Magnetic moment of an electron in hydrogen atom due to revolution around nuclons is `"hS"/(2pi)`. Here h is Planck's constant and S is specific charge of electron . Kinetic energy of this electron is

Options

A. a.4.53 eV
B. b.1.51 eV
C. c.3.4 eV
D. d.6.8 eV

Answer

To find the kinetic energy of the electron in a hydrogen atom, we start with the given magnetic moment and relate it to the angular momentum. Here’s a step-by-step solution:

### Step 1: Understand the Magnetic Moment
The magnetic moment \( \mu \) of the electron due to its revolution around the nucleus is given by:
\[ \mu = \frac{hS}{2\pi} \]
where \( h \) is Planck's constant and \( S \) is the specific charge of the electron.

### Step 2: Relate Magnetic Moment to Angular Momentum
The magnetic moment is also related to the angular momentum \( L \) of the electron:
\[ \mu = \frac{eL}{2m} \]
where \( e \) is the charge of the electron and \( m \) is its mass.

### Step 3: Equate the Two Expressions for Magnetic Moment
From the two expressions for magnetic moment, we can equate them:
\[ \frac{hS}{2\pi} = \frac{eL}{2m} \]

### Step 4: Express Specific Charge
The specific charge \( S \) is defined as:
\[ S = \frac{e}{m} \]
Substituting this into the equation gives:
\[ \frac{h \cdot \frac{e}{m}}{2\pi} = \frac{eL}{2m} \]

### Step 5: Simplify the Equation
Cancelling \( e \) and \( m \) from both sides (assuming \( e \neq 0 \) and \( m \neq 0 \)):
\[ \frac{h}{2\pi} = \frac{L}{2} \]
Multiplying both sides by 2 gives:
\[ \frac{h}{\pi} = L \]

### Step 6: Relate Angular Momentum to Quantum Number
The angular momentum \( L \) for an electron in a hydrogen atom can also be expressed in terms of the principal quantum number \( n \):
\[ L = n \frac{h}{2\pi} \]
Equating the two expressions for \( L \):
\[ \frac{h}{\pi} = n \frac{h}{2\pi} \]
Cancelling \( h \) from both sides (assuming \( h \neq 0 \)):
\[ 2 = n \]

### Step 7: Calculate Kinetic Energy
The kinetic energy \( K \) of the electron in a hydrogen atom is given by:
\[ K = -\frac{13.6 Z^2}{n^2} \text{ eV} \]
For hydrogen (\( Z = 1 \)) and \( n = 2 \):
\[ K = -\frac{13.6 \cdot 1^2}{2^2} = -\frac{13.6}{4} = -3.4 \text{ eV} \]

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