For the Balmer series in the spectrum of H atom, v = R_H { 1/n_1^2 - 1/n_2^2 }, the correct statemen

Q: Why do the spectral lines converge?

The energy levels in a hydrogen atom are given by $E_n = -13.6/n^2$ eV. As $n$ gets larger, the difference between consecutive energy levels becomes very small. Therefore, the energy of the emitted photons approaches a maximum limit, meaning the wavelengths approach a minimum limit, causing the lines to bunch up or converge.

Q: Can we find the ionization energy if we know the series limit of the Balmer series?

No, not directly. The series limit of the Balmer series (transition from $n=\infty$ to $n=2$) gives you the binding energy of the electron in the 2nd orbit. To find the true ionization energy, you need the energy required to remove the electron from the ground state ($n=1$).

Q: Which series has the line with the absolute longest wavelength in the hydrogen spectrum?

The longest wavelength corresponds to the smallest energy transition. As you go to higher series (Paschen, Brackett, Pfund), the energy gaps between adjacent levels become smaller. For example, the $n=6$ to $n=5$ transition (Pfund series) has a much longer wavelength than the $n=3$ to $n=2$ transition (Balmer series).

Q: Is the Balmer series only in the visible region?

Most of the prominent lines of the Balmer series (like $H_\alpha, H_\beta, H_\gamma, H_\delta$) are in the visible region. However, as the lines converge near the series limit, they actually fall into the near-ultraviolet region.

Q: What does $R_H$ represent?

$R_H$ is the Rydberg constant for hydrogen. It has a value of approximately $109677 \text{ cm}^{-1}$. It represents the limiting wave number of the Lyman series, which is directly related to the ionization energy of hydrogen.

Question

For the Balmer series in the spectrum of H atom,
$\bar{
u} = R_H \left\{ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight\}$, the correct statements among (I) to (IV) are :
(I) As wavelength decreases, the lines in the series converge
(II) The integer $n_1$ is equal to 2
(III) The lines of longest wavelength corresponds to $n_2 = 3$
(IV) The ionisation energy of hydrogen can be calculated from wave number of these lines

Options:

(A) I, II, IV

(B) II, III, IV

(C) I, III, IV

(D) I, II, III

Sigma Percentile · Accepted Answer

### Decoding the Hydrogen Spectrum

The hydrogen emission spectrum is one of the most beautiful and foundational concepts in quantum mechanics. It provides direct evidence for the quantized nature of energy levels in an atom. In this problem, we are diving deep into the **Balmer series**, which is the only series of hydrogen that falls in the visible region of the electromagnetic spectrum.

Let's break down the given statements one by one to see which ones hold true.

### Statement I: Convergence of Spectral Lines

The wave number of any spectral line in the hydrogen spectrum is given by the Rydberg formula:

$$ \bar{
u} = \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) $$

As we look at transitions from higher and higher energy levels (i.e., as $n_2$ increases), the energy difference between consecutive levels becomes smaller and smaller. For example, the gap between $n=3$ and $n=4$ is much larger than the gap between $n=10$ and $n=11$.

Because the energy levels crowd together near the ionization limit ($n 	o \infty$), the energy of the emitted photons approaches a maximum limit. Since energy is inversely proportional to wavelength ($E = \frac{hc}{\lambda}$), the wavelength approaches a minimum limit. This causes the spectral lines to bunch up or **converge** at the shorter wavelength (higher frequency) end of the series. Thus, Statement I is absolutely correct.

### Statement II: The Balmer Series Definition

By definition, the spectral series are named based on the lower energy level ($n_1$) to which the electron transitions:
- Lyman series: $n_1 = 1$
- **Balmer series: $n_1 = 2$**
- Paschen series: $n_1 = 3$

Since the question specifically mentions the Balmer series, the integer $n_1$ must be exactly 2. Statement II is correct.

### Statement III: The Longest Wavelength

We know that the energy of a photon is given by $E = \frac{hc}{\lambda}$. This tells us that energy and wavelength are inversely related.

To find the **longest wavelength** ($\lambda_{max}$), we need the transition that releases the **minimum energy**. In the Balmer series, the electron falls to $n_1 = 2$. The smallest possible energy jump to $n=2$ is from the immediately adjacent higher level, which is $n_2 = 3$.

Therefore, the line with the longest wavelength corresponds to the transition from $n_2 = 3$ to $n_1 = 2$ (often called the $H_\alpha$ line). Statement III is correct.

### Statement IV: Ionisation Energy

**Ionisation energy** is defined as the energy required to completely remove an electron from the atom in its ground state. For hydrogen, the ground state is $n_1 = 1$. The electron must be taken to $n_2 = \infty$.

The wave number corresponding to the ionisation energy would be:

$$ \bar{
u}_{IE} = R_H \left( \frac{1}{1^2} - \frac{1}{\infty^2} ight) = R_H $$

This transition belongs to the **Lyman series**, not the Balmer series. The Balmer series only gives us information about transitions ending at $n=2$. We cannot directly calculate the ionisation energy from the Balmer series lines alone without knowing the energy gap between $n=1$ and $n=2$. Thus, Statement IV is incorrect.

### Final Conclusion

Statements I, II, and III are correct, while Statement IV is incorrect. This makes option (d) the right choice. Understanding these fundamental definitions is crucial for mastering atomic structure!

Select Answer:

Visualized Solution

$Analyzing Statement I: Convergence of Lines$

$Analyzing Statement II: Balmer Series$

$Analyzing Statement III: Longest Wavelength$

$Analyzing Statement IV: Ionisation Energy$

$Final Conclusion$

The Sigma Insight: Bohr's Model

Decoding the Hydrogen Spectrum

Statement I

Statement II

Statement III

Statement IV

Final Conclusion

Similar Questions

The shortest wavelength of H atom in the Lyman series is $λ_{1}$ . The longest wavelength in the Balmer series of $He^{+}$ is

For any given series of spectral lines of atomic hydrogen, let $Δ \overset{ν}{ˉ} = \overset{ν}{ˉ}_{m a x} - \overset{ν}{ˉ}_{m i n}$ be the difference in maximum and minimum frequencies in $cm^{- 1}$ . The ratio $Δ \overset{ν}{ˉ}_{Lyman} /Δ \overset{ν}{ˉ}_{Balmer}$ is

The region in the electromagnetic spectrum where the Balmer series lines appear is

The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to be about 9. The spectral series are

For emission line of atomic hydrogen from $n_{i} = 8$ to $n_{f} = n$ , the plot of wave number ( $\overset{ν}{ˉ}$ ) against $(\frac{1}{n ^{2}})$ will be (The Rydberg constant, $R_{H}$ is in wave number unit)

Energy of an electron is given by $E = - 2.178 \times 1 0^{- 18} J (\frac{Z ^{2}}{n ^{2}})$ . Wavelength of light required to excite an electron in an hydrogen atom from level $n = 1$ to $n = 2$ will be ( $h = 6.62 \times 1 0^{- 34} J s$ and $c = 3.0 \times 1 0^{8} ms^{- 1}$ )

The frequency of light emitted for the transition $n = 4$ to $n = 2$ of $He^{+}$ is equal to the transition in H atom corresponding to which of the following?

Comprehension Passage

Which of the following options has the correct combination considering List-I and List-II ?

Which of the following options has the correct combination considering List-I and List-II ?

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Statement I: Convergence of Lines

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Statement II: Balmer Series

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Statement III: Longest Wavelength

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Statement IV: Ionisation Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Conclusion

The Sigma Insight: Bohr's Model

Decoding the Hydrogen Spectrum

Statement I

Statement II

Statement III

Statement IV

Final Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The shortest wavelength of H atom in the Lyman series is λ1​. The longest wavelength in the Balmer series of He+ is

The region in the electromagnetic spectrum where the Balmer series lines appear is

The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to be about 9. The spectral series are

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For emission line of atomic hydrogen from ni​=8 to nf​=n, the plot of wave number (νˉ) against (n21​) will be (The Rydberg constant, RH​ is in wave number unit)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The frequency of light emitted for the transition n=4 to n=2 of He+ is equal to the transition in H atom corresponding to which of the following?

Comprehension Passage

Which of the following options has the correct combination considering List-I and List-II ?

Which of the following options has the correct combination considering List-I and List-II ?

$Analyzing Statement I: Convergence of Lines$

$Analyzing Statement II: Balmer Series$

$Analyzing Statement III: Longest Wavelength$

$Analyzing Statement IV: Ionisation Energy$

$Final Conclusion$

The shortest wavelength of H atom in the Lyman series is $λ_{1}$ . The longest wavelength in the Balmer series of $He^{+}$ is

For emission line of atomic hydrogen from $n_{i} = 8$ to $n_{f} = n$ , the plot of wave number ( $\overset{ν}{ˉ}$ ) against $(\frac{1}{n ^{2}})$ will be (The Rydberg constant, $R_{H}$ is in wave number unit)

The frequency of light emitted for the transition $n = 4$ to $n = 2$ of $He^{+}$ is equal to the transition in H atom corresponding to which of the following?