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) kg ms s^(-1)`
C. `9.2 xx 10^(-37) kgm^(2) S^(-1)`
D. `1.2 xx10^(-32) kgm^(2) s^(-1)`

Answer

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

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

### Step 2: Substitute the known values
Given that the energy \(E_n = -3.4 \, \text{eV}\), we can substitute \(Z = 1\) into the equation:
\[ -3.4 = -\frac{13.6 \cdot 1^2}{n^2} \]

### Step 3: Simplify the equation
Removing the negative signs gives:
\[ 3.4 = \frac{13.6}{n^2} \]

### Step 4: Solve for \(n^2\)
Rearranging the equation to find \(n^2\):
\[ n^2 = \frac{13.6}{3.4} \]

### Step 5: Calculate \(n^2\)
Perform the division:
\[ n^2 = 4 \]
Thus, \(n = 2\).

### Step 6: Use the angular momentum formula
The angular momentum \(L\) of an electron in a hydrogen atom is given by:
\[ L = \frac{n h}{2\pi} \]
where \(h\) is Planck's constant.

### Step 7: Substitute \(n\) and \(h\)
Substituting \(n = 2\) into the formula:
\[ L = \frac{2h}{2\pi} = \frac{h}{\pi} \]

### Step 8: Substitute the value of Planck's constant
Using \(h = 6.626 \times 10^{-34} \, \text{J s}\):
\[ L = \frac{6.626 \times 10^{-34}}{\pi} \]

### Step 9: Calculate the angular momentum
Using \(\pi \approx 3.14\):
\[ L \approx \frac{6.626 \times 10^{-34}}{3.14} \approx 2.11 \times 10^{-34} \, \text{kg m}^2/\text{s} \]

### Final Result
Thus, the angular momentum of the electron in the hydrogen atom in the excited state is approximately:
\[ L \approx 2.1 \times 10^{-34} \, \text{kg m}^2/\text{s} \]