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

Options

A. `h/pi`
B. `"3h"/(2pi)`
C. `"5h"/(2pi)`
D. `"2h"/(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: Use the formula for the energy of an electron in a Bohr orbit.
The energy of an electron in the nth orbit of 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 in the nth orbit.

### Step 2: Set the energy equal to -0.544 eV.
We know from the problem that:
\[ E_n = -0.544 \, \text{eV} \]
Setting the two equations equal gives us:
\[ -\frac{13.6 \, \text{eV}}{n^2} = -0.544 \, \text{eV} \]

### Step 3: Solve for \(n^2\).
Removing the negative signs and rearranging gives:
\[ \frac{13.6 \, \text{eV}}{n^2} = 0.544 \, \text{eV} \]
Multiplying both sides by \(n^2\) and then dividing by 0.544 eV gives:
\[ n^2 = \frac{13.6 \, \text{eV}}{0.544 \, \text{eV}} \approx 25 \]
Taking the square root of both sides gives:
\[ n = 5 \]

### Step 4: Use the formula for angular momentum.
The angular momentum \(L\) of an electron in the nth orbit of a hydrogen atom is given by:
\[ L = n \frac{h}{2\pi} \]
where \(h\) is Planck's constant.

### Step 5: Substitute \(n\) into the angular momentum formula.
Substituting \(n = 5\) into the angular momentum formula gives:
\[ L = 5 \frac{h}{2\pi} \]

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

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