JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Atomic Structure

WhatsApp Email

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)

Answer: 492

Solution

Every Paschen-series transition ends at

n = 3

. The first line is the

4 \to 3

transition,

\frac{1}{λ _{1}} = R_{H} (\frac{1}{9} - \frac{1}{16}) = R_{H} \frac{7}{144}

, and the second line is the

5 \to 3

transition,

\frac{1}{λ _{2}} = R_{H} (\frac{1}{9} - \frac{1}{25}) = R_{H} \frac{16}{225}

. Dividing,

\frac{λ _{2}}{λ _{1}} = \frac{7/144}{16/225} = \frac{1575}{2304}

, so

λ_{2} = 720 \times \frac{1575}{2304} \approx 492 nm

. The second line of the series lies at

492 nm

JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Integer answer

Practise more: all JEE Main PYQs on Atomic Structure →

More JEE Main questions on Atomic Structure

Given below are two statements:

Statement I: When an electric discharge is passed through gaseous hydrogen, the hydrogen molecules dissociate and the energetically excited hydrogen atoms produce electromagnetic radiation of discrete frequencies.

Statement II: The frequency of s …
JEE Main 2026View solution →
The energy required by electrons, present in the first Bohr orbit of hydrogen atom to be excited to second Bohr orbit is ____ $J mol^{- 1}$ .

Given : $R_{H} = 2.18 \times 1 0^{- 11} ergs$ .
JEE Main 2026View solution →
Consider the following spectral lines for atomic hydrogen:
A. First line of Paschen series
B. Second line of Balmer series
C. Third line of Paschen series
D. Fourth line of Bracket series

The correct arrangement of the above lines in ascending order of energy is:
JEE Main 2026View solution →
The energy of first (lowest) Balmer line of H atom is x J. The energy (in J) of second Balmer line of H atom is :
JEE Main 2026View solution →
Which of the following statements regarding the energy of the stationary state is true in the following one-electron system ?
JEE Main 2026View solution →
Given,
(A) $n = 5, m_{1} = - 1$
(B) $n = 3, 1 = 2, m_{1} = - 1, m_{s} = + \frac{1}{2}$

The maximum number of electron(s) in an atom that can have the quantum numbers as given in (A) and (B) respectively are :
JEE Main 2026View solution →