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 $\mathrm{r}_{\mathrm{i}}: \mathrm{r}_{f}=16: 4$. The wavelength of photon emitted due to this transition is ______ nm. (Given Rydberg constant $=1.0973 \times 10^{7}$ /m)

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Modern 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 1 0^{7}

/m)

(A)121
(B)242
(C)486✓ Correct
(D)974

Answer: (C)

Solution

The orbit radius varies as

r_{n} \propto n^{2}

, so

\frac{r _{i}}{r _{f}} = \frac{16}{4} = \frac{n _{i}^{2}}{n _{f}^{2}}

gives

\frac{n _{i}}{n _{f}} = 2

with

n_{i} = 4

and

n_{f} = 2

— a transition ending in the second orbit (Balmer series). The Rydberg formula gives

\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} \times \frac{3}{16} = 2.057 \times 1 0^{6} m^{- 1}

. Hence

λ = 4.86 \times 1 0^{- 7}

m, that is

486

nm.

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Single correct option

Practise more: all JEE Main PYQs on Modern Physics →

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Modern Physics

Solution

More JEE Main questions on Modern Physics