Q61 / 75 · JEE Main 2026 (05 April Shift 1) · Physics

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)

  • A
    $121$
  • B
    $242$
  • Correct answer
    C
    $486$
  • D
    $974$

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: $486$

Hence, the correct answer is option: 'C'.

Relevant Questions