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 the electron in a hydrogen atom given that its angular momentum is \( \frac{h}{\pi} \), we can follow these steps:

### Step 1: Relate Angular Momentum to Quantum Number
According to Bohr's model of the atom, the angular momentum \( L \) of an electron in a hydrogen atom is given by the formula:
\[ L = n \frac{h}{2\pi} \]
where \( n \) is the principal quantum number.

Given that \( L = \frac{h}{\pi} \), we can set up the equation:
\[ \frac{h}{\pi} = n \frac{h}{2\pi} \]

### Step 2: Solve for the Principal Quantum Number \( n \)
To find \( n \), we can cancel \( h \) from both sides (assuming \( h \neq 0 \)):
\[ 1 = n \frac{1}{2} \]
Multiplying both sides by 2 gives:
\[ n = 2 \]

### Step 3: Use the Kinetic Energy Formula
The kinetic energy \( K \) of an electron in a hydrogen atom is given by the formula:
\[ K = \frac{13.6 Z^2}{n^2} \text{ eV} \]
For hydrogen, the atomic number \( Z = 1 \). Substituting \( Z \) and \( n \) into the formula:
\[ K = \frac{13.6 \times 1^2}{2^2} \text{ eV} \]

### Step 4: Calculate the Kinetic Energy
Now, we can calculate:
\[ K = \frac{13.6}{4} \text{ eV} = 3.4 \text{ eV} \]

### Final Answer
The kinetic energy of the electron in the hydrogen atom is:
\[ \boxed{3.4 \text{ eV}} \]

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