Question

If shortest wavelength of hydrogen atom in Lyman series is $x$, then longest wavelength in Balmer series of $He^{+}$ is:

• A. $\frac{9x}{5}$
• B. $\frac{36x}{5}$
• C. $\frac{x}{4}$
• D. $\frac{5x}{9}$

Hint:

Use the Rydberg formula $\frac{1}{\lambda} = RZ^2(\frac{1}{n_1^2} - \frac{1}{n_2^2})$. Shortest wavelength corresponds to $n_2 = \infty$, and longest wavelength corresponds to the first line of the series.

Answer 1

The Correct Option is A

The Rydberg formula allows us to calculate the wavelength ($\lambda$) of electromagnetic radiation emitted during electronic transitions in hydrogen-like atoms:
$$\frac{1}{\lambda} = R Z^{2} \left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)$$
where $R$ is the Rydberg constant, $Z$ is the atomic number, and $n_1, n_2$ are the principal quantum numbers of the lower and higher energy levels, respectively.

Step 1: Calculate the shortest wavelength ($x$) for Hydrogen in the Lyman series.
For Hydrogen, $Z = 1$. The Lyman series involves transitions to the $n_1 = 1$ level. The shortest wavelength corresponds to the highest energy transition, which occurs from $n_2 = \infty$ to $n_1 = 1$.
$$\frac{1}{x} = R \cdot (1)^{2} \left( \frac{1}{1^{2}} - \frac{1}{\infty^{2}} \right) = R (1 - 0) = R$$
So, $R = \frac{1}{x}$.

Step 2: Calculate the longest wavelength ($\lambda'$) for $He^{+}$ in the Balmer series.
For $He^{+}$, $Z = 2$. The Balmer series involves transitions to the $n_1 = 2$ level. The longest wavelength corresponds to the lowest energy transition, which is the first line of the series, from $n_2 = 3$ to $n_1 = 2$.
$$\frac{1}{\lambda'} = R \cdot (2)^{2} \left( \frac{1}{2^{2}} - \frac{1}{3^{2}} \right)$$
Substitute $R = \frac{1}{x}$ into the equation:
$$\frac{1}{\lambda'} = \frac{1}{x} \cdot 4 \cdot \left( \frac{1}{4} - \frac{1}{9} \right)$$
$$\frac{1}{\lambda'} = \frac{4}{x} \cdot \left( \frac{9 - 4}{36} \right) = \frac{4}{x} \cdot \frac{5}{36} = \frac{5}{9x}$$

Step 3: Solve for $\lambda'$.
Inverting both sides, we get $\lambda' = \frac{9x}{5}$.