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

### Step 1: Use the formula for energy levels in the Bohr model
The energy levels of a hydrogen atom in the Bohr model are given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \(E_n\) is the energy of the nth level and \(n\) is the principal quantum number.

### Step 2: Set up the equation for the given energy
We know the energy of the state is -1.51 eV. Therefore, we can set up the equation:
\[ -1.51 = -\frac{13.6}{n^2} \]

### Step 3: Solve for \(n^2\)
Rearranging the equation gives us:
\[ 1.51 = \frac{13.6}{n^2} \]
Multiplying both sides by \(n^2\):
\[ 1.51 n^2 = 13.6 \]
Now, divide both sides by 1.51:
\[ n^2 = \frac{13.6}{1.51} \]
Calculating this gives:
\[ n^2 \approx 9.01 \]
Taking the square root gives:
\[ n \approx 3 \]

### Step 4: Calculate the angular momentum
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 \(6.626 \times 10^{-34} \, \text{Js}\).

Substituting \(n = 3\):
\[ L_n = 3 \cdot \frac{6.626 \times 10^{-34}}{2\pi} \]

### Step 5: Calculate the value
Calculating the value:
\[ L_n = 3 \cdot \frac{6.626 \times 10^{-34}}{6.2832} \approx 3.16 \times 10^{-34} \, \text{kg m}^2/\text{s} \]

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

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