The electron in hydrogen atom in a sample is in `n^(th)` excited state, then the number of differrent spectrum lines obtained in its emission spectrum will be

Options

A. `1+2+3+"……….."+(n-1)`
B. `1+2+3+"…………."+(n)`
C. `1+2+3+"…………."+(n+1)`
D. `1xx2xx3xx"…………"xx(n-1)`

Explanation

To find the number of different spectral lines obtained in the emission spectrum of a hydrogen atom when the electron is in the \( n^{th} \) excited state, we can follow these steps:

### Step 1: Understand the excited state
When an electron is in the \( n^{th} \) excited state, it means it is in the \( n+1 \) energy level (since the ground state is considered as \( n=1 \)). Therefore, the total number of energy levels available for the electron is \( n+1 \).

### Step 2: Determine the number of transitions
The number of different spectral lines corresponds to the number of possible transitions between these energy levels. An electron can transition from a higher energy level to a lower energy level, resulting in the emission of a photon.

### Step 3: Calculate the number of transitions
To find the number of different transitions (or spectral lines), we need to select 2 energy levels from the \( n+1 \) levels. The number of ways to choose 2 levels from \( n+1 \) is given by the combination formula:

\[ \text{Number of spectral lines} = \binom{n+1}{2} = \frac{(n+1)!}{2!(n-1)!} \]

### Step 4: Simplify the combination
This can be simplified as follows:

\[ \binom{n+1}{2} = \frac{(n+1) \cdot n}{2} \]

### Step 5: Conclusion
Thus, the total number of different spectral lines in the emission spectrum when the electron is in the \( n^{th} \) excited state is:

\[ \frac{n(n+1)}{2} \]

### Final Answer
The number of different spectral lines obtained in its emission spectrum will be \( \frac{n(n+1)}{2} \).

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