The angular momentum of the electron in the third orbit of H atom is

Options

A. `2.3 xx10^(-34) Js`
B. `3.2 xx10^(-34) Js`
C. `5.2 xx10^(-34) Js`
D. `4.2xx10^(-34) Js`

Explanation

To find the angular momentum of the electron in the third orbit of a hydrogen atom, we can follow these steps:

### Step 1: Understand the formula for angular momentum
The angular momentum \( L \) of an electron in an orbit is given by the formula:
\[ L = n \frac{h}{2\pi} \]
where:
- \( n \) is the principal quantum number (which indicates the orbit),
- \( h \) is Planck's constant, approximately \( 6.63 \times 10^{-34} \) joule-seconds,
- \( \pi \) is approximately \( 3.14 \).

### Step 2: Identify the values
For the third orbit, we have:
- \( n = 3 \)
- \( h = 6.63 \times 10^{-34} \) joule-seconds
- \( \pi \approx 3.14 \)

### Step 3: Substitute the values into the formula
Now we can substitute the values into the formula:
\[ L = 3 \times \frac{6.63 \times 10^{-34}}{2 \times 3.14} \]

### Step 4: Calculate the denominator
First, calculate the denominator:
\[ 2 \times 3.14 = 6.28 \]

### Step 5: Calculate the angular momentum
Now substitute this back into the equation:
\[ L = 3 \times \frac{6.63 \times 10^{-34}}{6.28} \]

### Step 6: Perform the division
Now we perform the division:
\[ \frac{6.63 \times 10^{-34}}{6.28} \approx 1.055 \times 10^{-34} \]

### Step 7: Multiply by 3
Now multiply by 3:
\[ L \approx 3 \times 1.055 \times 10^{-34} \approx 3.165 \times 10^{-34} \text{ joule-seconds} \]

### Step 8: Round off the answer
Rounding off to two significant figures, we get:
\[ L \approx 3.2 \times 10^{-34} \text{ joule-seconds} \]

### Final Answer
Thus, the angular momentum of the electron in the third orbit of the hydrogen atom is approximately:
\[ \boxed{3.2 \times 10^{-34}} \text{ joule-seconds} \]

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