In the hydrogen atom, the electron makes a transition from the higher orbit to a lower orbit . The ratio

Question

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 = $1.0973 \times 10^{7} /m$ ) [2026]

Smart Achievers · Accepted Answer

(3)

$Given : \frac{r_{i}}{r_{f}} = \frac{16}{4} = \frac{n_{i}^{2}}{n_{f}^{2}}$

$\frac{n_{i}}{n_{f}} = \frac{4}{2}$

$\frac{1}{λ} = R (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}}) = 1.0973 \times 10^{7} (\frac{1}{{(2)}^{2}} - \frac{1}{{(4)}^{2}})$

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

Solution

1 121

2 242

3 486

4 974

Ans.
(3)

$Given : \frac{r_{i}}{r_{f}} = \frac{16}{4} = \frac{n_{i}^{2}}{n_{f}^{2}}$

$\frac{n_{i}}{n_{f}} = \frac{4}{2}$

$\frac{1}{λ} = R (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}}) = 1.0973 \times 10^{7} (\frac{1}{{(2)}^{2}} - \frac{1}{{(4)}^{2}})$

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

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