The energy of an excited H-atom is `-3.4eV`. Calculate angular momentum of `e^(-)`

Explanation

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

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

### Step 2: Set up the equation with the given energy
We know the energy of the excited hydrogen atom is -3.4 eV. Therefore, we can set up the equation:
\[ -3.4 = -\frac{13.6}{n^2} \]

### Step 3: Solve for \( n^2 \)
Removing the negative signs, we have:
\[ 3.4 = \frac{13.6}{n^2} \]
Now, we can rearrange this to find \( n^2 \):
\[ n^2 = \frac{13.6}{3.4} \]

### Step 4: Calculate \( n^2 \)
Calculating the right-hand side:
\[ n^2 = \frac{13.6}{3.4} = 4 \]
Thus, we find:
\[ n = 2 \]

### Step 5: Use the value of \( n \) to calculate angular momentum
The angular momentum \( L \) of an electron in a hydrogen atom is given by:
\[ L = n \frac{h}{2\pi} \]
Substituting \( n = 2 \):
\[ L = 2 \frac{h}{2\pi} = \frac{h}{\pi} \]

### Final Answer
The angular momentum of the electron in the excited state of the hydrogen atom is:
\[ L = \frac{h}{\pi} \]

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