The angular momentum of an electron in a given stationary state can be expressed as `m_(e)vr=n(h)/(2pi)`. Based on this expression an electron can move only in those orbits for which its angular momentum is

Options

A. equal to n
B. integral multiple of `h/(2pi)`
C. multiple of n
D. equal to `h/(2pi)` only.

Answer

To solve the problem regarding the angular momentum of an electron in a stationary state, we start with the given expression:

### Step 1: Understand the Angular Momentum Expression
The angular momentum \( L \) of an electron in a stationary state is given by the formula:
\[ L = m_e v r = \frac{n h}{2\pi} \]
where:
- \( m_e \) is the mass of the electron,
- \( v \) is the velocity of the electron,
- \( r \) is the radius of the orbit,
- \( n \) is the principal quantum number (an integer: 1, 2, 3, ...),
- \( h \) is Planck's constant.

### Step 2: Analyze the Formula
From the formula, we see that the angular momentum \( L \) is quantized. This means that \( L \) can only take specific discrete values based on the integer \( n \).

### Step 3: Determine the Values of Angular Momentum
For different values of \( n \):
- When \( n = 1 \):
\[ L = \frac{1h}{2\pi} = \frac{h}{2\pi} \]
- When \( n = 2 \):
\[ L = \frac{2h}{2\pi} = \frac{2h}{2\pi} \]
- When \( n = 3 \):
\[ L = \frac{3h}{2\pi} = \frac{3h}{2\pi} \]

Thus, the angular momentum can be expressed as:
\[ L = n \left(\frac{h}{2\pi}\right) \]

### Step 4: Conclude the Possible Values of Angular Momentum
From the analysis above, we conclude that the angular momentum of the electron can only take values that are integral multiples of \( \frac{h}{2\pi} \). Therefore, the electron can move only in those orbits for which its angular momentum is:
\[ L = n \left(\frac{h}{2\pi}\right) \quad \text{where } n = 1, 2, 3, \ldots \]

### Final Answer
The angular momentum of an electron in a given stationary state can only be integral multiples of \( \frac{h}{2\pi} \).

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