JEE Main 2026 — Physics Modern Physics

2026mcqmedium

In the hydrogen atom, the electron makes a transition from the higher orbit ( $i$ ) to a lower orbit ( $f$ ). The ratio of the radius of the orbits in given by $r_i : r_f = 16 : 4$ . The wavelength of photon emitted due to this transition is _____ nm.

(Given Rydberg constant $= 1.0973 \times 10^7$ /m)

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Options

A
$121$
B
$242$
C
$486$
D
$974$

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Solution

The radius of the $n$ -th orbit in a hydrogen atom is proportional to the square of the principal quantum number $n$ , i.e., $r_n \propto n^2$ .

Given the ratio of the radii of the initial and final orbits:

$\dfrac{r_i}{r_f} = \dfrac{16}{4}$

This implies:

$\dfrac{n_i^2}{n_f^2} = \dfrac{16}{4} \Rightarrow n_i = 4 \text{ and } n_f = 2$

The wavelength $\lambda$ of the emitted photon is given by the Rydberg formula:

$\dfrac{1}{\lambda} = R \left( \dfrac{1}{n_f^2} - \dfrac{1}{n_i^2} \right)$

Substituting the values:

$\dfrac{1}{\lambda} = 1.0973 \times 10^7 \left( \dfrac{1}{2^2} - \dfrac{1}{4^2} \right)$

$\dfrac{1}{\lambda} = 1.0973 \times 10^7 \left( \dfrac{1}{4} - \dfrac{1}{16} \right)$

$\dfrac{1}{\lambda} = 1.0973 \times 10^7 \times \dfrac{3}{16}$

$\lambda = \dfrac{16}{3 \times 1.0973 \times 10^7} \text{ m}$

$\lambda \approx 4.86 \times 10^{-7} \text{ m} = 486 \text{ nm}$

Answer: $486$

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Solution

The radius of the $n$ -th orbit in a hydrogen atom is proportional to the square of the principal quantum number $n$ , i.e., $r_n \propto n^2$ .

Given the ratio of the radii of the initial and final orbits:

$\dfrac{r_i}{r_f} = \dfrac{16}{4}$

This implies:

$\dfrac{n_i^2}{n_f^2} = \dfrac{16}{4} \Rightarrow n_i = 4 \text{ and } n_f = 2$

The wavelength $\lambda$ of the emitted photon is given by the Rydberg formula:

$\dfrac{1}{\lambda} = R \left( \dfrac{1}{n_f^2} - \dfrac{1}{n_i^2} \right)$

Substituting the values:

$\dfrac{1}{\lambda} = 1.0973 \times 10^7 \left( \dfrac{1}{2^2} - \dfrac{1}{4^2} \right)$

$\dfrac{1}{\lambda} = 1.0973 \times 10^7 \left( \dfrac{1}{4} - \dfrac{1}{16} \right)$

$\dfrac{1}{\lambda} = 1.0973 \times 10^7 \times \dfrac{3}{16}$

$\lambda = \dfrac{16}{3 \times 1.0973 \times 10^7} \text{ m}$

$\lambda \approx 4.86 \times 10^{-7} \text{ m} = 486 \text{ nm}$

Answer: $486$

JEE Main 2026 — Physics Modern Physics

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JEE Main 2026 — Physics Modern Physics

Options

Did you get this right?

Solution

Your note

JEE Main Modern Physics in other years

More JEE Main Modern Physics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Modern Physics