Energy of an electron in an excited hydrogen atom is `-3.4eV`. Its angualr momentum will be: `h = 6.626 xx 10^(-34) J-s`.

Options

A. `1.11 xx 10^(34) J-s`
B. `1.51 xx 10^(-31)J-s`
C. `2.11 xx 10^(-34)J-s`
D. `3.72 xx 10^(-34)J-s`

Explanation

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

### Step 1: Identify the principal quantum number (n)
The energy of an electron in a hydrogen atom is given by the formula:

\[ E = -\frac{13.6 \, \text{eV}}{n^2} \]

Given that the energy \( E = -3.4 \, \text{eV} \), we can set up the equation:

\[ -3.4 = -\frac{13.6}{n^2} \]

### Step 2: Solve for n
Removing the negative signs and rearranging gives:

\[ 3.4 = \frac{13.6}{n^2} \]

Multiplying both sides by \( n^2 \):

\[ 3.4 n^2 = 13.6 \]

Now, divide both sides by 3.4:

\[ n^2 = \frac{13.6}{3.4} = 4 \]

Taking the square root gives:

\[ n = 2 \]

### Step 3: Calculate the angular momentum (L)
The angular momentum \( L \) of an electron in a hydrogen atom is given by the formula:

\[ L = \frac{n h}{2 \pi} \]

Substituting \( n = 2 \) and \( h = 6.626 \times 10^{-34} \, \text{J s} \):

\[ L = \frac{2 \times 6.626 \times 10^{-34}}{2 \pi} \]

### Step 4: Simplify the expression
The 2s in the numerator and denominator cancel out:

\[ L = \frac{6.626 \times 10^{-34}}{\pi} \]

Using \( \pi \approx 3.14 \):

\[ L \approx \frac{6.626 \times 10^{-34}}{3.14} \approx 2.11 \times 10^{-34} \, \text{J s} \]

### Final Answer
Thus, the angular momentum of the electron in the excited hydrogen atom is:

\[ L \approx 2.11 \times 10^{-34} \, \text{J s} \]