A hydrogen atom is in a state with energy `-1.51 eV.` in the Bohr model, what is the angular momentum of the electron in the atom with respect to an axis at the nucleus?

Explanation

To find the angular momentum of an electron in a hydrogen atom with an energy of -1.51 eV using the Bohr model, we can follow these steps:

### Step 1: Identify the Energy Formula
The energy of an electron in a hydrogen atom in the Bohr model is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( E_n \) is the energy of the electron in the nth orbit.

### Step 2: Set Up the Equation
We know the energy of the electron:
\[ -1.51 \, \text{eV} = -\frac{13.6 \, \text{eV}}{n^2} \]

### Step 3: Solve for \( n^2 \)
Rearranging the equation gives:
\[ n^2 = \frac{13.6 \, \text{eV}}{1.51 \, \text{eV}} \]

### Step 4: Calculate \( n^2 \)
Calculating the right-hand side:
\[ n^2 = \frac{13.6}{1.51} \approx 9.01 \]

### Step 5: Find \( n \)
Taking the square root gives:
\[ n \approx 3 \]

### Step 6: Use the Angular Momentum Formula
The angular momentum \( L_n \) of the electron in the nth orbit is given by:
\[ L_n = n \frac{h}{2\pi} \]
where \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \, \text{Js} \).

### Step 7: Substitute \( n \) into the Angular Momentum Formula
Substituting \( n = 3 \):
\[ L_n = 3 \cdot \frac{6.626 \times 10^{-34}}{2\pi} \]

### Step 8: Calculate \( L_n \)
Calculating the value:
\[ L_n = 3 \cdot \frac{6.626 \times 10^{-34}}{2 \cdot 3.14} \approx 3.16 \times 10^{-34} \, \text{kg m}^2/\text{s} \]

### Final Answer
The angular momentum of the electron in the hydrogen atom is approximately:
\[ L_n \approx 3.16 \times 10^{-34} \, \text{kg m}^2/\text{s} \]

---