The velocity of electron in hydrogen atom is `7.29 xx 10^(7)` cm/sec The potential energy of that electron is

Options

A. `-13.6 eV`
B. `-3.4 eV`
C. `-3.02 eV`
D. `-1.70 eV`

Answer

To find the potential energy of the electron in a hydrogen atom given its velocity, we can follow these steps:

### Step 1: Convert the velocity from cm/s to m/s
The given velocity of the electron is \( 7.29 \times 10^7 \) cm/s. To convert this to meters per second, we use the conversion factor \( 1 \text{ cm} = 0.01 \text{ m} \).

\[ \text{Velocity in m/s} = 7.29 \times 10^7 \text{ cm/s} \times 0.01 \text{ m/cm} = 7.29 \times 10^5 \text{ m/s} \]

### Step 2: Use the formula for the velocity of an electron in the nth orbit
The velocity of an electron in the nth orbit of a hydrogen atom is given by the formula:

\[ v = \frac{2.165 \times 10^6 \cdot Z}{n} \text{ m/s} \]

For hydrogen, \( Z = 1 \). Therefore, we can set up the equation:

\[ 7.29 \times 10^5 = \frac{2.165 \times 10^6}{n} \]

### Step 3: Solve for n
Rearranging the equation to solve for \( n \):

\[ n = \frac{2.165 \times 10^6}{7.29 \times 10^5} \]

Calculating this gives:

\[ n \approx 2.965 \approx 3 \]

So, the electron is in the third orbit.

### Step 4: Calculate the potential energy
The potential energy (PE) of an electron in the nth orbit of a hydrogen atom is given by the formula:

\[ PE = -\frac{Z e^2}{4 \pi \epsilon_0 n} \]

For hydrogen, \( Z = 1 \), and using the known values:
- \( e = 1.6 \times 10^{-19} \text{ C} \)
- \( \epsilon_0 \approx 8.85 \times 10^{-12} \text{ C}^2/\text{N m}^2 \)

Substituting these values into the formula:

\[ PE = -\frac{1 \cdot (1.6 \times 10^{-19})^2}{4 \pi (8.85 \times 10^{-12}) \cdot 3} \]

Calculating this gives:

\[ PE \approx -3.4 \text{ eV} \]

### Final Answer
The potential energy of the electron in the hydrogen atom is approximately \( -3.4 \text{ eV} \).

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