Question

If the wavelength of limiting line of Lyman series for $He^+$ ion is $x$ Å, then what will be the wavelength of limiting line of Balmer Series for $Li^{+2}$ ion.

• A. $\frac{9x}{4} A$
• B. $\frac{16x}{9} A$
• C. $\frac{5x}{4} A$
• D. $\frac{4x}{7} A$

Answer 1

Answer: (B)

$\frac{16x}{9} A$

Answer 2

The problem involves calculating wavelengths of spectral lines using the Rydberg formula for hydrogen-like atoms.

Key Concept:

The Rydberg formula for the wavelength ($\lambda$) of a spectral line in a hydrogen-like atom is given by:

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

where:

• $R$ is the Rydberg constant.
• $Z$ is the atomic number of the atom/ion.
• $n_1$ is the principal quantum number of the lower energy level.
• $n_2$ is the principal quantum number of the higher energy level.

For a "limiting line" in any series, the electron transition occurs from $n_2 = \infty$ to the specific $n_1$ for that series.

Step-by-step Derivations:

1. For $He^+$ ion (Lyman series, limiting line):

• Ion: $He^+$
• Atomic Number (Z): For Helium, $Z = 2$.
• Series: Lyman series, which means the electron transitions to the $n_1 = 1$ energy level.
• Line: Limiting line, so the electron originates from $n_2 = \infty$.
• Given Wavelength: $\lambda = x$ Å.

Applying the Rydberg formula:

$$\frac{1}{x} = R (2)^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right)$$ $$\frac{1}{x} = R \cdot 4 \left( \frac{1}{1} - 0 \right)$$ $$\frac{1}{x} = 4R$$

From this, we can express the Rydberg constant $R$ in terms of $x$:

$$R = \frac{1}{4x} \quad \text{(Equation 1)}$$

2. For $Li^{+2}$ ion (Balmer series, limiting line):

• Ion: $Li^{+2}$
• Atomic Number (Z): For Lithium, $Z = 3$.
• Series: Balmer series, which means the electron transitions to the $n_1 = 2$ energy level.
• Line: Limiting line, so the electron originates from $n_2 = \infty$.
• Required Wavelength: Let this be $\lambda'$.

Applying the Rydberg formula:

$$\frac{1}{\lambda'} = R (3)^2 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right)$$ $$\frac{1}{\lambda'} = R \cdot 9 \left( \frac{1}{4} - 0 \right)$$ $$\frac{1}{\lambda'} = \frac{9R}{4} \quad \text{(Equation 2)}$$

3. Substitute Equation 1 into Equation 2:

Now, substitute the value of $R$ from Equation 1 into Equation 2:

$$\frac{1}{\lambda'} = \frac{9}{4} \left( \frac{1}{4x} \right)$$ $$\frac{1}{\lambda'} = \frac{9}{16x}$$

Solving for $\lambda'$:

$$\lambda' = \frac{16x}{9}$$

Thus, the wavelength of the limiting line of the Balmer series for the $Li^{+2}$ ion is $\frac{16x}{9}$ Å.