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 solve the question regarding the difference in angular momentum associated with an electron in two successive orbits of a hydrogen atom, we can follow these steps:

### Step 1: Understand the formula for angular momentum in Bohr's model
According to Bohr's atomic model, the angular momentum \( L_n \) of an electron in the \( n \)-th orbit is given by the formula:
\[ L_n = n \frac{h}{2\pi} \]
where \( h \) is Planck's constant and \( n \) is the principal quantum number (energy level).

### Step 2: Write the angular momentum for the successive orbit
For the \( (n+1) \)-th orbit, the angular momentum \( L_{n+1} \) is given by:
\[ L_{n+1} = (n+1) \frac{h}{2\pi} \]

### Step 3: Calculate the difference in angular momentum
To find the difference in angular momentum between the \( (n+1) \)-th orbit and the \( n \)-th orbit, we calculate:
\[ \Delta L = L_{n+1} - L_n \]
Substituting the expressions we derived:
\[ \Delta L = \left( (n+1) \frac{h}{2\pi} \right) - \left( n \frac{h}{2\pi} \right) \]

### Step 4: Simplify the expression
Now, simplifying the expression:
\[ \Delta L = \frac{h}{2\pi} \left( (n+1) - n \right) = \frac{h}{2\pi} \cdot 1 = \frac{h}{2\pi} \]

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

### Final Answer
The correct option is \( \frac{h}{2\pi} \).

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