# Chapter 3, Problem 3.49QP

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### CHEMISTRY: ATOMS FIRST VOL 1 W/CON...

14th Edition
Burdge
ISBN: 9781259327933

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### CHEMISTRY: ATOMS FIRST VOL 1 W/CON...

14th Edition
Burdge
ISBN: 9781259327933
Interpretation Introduction

Interpretation:

The value of ni in which the photon emitted that has a wavelength of 434 nm and the involved transition from an energy state of principal quantum number ni to the = 2 state of an electron in the hydrogen atom should be calculated using the concept of Bohr’s theory.

Concept Introduction:

The emission of radiation given by an energized hydrogen atom to the electron falling from a higher-energy orbit to a lower orbit give a quantum of energy in the form of light.  Based on electrostatic interaction and law of motion, Bohr derived the following equation.

En = 2.18 × 1018 J (1n2)

Where, n gets an integer values such as = 1, 2, 3 and so on.  This is the energy of electron in nth orbital.

The electrons are excited thermally when the light is used by an object.  As a result, an emission spectrum comes.  Line spectra consist of light only at specific, discrete wavelengths.  In emission, the electron returns to a lower energy state from nf (the i and f subscripts denote the initial and final energy states).  In most cases, the lower energy state corresponds to the ground state but it may be any energy state which is lower than the initial excited state.  The difference in the energies between the initial and final states is

ΔE = Ef  Ei

This transition results in the photon’s emission with frequency v and energy hv.  The following equation is resulted.

ΔE = hν = 2.18 × 1018 J (1nf21ni2)

When, ni > nf, a photon is emitted.  The term in parentheses is positive, making ΔE negative.  As a result, energy is lost to the surroundings.  When ni < nf, a photon is absorbed.  The term in parentheses is negative, so ΔE is positive.  As a result, energy is absorbed from the surroundings.

The speed, wavelength and frequency of a wave are interrelated by = λν where λ and ν are mentioned in meters (m) and reciprocal seconds (s1).    Hence, rearrange the equation for getting the wavelength is,

ν = cλ

Substitute the frequency formula and rearrange,

ΔE = hν = hcλ = 2.18 × 1018 J (1nf21ni2)1λ = 2.18 × 1018 Jhc (1nf21ni2)

Therefore, this formula is used to find the wavelength of the given photon in the emission line process.  For the absorption line process in which an electron is removed from the nucleus, the sign is changed as,

1λ = 2.18 × 1018 Jhc (1nf21ni2)

To find: Get the value of ni in which the photon emitted that has a wavelength of 434 nm and the involved transition from an energy state of principal quantum number ni to the = 2 state of an electron in the hydrogen atom

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