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. `2.11 xx 10^(-34)`
B. `3 xx 10^(-34)`
C. `1.055 xx 10^(-34)`
D. `0.5 xx 10^(-34)`

Explanation

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

### Step 1: Relate Energy to Principal Quantum Number
The energy of an electron in a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \text{ eV}}{n^2} \]
where \( E_n \) is the energy at the nth level.

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

### Step 3: Solve for \( n^2 \)
Removing the negative signs and rearranging gives:
\[ 3.4 = \frac{13.6}{n^2} \]
Multiplying both sides by \( n^2 \) and then dividing by 3.4 gives:
\[ n^2 = \frac{13.6}{3.4} \]

### Step 4: Calculate \( n^2 \)
Calculating the right side:
\[ n^2 = \frac{13.6}{3.4} = 4 \]
Taking the square root gives:
\[ n = 2 \]

### Step 5: Calculate Angular Momentum
The angular momentum \( L \) of an electron in a hydrogen atom is given by:
\[ L = n \frac{h}{2\pi} \]
Substituting \( n = 2 \) into the equation gives:
\[ L = 2 \frac{h}{2\pi} = \frac{h}{\pi} \]

### Step 6: Substitute the Value of \( h \)
We know \( h = 6.626 \times 10^{-34} \text{ J s} \). Therefore:
\[ L = \frac{6.626 \times 10^{-34}}{\pi} \]

### Step 7: Calculate \( L \)
Using \( \pi \approx 3.14 \):
\[ L = \frac{6.626 \times 10^{-34}}{3.14} \approx 2.11 \times 10^{-34} \text{ J s} \]

### Final Answer
The angular momentum of the electron in the excited hydrogen atom is:
\[ L \approx 2.11 \times 10^{-34} \text{ J s} \]

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