When an electron in hydrogen atom is excited, from its `4`th to `5`th stationary orbit, the change in angular momentum of electron is (Planck's constant: `h = 6.6 xx 10^(-34) J-s`)

Options

A. `4.16xx10^(-34) J-s`
B. `3.32xx10^(-34)J-s`
C. `1.05xx10^(-34) J-s`
D. None of these

Explanation

To find the change in angular momentum of an electron in a hydrogen atom when it is excited from the 4th to the 5th stationary orbit, we can use the formula for angular momentum in a stationary orbit.

### Step-by-Step Solution:

1. **Understand the formula for angular momentum**:
The angular momentum \( L \) of an electron in the \( n \)th orbit is given by:
\[ L_n = \frac{n h}{2 \pi} \]
where \( n \) is the principal quantum number and \( h \) is Planck's constant.

2. **Calculate angular momentum for the 4th orbit**:
For the 4th orbit (\( n = 4 \)):
\[ L_4 = \frac{4h}{2\pi} \]

3. **Calculate angular momentum for the 5th orbit**:
For the 5th orbit (\( n = 5 \)):
\[ L_5 = \frac{5h}{2\pi} \]

4. **Find the change in angular momentum**:
The change in angular momentum \( \Delta L \) when the electron moves from the 4th to the 5th orbit is given by:
\[ \Delta L = L_5 - L_4 = \left(\frac{5h}{2\pi}\right) - \left(\frac{4h}{2\pi}\right) \]
Simplifying this gives:
\[ \Delta L = \frac{(5 - 4)h}{2\pi} = \frac{h}{2\pi} \]

5. **Substitute the value of Planck's constant**:
Given \( h = 6.6 \times 10^{-34} \, \text{J-s} \):
\[ \Delta L = \frac{6.6 \times 10^{-34}}{2\pi} \]

6. **Calculate the numerical value**:
Using \( \pi \approx 3.14 \):
\[ \Delta L = \frac{6.6 \times 10^{-34}}{2 \times 3.14} = \frac{6.6 \times 10^{-34}}{6.28} \approx 1.05 \times 10^{-34} \, \text{J-s} \]

### Final Answer:
The change in angular momentum of the electron when it is excited from the 4th to the 5th stationary orbit is:
\[ \Delta L \approx 1.05 \times 10^{-34} \, \text{J-s} \]