An electron in the ground state of hydrogen has an angular momentum `L_(1)` and an electron in the fist excited state of lithium has an angular momentum `L_(2)`. Then,

Options

A. `L_(1) = L_(2)`
B. `L_(1) = 4 L_(2)`
C. `L_(2) =2 L_(1)`
D. `L_(1) = 2 L_(2)`

Explanation

To solve the problem, we need to calculate the angular momentum of an electron in the ground state of hydrogen and the first excited state of lithium using Bohr's quantization rule.

### Step 1: Calculate Angular Momentum for Hydrogen (L1)
According to Bohr's quantization rule, the angular momentum \( L \) of an electron in a hydrogen atom is given by the formula:

\[ L = n \frac{h}{2\pi} \]

For the ground state of hydrogen, the principal quantum number \( n \) is 1. Therefore, we can substitute \( n = 1 \) into the formula:

\[ L_1 = 1 \cdot \frac{h}{2\pi} = \frac{h}{2\pi} \]

### Step 2: Calculate Angular Momentum for Lithium (L2)
For lithium, we need to find the angular momentum of the electron in the first excited state. In this case, the principal quantum number \( n \) is 2. Thus, we can use the same formula:

\[ L_2 = n \frac{h}{2\pi} \]

Substituting \( n = 2 \):

\[ L_2 = 2 \cdot \frac{h}{2\pi} = \frac{2h}{2\pi} = \frac{h}{\pi} \]

### Step 3: Compare L1 and L2
Now we have both angular momenta:

- \( L_1 = \frac{h}{2\pi} \)
- \( L_2 = \frac{h}{\pi} \)

To compare \( L_2 \) and \( L_1 \), we can express \( L_2 \) in terms of \( L_1 \):

\[ L_2 = \frac{h}{\pi} = 2 \cdot \frac{h}{2\pi} = 2L_1 \]

### Conclusion
From the calculations, we find that:

\[ L_2 = 2L_1 \]

Thus, the correct option is that \( L_2 \) is equal to \( 2L_1 \).

### Final Answer
The correct option is \( L_2 = 2L_1 \).

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