Atoms and Nuclei JEE 2021: Hydrogen Spectrum Solution

What is the hydrogen-spectrum question in Atoms and Nuclei JEE 2021?

A and D are correct in this Atoms and Nuclei JEE 2021 question from JEE Advanced 2021, Paper 1, Physics. It is a multiple-correct question, not a JEE Main single-correct MCQ, and is rated medium on this question bank’s scale.

The task is to decide which claims about hydrogen spectral wavelengths and series ranges hold:

A: The longest-to-shortest wavelength ratio for the Balmer series is:

\frac{λ_{max}}{λ_{min}} = \frac{9}{5}

B: The wavelength ranges of the Balmer and Paschen series overlap.
C: Lyman wavelengths obey the proposed relation below, where the index is an integer and the reference wavelength is the Lyman limiting shortest wavelength:

λ = (1 + \frac{1}{m^{2}}) λ_{0}, m \in Z

D: The wavelength ranges of the Lyman and Balmer series do not overlap.

How does the Balmer wavelength ratio prove option A?

A and D are correct. Verify each statement independently using the official solution’s method: apply the Rydberg formula, calculate the wavelength endpoints, then compare them.

\frac{1}{λ} = R (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}})

R : Rydberg constant

n_{i} : initial principal quantum number, n_{f} : final principal quantum number, n_{i} > n_{f}

The smallest transition energy gives the longest wavelength. The shortest wavelength is the series limit, approached as the initial quantum number tends to infinity.

For Balmer:

n_{f} = 2, n_{i} = 3, 4, 5, \dots

The longest-wavelength transition is: $3 \to 2$

\frac{1}{λ_{max, B}} = R (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = R (\frac{1}{4} - \frac{1}{9}) = \frac{5 R}{36}

λ_{max, B} = \frac{36}{5 R}

At the series limit:

n_{i} \to \infty, \frac{1}{λ_{min, B}} = R (\frac{1}{4} - 0) = \frac{R}{4}

$λ_{min, B} = \frac{4}{R}$ The required ratio is:

\frac{λ_{max, B}}{λ_{min, B}} = \frac{36}{5 R} \times \frac{R}{4} = \frac{9}{5}

A is correct. No numerical value of the Rydberg constant is needed because it cancels in the ratio.

Do the Balmer and Paschen wavelength ranges overlap?

No. Even Balmer’s longest wavelength is shorter than Paschen’s limiting shortest wavelength, so B is incorrect. Calculate the endpoints rather than substitute memorised wavelengths.

For Paschen:

n_{f} = 3, n_{i} = 4, 5, 6, \dots

Its shortest-wavelength limit is:

\frac{1}{λ_{min, P}} = R (\frac{1}{3^{2}} - 0) = \frac{R}{9}, λ_{min, P} = \frac{9}{R}

Its longest wavelength comes from the first transition: $4 \to 3$

\frac{1}{λ_{max, P}} = R (\frac{1}{3^{2}} - \frac{1}{4^{2}}) = R (\frac{1}{9} - \frac{1}{16}) = \frac{7 R}{144}

λ_{max, P} = \frac{144}{7 R}

The limiting spans are:

Balmer: \frac{4}{R} to \frac{36}{5 R}, Paschen: \frac{9}{R} to \frac{144}{7 R}

Compare the nearest endpoints:

\frac{36}{5} < 9 \Rightarrow \frac{36}{5 R} < \frac{9}{R}

This strict inequality proves a gap between the ranges. These spans contain discrete spectral lines, not continuous bands; the shortest endpoint is a series limit, not a finite transition.

Why is option C wrong even if we change the integer index?

The exact Lyman expression has the square of the initial quantum number minus one in its denominator. That denominator cannot be an integer square, so changing the index cannot make C exact.

For Lyman, the final level and limiting wavelength satisfy:

n_{f} = 1, n_{i} \to \infty

\frac{1}{λ_{0}} = R (1 - 0) = R, λ_{0} = \frac{1}{R}

For any finite Lyman transition:

n_{i} = n, n \in Z, n \geq 2

\frac{1}{λ} = R (1 - \frac{1}{n^{2}}) = \frac{1}{λ_{0}} (1 - \frac{1}{n^{2}})

Invert the complete expression:

λ = \frac{λ_{0}}{1 - \frac{1}{n^{2}}} = \frac{n^{2}}{n^{2} - 1} λ_{0} = (1 + \frac{1}{n^{2} - 1}) λ_{0}

For option C to match this exactly, its integer index would have to satisfy:

1 + \frac{1}{m^{2}} = 1 + \frac{1}{n^{2} - 1} \Rightarrow m^{2} = n^{2} - 1

But:

(n - 1)^{2} < n^{2} - 1 < n^{2} (n \geq 2)

The required number lies strictly between consecutive integer squares. It cannot be an integer square, even if negative integer indices are allowed. C is incorrect, not merely written differently.

Why are the Lyman and Balmer ranges separate?

Lyman’s longest wavelength is below Balmer’s limiting shortest wavelength, so D is correct. Use the Lyman series limit already derived and calculate its longest wavelength from the first transition. $λ_{min, L} = \frac{1}{R}$

For the longest Lyman wavelength: $2 \to 1$

\frac{1}{λ_{max, L}} = R (1 - \frac{1}{2^{2}}) = R (1 - \frac{1}{4}) = \frac{3 R}{4}

λ_{max, L} = \frac{4}{3 R} < \frac{4}{R} = λ_{min, B}

The ranges do not overlap. The final answer is A and D only.

The endpoint reference below lists final level, limiting shortest wavelength and longest wavelength:

Lyman:

n_{f} = 1, λ_{min} = \frac{1}{R}, λ_{max} = \frac{4}{3 R}

Balmer:

n_{f} = 2, λ_{min} = \frac{4}{R}, λ_{max} = \frac{36}{5 R}

Paschen:

n_{f} = 3, λ_{min} = \frac{9}{R}, λ_{max} = \frac{144}{7 R}

What approximation could incorrectly make option C look exact?

The error is treating a truncated expansion as an identity. Replacing the reciprocal expression by just its first two terms drops nonzero terms:

\frac{1}{1 - x} \neq 1 + x in general

\frac{λ}{λ_{0}} = \frac{1}{1 - \frac{1}{n^{2}}} = 1 + \frac{1}{n^{2}} + \frac{1}{n^{4}} + \dots

Dropping all higher terms produces C’s form when: $m = n$

That is a large-quantum-number approximation, not an exact formula for every Lyman line. Check the smallest allowed initial level:

n = 2, m = 2

λ_{exact} = \frac{4}{3} λ_{0}, λ_{truncated} = (1 + \frac{1}{4}) λ_{0} = \frac{5}{4} λ_{0}

The values differ, so the approximation cannot prove the claimed identity. Invert first, preserve the denominator, and test exact claims at the smallest allowed quantum number.

Can you apply the endpoint method to two more questions?

Calculate each ratio from the first transition and the series limit. These are original practice questions based on the same chapter, not additional verified PYQs. Before checking either answer, identify the longest-wavelength transition and the series limit yourself.

What is the longest-to-shortest wavelength ratio for the Lyman series?

The relevant transitions and endpoints are:

2 \to 1, n_{i} \to \infty, n_{f} = 1

λ_{max} = \frac{4}{3 R}, λ_{min} = \frac{1}{R}, \frac{λ_{max}}{λ_{min}} = \frac{4}{3 R} \times R = \frac{4}{3}

What is the longest-to-shortest wavelength ratio for the Paschen series?

Use the first transition and the limit:

4 \to 3, n_{i} \to \infty, n_{f} = 3

λ_{max} = \frac{144}{7 R}, λ_{min} = \frac{9}{R}

\frac{λ_{max}}{λ_{min}} = \frac{144}{7 R} \times \frac{R}{9} = \frac{16}{7}

For the next question, fix the final quantum number, calculate the first transition and series limit, then compare wavelengths, not their reciprocals.

Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).

For a worked example of the same idea, see JEE Physics Preparation: Daily Study Plan and Practice.

Frequently asked questions

What is the answer to the JEE Advanced 2021 hydrogen-spectrum question?

A and D only are correct in this multiple-correct question from JEE Advanced 2021, Paper 1, Physics. The Balmer longest-to-shortest wavelength ratio is 9/5, and the Lyman and Balmer wavelength ranges do not overlap.

How do you calculate the Balmer series maximum-to-minimum wavelength ratio?

Using the Rydberg formula, the longest Balmer wavelength comes from the 3-to-2 transition and equals 36/(5R). The limiting shortest wavelength is 4/R as the initial quantum number tends to infinity. Their ratio is 9/5, with the Rydberg constant R cancelling.

Do the Balmer and Paschen wavelength ranges overlap?

No, the Balmer and Paschen wavelength ranges do not overlap. Balmer's longest wavelength is 36/(5R), which is less than Paschen's limiting shortest wavelength, 9/R. Comparing these nearest endpoints proves a gap between the ranges.

Why is option C wrong in the JEE Advanced 2021 hydrogen-spectrum question?

The exact Lyman relation is λ = [1 + 1/(n² − 1)]λ₀, where λ₀ = 1/R and n is an integer at least 2. Matching option C would require an integer m satisfying m² = n² − 1, but this value lies strictly between consecutive integer squares. Replacing the exact reciprocal expression with 1 + 1/n² is only a large-n approximation, not an identity.

Atoms and Nuclei JEE 2021: Hydrogen Spectrum Solution

What is the hydrogen-spectrum question in Atoms and Nuclei JEE 2021?

How does the Balmer wavelength ratio prove option A?

Do the Balmer and Paschen wavelength ranges overlap?

Why is option C wrong even if we change the integer index?

Why are the Lyman and Balmer ranges separate?

What approximation could incorrectly make option C look exact?

Can you apply the endpoint method to two more questions?

Frequently asked questions

What is the answer to the JEE Advanced 2021 hydrogen-spectrum question?

How do you calculate the Balmer series maximum-to-minimum wavelength ratio?

Do the Balmer and Paschen wavelength ranges overlap?

Why is option C wrong in the JEE Advanced 2021 hydrogen-spectrum question?

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