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Chapter 18: Problem 2557

excited hydrogen atom emits a Photon of wave length \(\lambda\) in returning to the ground state The quantum number \(\mathrm{n}\) of excited state is given by (A) \(\sqrt{[}(\lambda . \mathrm{R}-1) /(\lambda \mathrm{R})]\) (B) \(\sqrt{[}(\lambda \mathrm{R}) /(\lambda \mathrm{R}-1)]\) (C) \(\sqrt{[\lambda R}(\lambda \mathrm{R}-1)]\) (D) \(\lambda \mathrm{R}(\mathrm{R}-1)\)

Short Answer

Expert verified
The correct expression for the quantum number n of an excited hydrogen atom that returns to its ground state by emitting a photon with a wavelength λ is (B) \(n = \sqrt{\frac{(\lambda R)}{(\lambda R - 1)}}\).

Step by step solution

01

Recall the Rydberg formula for hydrogen

The Rydberg formula for hydrogen can be expressed as: \[ \frac{1}{\lambda} = R \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) \] where: - \(λ\) is the wavelength of the photon emitted, - \(R\) is the Rydberg constant for hydrogen, approximately equal to \(1.097 \times 10^7 \, \text{m}^{-1}\), - \(n_f\) is the final principal quantum number of the electron, - \(n_i\) is the initial principal quantum number of the electron. In this case, the hydrogen atom returns to its ground state (i.e., \(n_f = 1\)), and we need to find the expression for \(n_i\), referred to as \(n\).
02

Substitute the values in the Rydberg formula

Substitute \(n_f = 1\) into the Rydberg formula: \[ \frac{1}{\lambda} = R \left(\frac{1}{1^2} - \frac{1}{n^2}\right) \]
03

Simplify and solve for n

Now simplify the formula and solve for \(n\): \[ \frac{1}{\lambda} = R \left(1 - \frac{1}{n^2}\right) \] Divide both sides of the equation by R: \[ \frac{1}{\lambda R} = 1 - \frac{1}{n^2} \] Rearrange to isolate the term \(\frac{1}{n^2}\): \[ \frac{1}{n^2} = 1 - \frac{1}{\lambda R} \] Now, take the reciprocal of both sides: \[ n^2 = \frac{1}{1 - \frac{1}{\lambda R}} \] Finally, to obtain the expression for \(n\), take the square root of both sides: \[ n = \sqrt{\frac{1}{1 - \frac{1}{\lambda R}}} \] Comparing the given options, we find the correct expression is:
04

Answer

(B) \(n = \sqrt{\frac{(\lambda R)}{(\lambda R - 1)}}\)

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Most popular questions from this chapter

Plutonium decays with half life time 24000 yrs. if Plutonium is stored after 72000 yrs, the fraction of it that remain (A) \((1 / 2)\) (B) \((1 / 9)\) (C) \((1 / 12)\) (D) \((1 / 8)\)

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A radioactive sample has \(\mathrm{n}_{0}\) active atom at \(\mathrm{t}=\mathrm{o}\), at the rate of disintegration at any time is \(\mathrm{R}\) and the number of atom is \(\mathrm{N}\), then ratio. $(\mathrm{R} / \mathrm{N})\( varies with time \)(\mathrm{t})$ as.

A gamma ray Photon creates an electron-Positron Pair. If the rest mass energy of an electron is \(0.5 \mathrm{MeV}\). and the total kinetic energy $0.7 \mathrm{MeV}$, then the energy of the gamma ray Photon must be (A) \(3.9 \mathrm{MeV}\) (B) \(1.78 \mathrm{MeV}\) (C) \(0.78 \mathrm{MeV}\) (D) \(0.16 \mathrm{MeV}\)

Match column I and II and chose correct Answer form the given below. (a) Nuclear fusion (p) converts some matter into energy (b) Nuclear fission (q) generally Possible for nuclei with low atomic number (c) \(\beta\) decay (r) generally Possible for nuclei with high atomic number (d) Exothermic nuclear (s) Essentially Proceeds by weak reaction nuclear force(c) (A) $\mathrm{a} \rightarrow \mathrm{p}, \mathrm{b} \rightarrow \mathrm{r}, \mathrm{c} \rightarrow \mathrm{s}, \mathrm{d} \rightarrow \mathrm{q}$ (B) $\mathrm{a} \rightarrow \mathrm{q}, \mathrm{b} \rightarrow \mathrm{r}, \mathrm{c} \rightarrow \mathrm{p}, \mathrm{d} \rightarrow \mathrm{s}$ (C) $\mathrm{a} \rightarrow \mathrm{q}, \mathrm{b} \rightarrow \mathrm{r}, \mathrm{c} \rightarrow \mathrm{s}, \mathrm{d} \rightarrow \mathrm{p}$ (D) $\mathrm{a} \rightarrow \mathrm{r}, \mathrm{b} \rightarrow \mathrm{q}, \mathrm{c} \rightarrow \mathrm{p}, \mathrm{d} \rightarrow \mathrm{s}$
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