An electron in nth excited state in a hydrogen atom comes down to first excited state by emitting ten different wavelength. Find value of `n` (an integer).

Options

A. 6
B. 7
C. 8
D. 9

Answer

To solve the problem of finding the value of \( n \) for an electron in the nth excited state of a hydrogen atom that emits 10 different wavelengths while transitioning to the first excited state, we can follow these steps:

### Step 1: Understand the Energy Levels
In a hydrogen atom, the energy levels are given by the principal quantum number \( n \). The first excited state corresponds to \( n = 2 \).

### Step 2: Identify the Initial and Final States
- The final state (after emission) is the first excited state, which is \( n_1 = 2 \).
- The initial state is the nth excited state, which is \( n_2 \).

### Step 3: Use the Formula for Spectral Lines
The number of different wavelengths (spectral lines) emitted when an electron transitions from \( n_2 \) to \( n_1 \) can be calculated using the formula:
\[ \text{Number of spectral lines} = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2} \]
In our case, we know that the number of spectral lines is 10, so we set up the equation:
\[ 10 = \frac{(n_2 - 2)(n_2 - 2 + 1)}{2} \]

### Step 4: Simplify the Equation
Multiply both sides by 2 to eliminate the fraction:
\[ 20 = (n_2 - 2)(n_2 - 1) \]

### Step 5: Expand and Rearrange
Expanding the right-hand side gives:
\[ 20 = n_2^2 - 3n_2 + 2 \]
Rearranging this equation leads to:
\[ n_2^2 - 3n_2 - 18 = 0 \]

### Step 6: Factor the Quadratic Equation
Now we can factor the quadratic:
\[ (n_2 - 6)(n_2 + 3) = 0 \]
This gives us two possible solutions:
\[ n_2 - 6 = 0 \quad \text{or} \quad n_2 + 3 = 0 \]
Thus, \( n_2 = 6 \) or \( n_2 = -3 \). Since \( n \) must be a positive integer, we discard \( n_2 = -3 \).

### Step 7: Conclusion
The only valid solution is:
\[ n_2 = 6 \]
Thus, the value of \( n \) is 6.