A hydrogen atom in ground state absorbs 12.09 eV of energy . The orbital angular momentum of the electron

Options

A. Is doubles
B. Is halved
C. Remains same
D. Becomes three times

Answer

To find the orbital angular momentum of the electron in a hydrogen atom after it absorbs 12.09 eV of energy, we can follow these steps:

### Step 1: Identify the initial state energy
The energy of the hydrogen atom in the ground state (n=1) is given by:
\[ E_1 = -13.6 \, \text{eV} \]

### Step 2: Determine the energy absorbed
The energy absorbed by the hydrogen atom is:
\[ \Delta E = 12.09 \, \text{eV} \]

### Step 3: Calculate the final energy
The final energy after absorption can be calculated as:
\[ E_f = E_1 + \Delta E \]
Substituting the values:
\[ E_f = -13.6 \, \text{eV} + 12.09 \, \text{eV} = -1.51 \, \text{eV} \]

### Step 4: Use the energy formula to find n
The energy of the hydrogen atom in a state n is given by:
\[ E_n = -\frac{13.6}{n^2} \, \text{eV} \]
Setting \( E_n = -1.51 \, \text{eV} \):
\[ -\frac{13.6}{n^2} = -1.51 \]
This simplifies to:
\[ \frac{13.6}{n^2} = 1.51 \]
Rearranging gives:
\[ n^2 = \frac{13.6}{1.51} \]
Calculating this gives:
\[ n^2 \approx 9 \]
Thus:
\[ n = 3 \]

### Step 5: Calculate the orbital angular momentum
The orbital angular momentum \( L \) of an electron in a hydrogen atom is given by the formula:
\[ L = n \frac{h}{2\pi} \]
For \( n = 3 \):
\[ L = 3 \frac{h}{2\pi} \]

### Conclusion
The orbital angular momentum of the electron in the new state (after absorbing 12.09 eV) is:
\[ L = \frac{3h}{2\pi} \]