The velocity of an electron in the orbit of hydrogen atom is `5.47xx10^(5)`m/sec. Total number of waves formed by the electron in one complete revolution is

Explanation

To find the total number of waves formed by an electron in one complete revolution in a hydrogen atom, we can use the de Broglie wavelength concept and the relationship between velocity, wavelength, and frequency. Here’s a step-by-step solution:

### Step 1: Understand the relationship
The number of waves \( N \) formed in one complete revolution can be calculated using the formula:
\[ N = \frac{v}{\lambda} \]
where \( v \) is the velocity of the electron and \( \lambda \) is the wavelength.

### Step 2: Calculate the de Broglie wavelength
The de Broglie wavelength \( \lambda \) of an electron can be calculated using the formula:
\[ \lambda = \frac{h}{mv} \]
where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( m \) is the mass of the electron (\( 9.11 \times 10^{-31} \, \text{kg} \)),
- \( v \) is the velocity of the electron (\( 5.47 \times 10^{5} \, \text{m/s} \)).

### Step 3: Substitute the values
Substituting the known values into the wavelength formula:
\[ \lambda = \frac{6.626 \times 10^{-34}}{(9.11 \times 10^{-31})(5.47 \times 10^{5})} \]

### Step 4: Calculate the wavelength
Calculating the denominator:
\[ m \cdot v = 9.11 \times 10^{-31} \times 5.47 \times 10^{5} \approx 4.975 \times 10^{-25} \, \text{kg m/s} \]
Now substituting back to find \( \lambda \):
\[ \lambda = \frac{6.626 \times 10^{-34}}{4.975 \times 10^{-25}} \approx 1.33 \times 10^{-9} \, \text{m} \]

### Step 5: Calculate the number of waves
Now we can calculate \( N \):
\[ N = \frac{v}{\lambda} = \frac{5.47 \times 10^{5}}{1.33 \times 10^{-9}} \approx 4.12 \times 10^{14} \]

### Step 6: Conclusion
The total number of waves formed by the electron in one complete revolution is approximately \( 4.12 \times 10^{14} \).