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Atomic hydrogen is excited from the grou...

Atomic hydrogen is excited from the grounds state to the `n^(th)` state. The number of lines in the emission spectrum will be:

A

`(n(n+1))/(2)`

B

`(n(n-1))/(2)`

C

`((n-1)^(2))/(2)`

D

`((n+1)^(2))/(2)`

Text Solution

AI Generated Solution

The correct Answer is:
To find the number of lines in the emission spectrum when atomic hydrogen is excited from the ground state to the \( n^{th} \) state, we can follow these steps: ### Step 1: Understand the Concept of Emission Spectrum When an electron in a hydrogen atom is excited to a higher energy level (from the ground state to the \( n^{th} \) state), it can return to the ground state by emitting photons. Each transition from a higher energy level to a lower one corresponds to a spectral line in the emission spectrum. **Hint:** Remember that each transition corresponds to a unique line in the spectrum. ### Step 2: Identify the Transitions When the electron is in the \( n^{th} \) state, it can transition to any of the lower energy levels (from \( n-1 \) down to 1). The possible transitions are: - From \( n \) to \( n-1 \) - From \( n \) to \( n-2 \) - ... - From \( n \) to \( 1 \) **Hint:** Count how many levels the electron can transition to from the \( n^{th} \) state. ### Step 3: Calculate the Number of Unique Transitions To find the total number of unique transitions, we can use the formula for the number of lines in the emission spectrum: \[ \text{Number of lines} = \frac{n(n-1)}{2} \] Where \( n \) is the final excited state. **Hint:** This formula counts all possible pairs of transitions from the \( n^{th} \) state to lower states. ### Step 4: Apply the Formula For the \( n^{th} \) state: - If \( n = 4 \): \[ \text{Number of lines} = \frac{4(4-1)}{2} = \frac{4 \times 3}{2} = \frac{12}{2} = 6 \] **Hint:** Substitute the value of \( n \) into the formula carefully to avoid mistakes. ### Step 5: Conclusion Thus, the total number of lines in the emission spectrum when atomic hydrogen is excited from the ground state to the \( n^{th} \) state is 6. **Final Answer:** The number of lines in the emission spectrum will be \( 6 \).
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