The angular momentum of an electron in hydrogen atom is `h/pi` The kinetic energy of the electron is

Options

A. 13.6 eV
B. 3.4 eV
C. 1.51 eV
D. 0.85 eV

Answer

To find the kinetic energy of an electron in a hydrogen atom given its angular momentum, we can follow these steps:

### Step 1: Understand the relationship between angular momentum and quantum number
The angular momentum \( L \) of an electron in a hydrogen atom is quantized and is given by the formula:
\[ L = n \frac{h}{2\pi} \]
where \( n \) is the principal quantum number and \( h \) is Planck's constant.

### Step 2: Set up the equation with the given angular momentum
We are given that the angular momentum \( L \) is \( \frac{h}{\pi} \). Setting this equal to the quantized angular momentum gives us:
\[ \frac{h}{\pi} = n \frac{h}{2\pi} \]

### Step 3: Simplify the equation
We can simplify the equation by canceling \( h \) from both sides:
\[ \frac{1}{\pi} = n \frac{1}{2\pi} \]
Multiplying both sides by \( 2\pi \) gives:
\[ 2 = n \]
Thus, the principal quantum number \( n \) is 2.

### Step 4: Use the formula for kinetic energy
The kinetic energy \( K \) of an electron in a hydrogen atom can be calculated using the formula:
\[ K = \frac{13.6 Z^2}{n^2} \text{ eV} \]
where \( Z \) is the atomic number. For hydrogen, \( Z = 1 \).

### Step 5: Substitute the values into the kinetic energy formula
Substituting \( Z = 1 \) and \( n = 2 \) into the kinetic energy formula gives:
\[ K = \frac{13.6 \cdot 1^2}{2^2} \text{ eV} = \frac{13.6}{4} \text{ eV} = 3.4 \text{ eV} \]

### Conclusion
The kinetic energy of the electron in the hydrogen atom is \( 3.4 \text{ eV} \).

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