One of the lines in the emission spectrum of `Li^(2+)` has the same wavelength as that of the 2nd line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is n = 12 `to`n=x. Find the value of x.

Options

A. 8
B. 6
C. 7
D. 5

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the emission spectrum of Li²⁺ and the Balmer series of hydrogen.
The emission spectrum of Li²⁺ has transitions similar to those of hydrogen, but with a different atomic number (Z=3 for Li²⁺). The second line of the Balmer series in hydrogen corresponds to a transition from n=4 to n=2.

### Step 2: Write down the formula for the wavelength of emitted light.
The formula for the wavelength (λ) of emitted light in a hydrogen-like atom is given by:

\[ \frac{1}{\lambda} = Z^2 R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( Z \) is the atomic number (3 for Li²⁺),
- \( R \) is the Rydberg constant,
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the electron transition.

### Step 3: Set up the equation for Li²⁺.
For the transition from n=12 to n=x in Li²⁺, we have:

\[ \frac{1}{\lambda} = 3^2 R \left( \frac{1}{x^2} - \frac{1}{12^2} \right) \]

This simplifies to:

\[ \frac{1}{\lambda} = 9R \left( \frac{1}{x^2} - \frac{1}{144} \right) \]

### Step 4: Set up the equation for the second line of the Balmer series in hydrogen.
For the second line of the Balmer series (n=4 to n=2), we have:

\[ \frac{1}{\lambda} = 1^2 R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]

This simplifies to:

\[ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{16} \right) \]

Calculating the right-hand side:

\[ \frac{1}{\lambda} = R \left( \frac{4 - 1}{16} \right) = R \left( \frac{3}{16} \right) \]

### Step 5: Equate the two expressions for \(\frac{1}{\lambda}\).
Now we can equate the two expressions for \(\frac{1}{\lambda}\):

\[ 9R \left( \frac{1}{x^2} - \frac{1}{144} \right) = R \left( \frac{3}{16} \right) \]

### Step 6: Simplify the equation.
Dividing both sides by \( R \):

\[ 9 \left( \frac{1}{x^2} - \frac{1}{144} \right) = \frac{3}{16} \]

### Step 7: Solve for \(\frac{1}{x^2}\).
Rearranging gives:

\[ \frac{1}{x^2} - \frac{1}{144} = \frac{3}{144} \]

This simplifies to:

\[ \frac{1}{x^2} = \frac{3}{144} + \frac{1}{144} = \frac{4}{144} = \frac{1}{36} \]

### Step 8: Find the value of \( x \).
Taking the reciprocal gives:

\[ x^2 = 36 \implies x = 6 \]

### Final Answer:
The value of \( x \) is \( 6 \).

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