JEE Main 2026 — Apr 5, Shift 1 — Physics solution

JEE Main2026Apr 5, Shift 1Physics

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 is given by $r_{i} : r_{f} = 16 : 4$ . The wavelength of photon emitted due to this transition is ___ nm. (Given Rydberg constant $= 1.0973 \times 1 0^{7}$ /m)

A
121
B
242
C
486
D
974

Solution

Bohr radius scales as $r_{n} \propto n^{2}$ , so $\frac{r _{i}}{r _{f}} = \frac{n _{i}^{2}}{n _{f}^{2}} = \frac{16}{4} = 4$ , giving $n_{i} = 4$ and $n_{f} = 2$ .

Rydberg formula: $\frac{1}{λ} = R (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}}) = 1.0973 \times 1 0^{7} (\frac{1}{4} - \frac{1}{16}) = 1.0973 \times 1 0^{7} \cdot \frac{3}{16}$ .

Hence $λ \approx 4.86 \times 1 0^{- 7}$ m $\approx 486$ nm.

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