The difference in angular momentum associated with electron in two successive orbits of hydrogen atom is:

Options

A. `(h)/(pi)`
B. `(h)/(2pi)`
C. `(h)/(2)`
D. `((n-1)h)/(2pi)`

Answer

To find the difference in angular momentum associated with the electron in two successive orbits of a hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

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

2. **Determine the Angular Momentum for Successive Orbits**:
- For the nth orbit, the angular momentum is:
\[ L_n = \frac{n h}{2\pi} \]
- For the (n+1)th orbit, the angular momentum is:
\[ L_{n+1} = \frac{(n+1) h}{2\pi} \]

3. **Calculate the Difference in Angular Momentum**:
To find the difference in angular momentum between the (n+1)th orbit and the nth orbit, we subtract \( L_n \) from \( L_{n+1} \):
\[ \Delta L = L_{n+1} - L_n \]
Substituting the expressions we have:
\[ \Delta L = \frac{(n+1) h}{2\pi} - \frac{n h}{2\pi} \]

4. **Simplify the Expression**:
Combine the terms:
\[ \Delta L = \frac{(n+1) h - n h}{2\pi} = \frac{h}{2\pi} \]

5. **Final Result**:
Thus, the difference in angular momentum associated with the electron in two successive orbits of a hydrogen atom is:
\[ \Delta L = \frac{h}{2\pi} \]

### Conclusion:
The correct answer is \( \frac{h}{2\pi} \), which corresponds to option number 2.

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