Atoms & Nuclei

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28 of 239

Question 28 of 239

In the hydrogen atom, the electron makes a transition from the higher orbit (ii) to a lower orbit (ff). The ratio of the radius of the orbits in given by ri:rf=16:4r_i : r_f = 16 : 4. The wavelength of photon emitted due to this transition is _____ nm.
(Given Rydberg constant =1.0973×107= 1.0973 \times 10^7 /m)

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Answer: (C) 486486

The radius of the nn-th orbit in a hydrogen atom is proportional to the square of the principal quantum number nn, i.e., rn∝n2r_n \propto n^2.

Given the ratio of the radii of the initial and final orbits:
rirf=164\dfrac{r_i}{r_f} = \dfrac{16}{4}

This implies:
ni2nf2=164⇒ni=4 and nf=2\dfrac{n_i^2}{n_f^2} = \dfrac{16}{4} \Rightarrow n_i = 4 \text{ and } n_f = 2

The wavelength λ\lambda of the emitted photon is given by the Rydberg formula:
1λ=R(1nf2−1ni2)\dfrac{1}{\lambda} = R \left( \dfrac{1}{n_f^2} - \dfrac{1}{n_i^2} \right)

Substituting the values:
1λ=1.0973×107(122−142)\dfrac{1}{\lambda} = 1.0973 \times 10^7 \left( \dfrac{1}{2^2} - \dfrac{1}{4^2} \right)

1λ=1.0973×107(14−116)\dfrac{1}{\lambda} = 1.0973 \times 10^7 \left( \dfrac{1}{4} - \dfrac{1}{16} \right)

1λ=1.0973×107×316\dfrac{1}{\lambda} = 1.0973 \times 10^7 \times \dfrac{3}{16}

λ=163×1.0973×107 m\lambda = \dfrac{16}{3 \times 1.0973 \times 10^7} \text{ m}

λ≈4.86×10−7 m=486 nm\lambda \approx 4.86 \times 10^{-7} \text{ m} = 486 \text{ nm}

Answer: 486486

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