In the Bohr model of the hydrogen atom, what is the de-Broglie wavelength for the electron when it is in (a) the n=1 level?
(b) Then n=4 level? In each case, compare the de-Broglie wavelength to the circumference `2pir_n` of the orbit.

Explanation

To solve the problem of finding the de-Broglie wavelength of an electron in the hydrogen atom using the Bohr model, we'll follow these steps for both the n=1 and n=4 levels.

### Step-by-Step Solution:

### Part (a): For n=1 Level

1. **Determine the velocity of the electron (V1)**:
The formula for the velocity of the electron in the nth orbit is given by:
\[ V_n = \frac{e^2}{2 \epsilon_0 n h} \]
For n=1:
\[ V_1 = \frac{e^2}{2 \epsilon_0 \cdot 1 \cdot h} \]
Substituting the known constants:
\[ V_1 = 2.19 \times 10^6 \text{ m/s} \]

2. **Calculate the de-Broglie wavelength (λ1)**:
The de-Broglie wavelength is given by:
\[ \lambda = \frac{h}{m_e V} \]
Where \( h = 6.63 \times 10^{-34} \text{ J s} \) and \( m_e = 9.11 \times 10^{-31} \text{ kg} \).
Substituting the values:
\[ \lambda_1 = \frac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \cdot 2.19 \times 10^6} \]
Performing the calculation gives:
\[ \lambda_1 \approx 3.32 \times 10^{-10} \text{ m} \]

3. **Calculate the circumference of the orbit (C1)**:
The radius of the first orbit (r1) is given by:
\[ r_1 = 0.529 \times 10^{-10} \text{ m} \]
The circumference is:
\[ C_1 = 2 \pi r_1 = 2 \cdot 3.14 \cdot 0.529 \times 10^{-10} \approx 3.32 \times 10^{-10} \text{ m} \]

4. **Comparison**:
\[ \lambda_1 = C_1 \Rightarrow 3.32 \times 10^{-10} \text{ m} = 3.32 \times 10^{-10} \text{ m} \]

### Part (b): For n=4 Level

1. **Determine the velocity of the electron (V4)**:
The velocity for n=4 is:
\[ V_4 = \frac{V_1}{4} = \frac{2.19 \times 10^6}{4} = 5.475 \times 10^5 \text{ m/s} \]

2. **Calculate the de-Broglie wavelength (λ4)**:
\[ \lambda_4 = \frac{h}{m_e V_4} \]
Substituting the values:
\[ \lambda_4 = \frac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \cdot 5.475 \times 10^5} \]
Performing the calculation gives:
\[ \lambda_4 \approx 1.328 \times 10^{-10} \text{ m} \]

3. **Calculate the circumference of the orbit (C4)**:
The radius of the fourth orbit (r4) is:
\[ r_4 = r_1 n^2 = 0.529 \times 10^{-10} \cdot 4^2 = 0.529 \times 10^{-10} \cdot 16 = 8.464 \times 10^{-10} \text{ m} \]
The circumference is:
\[ C_4 = 2 \pi r_4 = 2 \cdot 3.14 \cdot 8.464 \times 10^{-10} \approx 1.328 \times 10^{-10} \text{ m} \]

4. **Comparison**:
\[ \lambda_4 = C_4 \Rightarrow 1.328 \times 10^{-10} \text{ m} = 1.328 \times 10^{-10} \text{ m} \]

### Final Results:
- For n=1:
- De-Broglie wavelength \( \lambda_1 \approx 3.32 \times 10^{-10} \text{ m} \)
- Circumference \( C_1 \approx 3.32 \times 10^{-10} \text{ m} \)

- For n=4:
- De-Broglie wavelength \( \lambda_4 \approx 1.328 \times 10^{-10} \text{ m} \)
- Circumference \( C_4 \approx 1.328 \times 10^{-10} \text{ m} \)