Second Line of Paschen Series | JEE Main PYQ | Structure of Atom
Q. If wavelength of the first line of the Paschen series of hydrogen atom is 720 nm, then the wavelength of the second line of this series is ______ nm. (Nearest integer)

Correct Answer: 492 nm

Explanation

For Paschen series, final state nf = 3.

First line: n = 4 → 3 Second line: n = 5 → 3

Using Rydberg formula:

1/λ = R ( 1/nf2 − 1/ni2 )

First line:

1/λ₁ = R ( 1/3² − 1/4² )

Second line:

1/λ₂ = R ( 1/3² − 1/5² )

Taking ratio:

λ₂ / λ₁ = (1/3² − 1/4²) ÷ (1/3² − 1/5²)

λ₂ = 720 × (7/144) ÷ (16/225)

λ₂ ≈ 492 nm

Previous Year Questions – Structure of Atom
The total number of isoelectronic species from the given set is ________.
The orbital angular momentum of an electron in 3s orbital is (xh / 2π). The value of x is ________ (nearest integer).
The energy of an electron in the first Bohr orbit of hydrogen atom is −2.18 × 10⁻¹⁸ J. Its energy in the third Bohr orbit is ________.
The number of correct statements from the following is ________. (Atomic orbitals and probability density)
Assertion–Reason question based on photoelectric effect.
JEE Main Chapter-wise PYQs – Physics & Chemistry
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