Question

For the Balmer series in the spectrum of H atom, $\overline{v} = R_H \{\frac{1}{n_1^2} - \frac{1}{n_2^2}\}$, the correct statements among (I) to (IV) are:

• A. As wavelength decreases, the lines in the series converge.
• B. The lines in the Balmer series lie in the ultraviolet region.
• C. The transition $n=2 \to n=1$ is part of the Balmer series.
• D. The longest wavelength in the Balmer series corresponds to the transition from $n=3$ to $n=2$.

Answer 1

Answer: (A), (D)

(I) and (IV)

Answer 2

For the Balmer series, $n_1=2$. The wave number is given by $\overline{v} = R_H (\frac{1}{n_1^2} - \frac{1}{n_2^2}) = R_H (\frac{1}{4} - \frac{1}{n_2^2})$, where $n_2 = 3, 4, 5, \dots$.

(I) As $n_2$ increases, $\frac{1}{n_2^2}$ decreases, so $\overline{v}$ increases. Since $\overline{v} = 1/\lambda$, as $\overline{v}$ increases, $\lambda$ decreases. The wave number approaches a limit $\frac{R_H}{4}$ as $n_2 \to \infty$, which means the wavelength approaches a limit $\frac{4}{R_H}$. Thus, as wavelength decreases, the lines in the series converge towards this limit. Statement (I) is Correct.

(II) The longest wavelength in the Balmer series occurs for the smallest energy difference, i.e., the transition from $n_2=3$ to $n_1=2$. This transition corresponds to approximately 656 nm, which is in the visible region (red light). The shortest wavelength occurs as $n_2 \to \infty$, approaching 364 nm, which is in the ultraviolet region. Since the series includes lines in both visible and UV regions, the statement that the lines lie in the ultraviolet region is Incorrect.

(III) The Balmer series is defined by transitions where the final energy level is $n_1=2$. The transition $n=2 \to n=1$ ends at $n=1$, which is characteristic of the Lyman series, not the Balmer series. Statement (III) is Incorrect.

(IV) The longest wavelength corresponds to the smallest energy difference (smallest wave number). For the Balmer series ($n_1=2$), the wave number is $\overline{v} = R_H (\frac{1}{4} - \frac{1}{n_2^2})$. To minimize $\overline{v}$, $n_2$ must be minimized. The smallest integer value for $n_2$ greater than $n_1=2$ is $n_2=3$. Therefore, the transition $n=3 \to n=2$ yields the longest wavelength. Statement (IV) is Correct.

Based on this analysis, the correct statements are (I) and (IV).