If energy of the electron in hydrogen atom in some excited state is -3.4e V then what will be its angular momentum

Options

A. `1.8 xx10^(-30) kgm^(2)s^(-1)`
B. `2.1 xx10^(-34) kgms^(2) s^(-1)`
C. `9.2 xx 10^(-37) kgm^(2) S^(-1)`
D. `1.2 xx10^(-32) kgm^(2) s^(-1)`

Answer

To find the angular momentum of an electron in a hydrogen atom given its energy in an excited state, we can follow these steps:

### Step 1: Understand the Energy Formula
The energy of an electron in a hydrogen atom can be calculated using the formula:
\[ E_n = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
where \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)) and \( n \) is the principal quantum number.

### Step 2: Substitute the Given Energy
We are given that the energy \( E \) is -3.4 eV. We can set up the equation:
\[ -3.4 = -\frac{13.6 \cdot 1^2}{n^2} \]

### Step 3: Solve for \( n^2 \)
Removing the negative signs, we have:
\[ 3.4 = \frac{13.6}{n^2} \]
Now, cross-multiplying gives:
\[ 3.4 n^2 = 13.6 \]
Dividing both sides by 3.4:
\[ n^2 = \frac{13.6}{3.4} = 4 \]
Taking the square root:
\[ n = 2 \]

### Step 4: Calculate Angular Momentum
The angular momentum \( L \) of an electron in a hydrogen atom is given by the formula:
\[ L = n \frac{h}{2\pi} \]
where \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \, \text{J s} \).

### Step 5: Substitute \( n \) and Calculate
Substituting \( n = 2 \):
\[ L = 2 \cdot \frac{6.626 \times 10^{-34}}{2\pi} \]
Calculating \( 2\pi \):
\[ 2\pi \approx 6.283 \]
Thus,
\[ L = 2 \cdot \frac{6.626 \times 10^{-34}}{6.283} \approx \frac{13.252 \times 10^{-34}}{6.283} \approx 2.11 \times 10^{-34} \, \text{J s} \]

### Step 6: Final Answer
The angular momentum of the electron in the hydrogen atom in the excited state where its energy is -3.4 eV is:
\[ L \approx 2.11 \times 10^{-34} \, \text{J s} \]