An electron in Bohr's hydrogen atom has an energy of -3.4 eV. The angular momentum of the electron is

Options

A. `(h)/(pi)`
B. `(h)/(2pi)`
C. `(nh)/(2pi)`(n is an integer)
D. `(2h)/(pi)`

Explanation

To find the angular momentum of an electron in a hydrogen atom with an energy of -3.4 eV, we can follow these steps:

### Step 1: Identify the energy formula for the electron in a hydrogen atom.
The energy of an electron in the nth orbit of a hydrogen atom is given by the formula:
\[ E = -\frac{13.6 \, \text{eV}}{n^2} \]
where \(E\) is the energy and \(n\) is the principal quantum number (the orbit number).

### Step 2: Substitute the given energy into the formula.
We are given that the energy \(E = -3.4 \, \text{eV}\). We can set up the equation:
\[ -3.4 = -\frac{13.6}{n^2} \]

### Step 3: Solve for \(n^2\).
Removing the negative signs from both sides, we have:
\[ 3.4 = \frac{13.6}{n^2} \]
Now, cross-multiplying gives:
\[ 3.4n^2 = 13.6 \]
Dividing both sides by 3.4:
\[ n^2 = \frac{13.6}{3.4} = 4 \]

### Step 4: Find \(n\).
Taking the square root of both sides:
\[ n = 2 \]

### Step 5: Use the angular momentum formula.
According to Bohr's model, the angular momentum \(L\) of an electron in the nth orbit is given by:
\[ L = n \cdot \frac{h}{2\pi} \]
Substituting \(n = 2\):
\[ L = 2 \cdot \frac{h}{2\pi} = \frac{h}{\pi} \]

### Final Answer:
Thus, the angular momentum of the electron is:
\[ L = \frac{h}{\pi} \]