What is the de-Broglie wavelength associated with the hydrogen electron in its third orbit

Options

A. `9.96xx10^(-10)` cm
B. `9.96xx10^(-8)` cm
C. `9.96xx10^(4)` cm
D. `9.96xx10^(8)` cm

Explanation

To find the de-Broglie wavelength associated with the hydrogen electron in its third orbit, follow these steps:

### Step 1: Determine the velocity of the electron in the third orbit
The velocity of an electron in the nth orbit of a hydrogen atom is given by:
\[ v_n = \frac{2.18 \times 10^6 \, \text{m/s} \cdot Z}{n} \]
For hydrogen (Z = 1) in the third orbit (n = 3):
\[ v_3 = \frac{2.18 \times 10^6 \, \text{m/s} \cdot 1}{3} = \frac{2.18 \times 10^6}{3} \, \text{m/s} \]
\[ v_3 = 0.726 \times 10^6 \, \text{m/s} \]
\[ v_3 = 7.26 \times 10^5 \, \text{m/s} \]

### Step 2: Use the de-Broglie wavelength formula
The de-Broglie wavelength (\(\lambda\)) is given by:
\[ \lambda = \frac{h}{mv} \]
where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \))
- \( m \) is the mass of the electron (\( 9.1 \times 10^{-31} \, \text{kg} \))
- \( v \) is the velocity of the electron

### Step 3: Substitute the values into the formula
\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{Js}}{(9.1 \times 10^{-31} \, \text{kg}) \times (7.26 \times 10^5 \, \text{m/s})} \]

### Step 4: Perform the calculation
\[ \lambda = \frac{6.626 \times 10^{-34}}{9.1 \times 7.26 \times 10^{-31} \times 10^5} \]
\[ \lambda = \frac{6.626 \times 10^{-34}}{66.066 \times 10^{-26}} \]
\[ \lambda = \frac{6.626}{66.066} \times 10^{-34 + 26} \]
\[ \lambda = 0.1003 \times 10^{-8} \, \text{m} \]
\[ \lambda = 1.003 \times 10^{-10} \, \text{m} \]

### Final Answer
The de-Broglie wavelength associated with the hydrogen electron in its third orbit is:
\[ \lambda = 1.003 \times 10^{-10} \, \text{m} \]