The angular momentum of an electron in the hydrogen atom is `(3h)/(2pi)` Here. h is Planck's constant. The kinetic energy of this electron is

Options

A. `4.35 eV`
B. `1.51 eV`
C. `3.4 eV`
D. `6.8 eV`

Answer

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

### Step 1: Identify the given angular momentum
The angular momentum \( L \) of the electron is given as:
\[ L = \frac{3h}{2\pi} \]
where \( h \) is Planck's constant.

### Step 2: Relate angular momentum to the principal quantum number
The angular momentum 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.

### Step 3: Set the two expressions for angular momentum equal to each other
From the given information, we can equate the two expressions for angular momentum:
\[ n \frac{h}{2\pi} = \frac{3h}{2\pi} \]

### Step 4: Solve for the principal quantum number \( n \)
By simplifying the equation, we can find \( n \):
\[ n = 3 \]

### Step 5: Find the total energy for \( n = 3 \)
The total energy \( E \) of an electron in a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6}{n^2} \text{ eV} \]
For \( n = 3 \):
\[ E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} = -1.51 \text{ eV} \]

### Step 6: Relate total energy to kinetic energy
In a hydrogen atom, the total energy \( E \) is related to the kinetic energy \( K \) by the equation:
\[ E = K + V \]
where \( V \) is the potential energy. For a hydrogen atom, the kinetic energy is equal to half the magnitude of the total energy:
\[ K = -\frac{E}{2} \]
Thus, since the total energy \( E_3 = -1.51 \text{ eV} \):
\[ K = -\frac{-1.51}{2} = 0.755 \text{ eV} \]

### Step 7: Conclusion
The kinetic energy of the electron in the hydrogen atom when \( n = 3 \) is:
\[ K = 1.51 \text{ eV} \]

### Final Answer
The kinetic energy of this electron is \( 1.51 \text{ eV} \).

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