The energy of an electron in excited hydrogen atom is -3.4 eV . Then, according to Bohr's therory, the angular momentum of the electron of the electron is

Options

A. `2.1xx10^(-34) J-s`
B. `3xx10^(-34) J-s`
C. `2xx10^(-34) J-s`
D. `0.5xx10^(-34) J-s`

Explanation

To find the angular momentum of an electron in an excited hydrogen atom with energy -3.4 eV according to Bohr's theory, we can follow these steps:

### Step 1: Identify the Principal Quantum Number (n)
The energy of an electron in a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
Given that the energy \(E_n = -3.4 \, \text{eV}\), we can set up the equation:
\[ -3.4 = -\frac{13.6}{n^2} \]
This simplifies to:
\[ 3.4 = \frac{13.6}{n^2} \]
From this, we can solve for \(n^2\):
\[ n^2 = \frac{13.6}{3.4} \]
Calculating this gives:
\[ n^2 = 4 \quad \Rightarrow \quad n = 2 \]

### Step 2: Use Bohr's Angular Momentum Formula
According to Bohr's theory, the angular momentum \(L\) of an electron in an orbit is given by:
\[ L = n \frac{h}{2\pi} \]
Where:
- \(h\) is Planck's constant, approximately \(6.626 \times 10^{-34} \, \text{Js}\)
- \(n\) is the principal quantum number we found in Step 1, which is \(2\)

### Step 3: Substitute Values into the Formula
Now we can substitute \(n\) and \(h\) into the angular momentum formula:
\[ L = 2 \cdot \frac{6.626 \times 10^{-34}}{2\pi} \]
Calculating \(2\pi\):
\[ 2\pi \approx 6.28 \]
So we have:
\[ L = 2 \cdot \frac{6.626 \times 10^{-34}}{6.28} \]

### Step 4: Perform the Calculation
Now, we can calculate:
\[ L = \frac{2 \cdot 6.626 \times 10^{-34}}{6.28} \approx \frac{13.252 \times 10^{-34}}{6.28} \approx 2.11 \times 10^{-34} \, \text{Js} \]

### Final Answer
Thus, the angular momentum of the electron in the excited hydrogen atom is approximately:
\[ L \approx 2.11 \times 10^{-34} \, \text{Js} \]

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