
Substituting the kinetic energy we found:
\[ \lambda = \sqrt{\frac{150}{3.4}} \]

3. **Calculate the Wavelength:**
\[ \lambda = \sqrt{44.1176} \approx 6.64 \, \text{Å} \]

### Method 2: Using Circular Orbit and Radius

1. **Determine the Radius of the Second Orbit:**
The radius of the nth orbit in a hydrogen atom is given by:
\[ r_n = r_0 \cdot \frac{n^2}{Z} \]
For hydrogen (\(Z = 1\)), and for the second orbit (\(n = 2\)):
\[ r_2 = r_0 \cdot 2^2 = r_0 \cdot 4 \]
Given \(r_0 = 0.529 \, \text{Å}\):
\[ r_2 = 0.529 \cdot 4 = 2.116 \, \text{Å} \]

2. **Calculate the Circumference of the Orbit:**
The circumference of the orbit is given by:
\[ C = 2 \pi r \]
Therefore:
\[ C = 2 \pi r_2 = 2 \pi (2.116 \, \text{Å}) \approx 13.29 \, \text{Å} \]

3. **Relate the Wavelength to the Orbit:**
The de-Broglie wavelength is related to the circumference of the orbit:
\[ C = n \lambda \]
For \(n = 2\):
\[ 2 \lambda = C \implies \lambda = \frac{C}{2} = \frac{13.29 \, \text{Å}}{2} \approx 6.64 \, \text{Å} \]

### Conclusion
From both methods, we find that the de-Broglie wavelength of a single electron in the second orbit of a hydrogen atom is approximately \(6.64 \, \text{Å}\).