The ratio of angular momentum of electron in two successive orbit is a `(a gt1)` and their difference is b . Then `a/b` is equal to `

Options

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

Answer

To solve the problem, we need to find the ratio \( \frac{a}{b} \) where \( a \) is the ratio of angular momentum of an electron in two successive orbits and \( b \) is their difference.

### Step-by-Step Solution:

1. **Understanding Angular Momentum in Orbits**:
The angular momentum \( L \) of an electron in a given orbit \( n \) is given by the formula:
\[ L_n = n \frac{h}{2\pi} \]
where \( h \) is Planck's constant.

2. **Identifying Successive Orbits**:
Let’s denote the angular momentum of the electron in the first orbit (n) as \( L_1 \) and in the second orbit (n+1) as \( L_2 \):
\[ L_1 = n \frac{h}{2\pi} \]
\[ L_2 = (n+1) \frac{h}{2\pi} \]

3. **Calculating the Ratio \( a \)**:
The ratio \( a \) of angular momentum in two successive orbits is:
\[ a = \frac{L_2}{L_1} = \frac{(n+1) \frac{h}{2\pi}}{n \frac{h}{2\pi}} = \frac{n+1}{n} \]

4. **Calculating the Difference \( b \)**:
The difference \( b \) in angular momentum between the two orbits is:
\[ b = L_2 - L_1 = \left((n+1) \frac{h}{2\pi}\right) - \left(n \frac{h}{2\pi}\right) = \frac{h}{2\pi} \]

5. **Finding \( \frac{a}{b} \)**:
Now we can calculate \( \frac{a}{b} \):
\[ \frac{a}{b} = \frac{\frac{n+1}{n}}{\frac{h}{2\pi}} = \frac{(n+1) \cdot 2\pi}{n \cdot h} \]

6. **Final Result**:
Therefore, the value of \( \frac{a}{b} \) is:
\[ \frac{a}{b} = \frac{(n+1) \cdot 2\pi}{n \cdot h} \]