Answered: Calculate the wavelength (nm) and the…

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter28: The Atom

Section28.1: The Bohr Model Of The Atom

Problem 6PP

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Calculate the wavelength (nm) and the energy (J) of the first line of the hydrogen spectrum in the Brackett series. (Rydberg constant = 1.09x107m-1; Planck's constant = 6,626x10-34 Js)

Expert Solution

Step 1

This is a question from the Atomic and Molecular Physics. The first line in the bracket series arises when the transition is between the levels $n_{2} = 4 t o n_{1} = 5$ .

So, we will solve this is two parts,

(a) The wavelength of the first line

(b) The energy of the first line

Step 2

(a)

Data:

$R y d b e r g c o n s \tan t = R = 1.09 \times 10^{7} m^{- 1} n_{2} = 4 n_{1} = 5$

Formula:

$\frac{1}{λ} = R [\frac{1}{{n_{2}}^{2}} - \frac{1}{{n_{1}}^{2}}]$

To Find:

$λ = ?$

Solution:

Using the formula for Bracketseories in a hydrogen atom,

$\frac{1}{λ} = R [\frac{1}{{n_{2}}^{2}} - \frac{1}{{n_{1}}^{2}}]$

Substituting all the given values,

$\begin{array}{rcl} ∴ \frac{1}{λ} & = & 1.09 \times 10^{7} [\frac{1}{4^{2}} - \frac{1}{5^{2}}] \\ ∴ \frac{1}{λ} & = & 1.09 \times 10^{7} [\frac{1}{16} - \frac{1}{25}] \\ ∴ \frac{1}{λ} & = & 1.09 \times 10^{7} \times 2.25 \times 10^{- 2} \\ ∴ \frac{1}{λ} & = & 2.45 \times 10^{5} \\ ∴ λ & = & 4.082 \times 10^{- 6} m \\ ∴ λ & = & 4082 \times 10^{- 9} m \\ ∴ λ & = & 4082 n m \end{array}$

Answer:

$\begin{array}{rcl} λ \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} 4082 \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} n \end{array} \begin{array}{rcl} m \end{array}$

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