Physics Q36 – JEE-MAIN 2026

Home Physics Structure of AtomQuestion #483

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

)

Clue 1 of 3: Bohr's Model for Hydrogen Atom

Recall the relationship between the radius of an electron's orbit and its principal quantum number in a hydrogen atom.

🧠 Full Solution Path

Step 1: Determine Initial and Final Quantum Numbers (

n_{i}, n_{f}

)✦ Active

The radius of an electron's orbit in a hydrogen atom is given by $r_{n} = a_{0} n^{2}$ , where $a_{0}$ is the Bohr radius and $n$ is the principal quantum number. Given the ratio of radii $r_{i} : r_{f} = 16 : 4$ , we can write:

\frac{r_{i}}{r_{f}} = \frac{16}{4} = 4

Substituting the formula for radius:

\frac{a_{0} n_{i}^{2}}{a_{0} n_{f}^{2}} = 4 ⟹ \frac{n_{i}^{2}}{n_{f}^{2}} = 4

Taking the square root, we get $\frac{n_{i}}{n_{f}} = 2$ . The most direct interpretation of $r_{i} : r_{f} = 16 : 4$ for integer quantum numbers is that $n_{i}^{2} = 16$ and $n_{f}^{2} = 4$ . Thus, the initial principal quantum number is $n_{i} = 4$ and the final principal quantum number is $n_{f} = 2$ .

Step 2: Apply Rydberg Formula○ Expand

The wavelength ( $λ$ ) of the photon emitted during a transition from a higher orbit ( $n_{i}$ ) to a lower orbit ( $n_{f}$ ) is given by the Rydberg formula:

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

Substitute the given Rydberg constant $R = 1.0973 \times 10^{7} m^{- 1}$ , and the determined quantum numbers $n_{i} = 4$ and $n_{f} = 2$ :

\frac{1}{λ} = 1.0973 \times 10^{7} (\frac{1}{2^{2}} - \frac{1}{4^{2}})

\frac{1}{λ} = 1.0973 \times 10^{7} (\frac{1}{4} - \frac{1}{16})

\frac{1}{λ} = 1.0973 \times 10^{7} (\frac{4 - 1}{16})

\frac{1}{λ} = 1.0973 \times 10^{7} \times \frac{3}{16}

Step 3: Calculate Wavelength and Convert to Nanometers○ Expand

Now, calculate the wavelength $λ$ :

λ = \frac{16}{3 \times 1.0973 \times 10^{7}} m

λ = \frac{16}{3.2919 \times 10^{7}} m

λ \approx 4.860 \times 10^{- 7} m

To convert the wavelength from meters to nanometers (nm), multiply by $10^{9}$ :

λ \approx 4.860 \times 10^{- 7} \times 10^{9} nm

λ \approx 486.0 nm

The wavelength of the emitted photon is approximately $486 nm$ .

💡 Teacher's Secret Hint

Ensure correct unit conversion from meters to nanometers.

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Physics Question 36 – JEE-MAIN 2026

Clue 1 of 3: Bohr's Model for Hydrogen Atom

🧠 Full Solution Path