What is the angular momentum of an electron in Bohr's hydrogen atom whose energy is `-0.544 eV`?

Options

A. `(h)/(pi)`
B. `(2h)/(pi)`
C. `(5h)/(2pi)`
D. `(7h)/(2pi)`

Answer

To find the angular momentum of an electron in a Bohr hydrogen atom with an energy of -0.544 eV, we can follow these steps:

### Step 1: Understand the relationship between energy and the principal quantum number (n)
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 of the electron and \( n \) is the principal quantum number.

### Step 2: Set up the equation with the given energy
We know that the energy of the electron is -0.544 eV. Therefore, we can set up the equation:
\[ -0.544 = -\frac{13.6}{n^2} \]

### Step 3: Solve for n
Removing the negative signs from both sides gives:
\[ 0.544 = \frac{13.6}{n^2} \]
Now, rearranging the equation to solve for \( n^2 \):
\[ n^2 = \frac{13.6}{0.544} \]

Calculating the right-hand side:
\[ n^2 = 25 \]
Taking the square root gives:
\[ n = 5 \]

### Step 4: Calculate the angular momentum
The angular momentum \( L \) of an electron in a Bohr atom is given by:
\[ L = \frac{n h}{2 \pi} \]
Substituting the value of \( n \):
\[ L = \frac{5h}{2 \pi} \]

### Step 5: Conclusion
Thus, the angular momentum of the electron in the hydrogen atom whose energy is -0.544 eV is:
\[ L = \frac{5h}{2 \pi} \]